Differential equations
A differential equation links an unknown function with its derivatives, so it describes how a quantity changes. This unit shows how to classify a DE by order, degree and linearity, how to form one by eliminating arbitrary constants, and what general and particular solutions mean. It then solves first order, first degree equations by separating the variables or, for homogeneous equations, by the substitution . The same few methods model population growth, radioactive decay, Newton's law of cooling, falling bodies, electric circuits and continuous compound interest.
Learning outcomes (4)
- Identify and construct first order differential equations from practical situations.
- Give the concept of solution of a differential equation.
- Solve differential equations of first order and first degree of the form: separable variables equations and homogeneous equations.
- Solve real life problems related to differential equations, such as population growth and decay, the cooling/warming law, flow of electricity, series circuits, economics and finance, and radioactive decay.
Introduction and mathematical models
A differential equation (DE) is an equation that contains the derivative of an unknown function. It links a quantity with its rate of change. In real life we often know how fast something changes (a population grows by a fixed fraction each year, a hot cup of tea cools faster when it is much hotter than the room). A DE turns that knowledge into a formula for the quantity itself.
DEs are used wherever quantities change continuously in time or space:
- Physics: motion of bodies under forces (Newton's laws) and wave propagation.
- Biology: population growth, the spread of diseases, interaction between species.
- Engineering: electric circuits, control systems and mechanical vibrations.
- Chemistry: reaction rates and diffusion.
- Economics: growth of investments, market behaviour and optimisation.
Mathematical models
A mathematical model is a mathematical description, such as a differential equation, that imitates a real situation so that we can predict how it will behave. Most applications of DEs are mathematical models.
Example: a falling object. Newton's second law says that the net force on a body equals its mass times its acceleration. Acceleration is the rate of change of velocity with time , that is . So
This law is a differential equation because it contains a derivative. If we ignore air resistance, the only force on an object falling close to Earth's surface is its weight , directed downward, where . The model of the free-falling object is therefore
Translating words into a differential equation
The first SLO of this unit is to construct first order DEs from practical situations. Three phrases do most of the work: **rate of change of ** means ; is proportional to means "equals a constant times"; decreases or decays means the rate is negative.
| Statement | Differential equation |
|---|---|
| A population grows at a rate proportional to its size. | |
| A radioactive mass decays at a rate proportional to the mass remaining. | |
| A body cools at a rate proportional to the difference between its temperature and the room temperature . | |
| Money grows at a continuous interest rate . | |
| The velocity of a falling body increases at the constant rate . | |
| The slope of a curve at any point is . |
4.1Differential equation
A differential equation is an equation containing one or more derivatives of an unknown function. The book's definition: a mathematical equation containing the derivatives of one dependent variable with respect to one independent variable is called a differential equation (DE).
If the unknown function depends on only one variable, the DE is an ordinary differential equation (ODE). In this unit every DE is ordinary, so we simply call it a DE. Two things must be present: a derivative and an equation (an equals sign).
- Dependent variable
- The unknown function whose derivatives appear, usually (or , , in applications).
- Independent variable
- The variable with respect to which we differentiate, usually (or time ).
Where a differential equation comes from
Take the function , where is an arbitrary constant (it may have any fixed value). Differentiate with respect to . The constant disappears:
The result is a differential equation. Every curve , for every value of , satisfies it. So one DE describes a whole family of curves.
The simplest differential equations have the form
where is a known function of . Some more examples, in each of which is the dependent variable and the independent variable:
4.2Order and degree of a differential equationOften examined
Order. The order of a differential equation is the order of the highest order derivative present in the equation.
Types of DE with respect to order
- First order DE
- A DE whose highest derivative is the first derivative. It can be written . Example: .
- Second order DE
- A DE whose highest derivative is the second derivative . Example: .
Degree
Degree. The degree of a differential equation is the power (exponent) of the highest order derivative in the equation, after the equation has been made free of radicals and fractional powers of the derivatives.
| Equation | Highest derivative | Order | Degree |
|---|---|---|---|
| , power 1 | 1 | 1 | |
| , power 2 | 2 | 2 | |
| , power 3 | 2 | 3 |
- (i) The only derivative is , of first order, and it appears to the power 1.
- (ii) The highest derivative is , and it appears to the power 1.
- (iii) The highest derivative is , and it is squared. The lower derivatives do not affect the degree.
Some more types of differential equations are: ordinary DEs (ODE), partial DEs (PDE), linear DEs, non-linear DEs, homogeneous DEs and non-homogeneous DEs. They are described in the next section.
4.2.3Ordinary, linear and non-linear equations
Ordinary differential equation (ODE)
If only one independent variable is involved, the DE is called an ordinary differential equation. It contains one or more derivatives of the unknown function with respect to that single independent variable. The general form of an th order ODE, solved for its highest derivative, is
where is a known function. This is called an explicit ODE of order . Examples of ODEs:
Linear and non-linear DEs
An th order differential equation is linear if it can be written in the form
that is, it satisfies the following two conditions:
- (a) the dependent variable and all its derivatives are of power one (and are not multiplied together);
- (b) all the coefficients and the function are constants or depend only on the independent variable .
If any one of these two conditions is not satisfied, the DE is non-linear. Signs of non-linearity: a power of or of a derivative (, ), a product such as , or a function of such as or .
| Equation | Type | Reason |
|---|---|---|
| (i) | linear | has power 1; the right side is a constant. |
| (ii) | linear | , and all have power 1 with constant coefficients. |
| (iii) | non-linear | the coefficient of is (a product ), which breaks condition (b). |
| (iv) | non-linear | written as , it contains the product . |
| (v) | linear | the coefficient of depends only on . |
| (vi) | non-linear | is squared, which breaks condition (a). |
4.3Concept of solution of a differential equation
Differential equations are studied mainly through their solutions: the functions that satisfy them.
A solution of a differential equation is any function y=f(x), defined on some interval I, that reduces the equation to an identity (a statement true for every x in I) when y and its derivatives are substituted.
To solve a DE such as \dfrac{dy}{dx}-x=0 means to find the unknown function, either explicitly as y=f(x) or implicitly as a relation between x and y.
The simplest case: integrate
Consider the first order DE \dfrac{dy}{dx}=f(x). If f(x) is continuous, integrating both sides gives
where F(x) is an anti-derivative of f(x) and c is the constant of integration.
- The right side contains only x, so integrate both sides with respect to x.y=\int\left(1+e^{2x}\right)dx
- Use \int e^{ax}\,dx=\dfrac{1}{a}e^{ax} with a=2, and add one constant.y=x+\dfrac{1}{2}e^{2x}+c
- Differentiate the given relation implicitly with respect to x. The constant c gives 0.2x+2y\dfrac{dy}{dx}=0
- Divide by 2 and rearrange. This is exactly the given DE, so the relation is a solution.y\dfrac{dy}{dx}+x=0
- Graph. The solution depends on the arbitrary constant c. Choosing c=1,4,9,16,\dots gives different solutions.x^{2}+y^{2}=1^{2},\quad x^{2}+y^{2}=2^{2},\quad x^{2}+y^{2}=3^{2},\quad x^{2}+y^{2}=4^{2},\ \dots
- These are circles with centre (0,0) and radii 1,2,3,4,\dots. So the solution x^{2}+y^{2}=c represents a family of infinitely many circles.
This example shows that a DE has two types of solutions: the general solution and a particular solution.
4.3.1General and particular solutionsOften examined
- General solution
- The solution that contains as many arbitrary constants as the order of the differential equation. It is a relation between the independent variable x and the dependent variable y, obtained by removing the derivatives (by integration). In Example 3, x^{2}+y^{2}=c is the general solution.
- Particular solution
- A solution free from arbitrary constants. It is obtained by giving particular values to the arbitrary constants. In Example 3, x^{2}+y^{2}=4 and x^{2}+y^{2}=9 are particular solutions.
| General solution | Particular solution | |
|---|---|---|
| Arbitrary constants | as many as the order | none |
| Graph | a family of curves | one curve of the family |
| Example for \dfrac{dy}{dx}=2x | y=x^{2}+c | y=x^{2}+1 |
- Differentiate the given solution y=\dfrac{x^{4}}{16} with respect to x.\dfrac{dy}{dx}=\dfrac{4x^{3}}{16}=\dfrac{x^{3}}{4}
- Substitute y and \dfrac{dy}{dx} in the left side of the DE. Note \left(\dfrac{x^{4}}{16}\right)^{1/2}=\dfrac{x^{2}}{4}.\dfrac{dy}{dx}-xy^{1/2}=\dfrac{x^{3}}{4}-x\left(\dfrac{x^{4}}{16}\right)^{1/2}=\dfrac{x^{3}}{4}-x\left(\dfrac{x^{2}}{4}\right)=\dfrac{x^{3}}{4}-\dfrac{x^{3}}{4}=0
- The left side is 0 for every real x, which is the right side. So the function is a solution on \mathbb{R}.\sqrt{x^{4}/16}=x^{2}/4 is true for every real x because x^{2}/4\ge0.
- There is one arbitrary constant, so differentiate once. The constant c disappears.\dfrac{dy}{dx}=6x
- For each value of c, y=3x^{2}+c is the parabola y=3x^{2} moved c units up (or down if c<0). All of them have the y-axis as axis and open upward.
- Differentiate y=xe^{x} with the product rule.y'=xe^{x}+e^{x}
- Differentiate again.y''=xe^{x}+e^{x}+e^{x}=xe^{x}+2e^{x}
- Substitute in the left side of the DE and simplify.y''-2y'+y=xe^{x}+2e^{x}-2\left(xe^{x}+e^{x}\right)+xe^{x}=xe^{x}+2e^{x}-2xe^{x}-2e^{x}+xe^{x}=0
- This is true for every x\in\mathbb{R}, so y=xe^{x} is a solution on (-\infty,+\infty).
- Trivial solution
- A solution of a DE that is identically zero on an interval, such as y=0 in Examples 4 and 5.
4.4Formation of differential equationsOften examined
We can form a differential equation by eliminating the arbitrary constants from an algebraic equation (which is then a solution of the DE). Each differentiation gives one more equation, so to remove n constants we differentiate n times.
- Count the arbitrary constants in the given equation; call the number n.
- Differentiate the equation n times with respect to x.
- Use the n+1 equations (the given one and its n derivatives) to eliminate all n constants.
- The result is a DE of order n, the lowest possible order.
- Differentiate.y=e^{x}\quad\Longrightarrow\quad y'=e^{x}
- Both right sides are e^{x}, so y'=y.y'-y=0
- There are two constants, a and b, so differentiate twice.y'=-a\sin x+b\cos x
- Differentiate again.y''=-a\cos x-b\sin x=-\left(a\cos x+b\sin x\right)
- The bracket is y itself, so both constants disappear.y''=-y\qquad\text{or}\qquad y''+y=0
So y=a\cos x+b\sin x is a solution of the differential equation y''+y=0.
- Two constants A and B, so differentiate twice. Use the chain rule.\dfrac{dy}{dx}=2A\cos(2x-B)
- Differentiate again.\dfrac{d^{2}y}{dx^{2}}=-4A\sin(2x-B)=-4\left[A\sin(2x-B)\right]
- The bracket is y.\dfrac{d^{2}y}{dx^{2}}=-4y\qquad\Longrightarrow\qquad \dfrac{d^{2}y}{dx^{2}}+4y=0
This is a second order ODE. Its physical meaning: if y is a displacement and x is time, the acceleration \dfrac{d^{2}y}{dx^{2}} is proportional to the displacement and opposite to it. This is the differential equation of simple harmonic motion.
- Differentiate y^{2}=4a(x-b) implicitly. The constant b disappears.2y\dfrac{dy}{dx}=4a\quad\Longrightarrow\quad y\dfrac{dy}{dx}=2a
- Differentiate again with the product rule. The constant a disappears.y\dfrac{d^{2}y}{dx^{2}}+\dfrac{dy}{dx}\cdot\dfrac{dy}{dx}=0\quad\Longrightarrow\quad y\dfrac{d^{2}y}{dx^{2}}+\left(\dfrac{dy}{dx}\right)^{2}=0
- This is the given DE, so every parabola y^{2}=4a(x-b) is a solution of it.
4.4.1Explicit and implicit solutions
- Explicit solution
- A solution that can be written in the form y=f(x), with y alone on one side.
- Implicit solution
- A solution in the form of a relation f(x,y)=0, not solved for y.
- Differentiate the relation implicitly with respect to x.2x+2y\dfrac{dy}{dx}=0
- Solve for \dfrac{dy}{dx} (possible because y\neq0 when -2<x<2). This is the given DE.\dfrac{dy}{dx}=-\dfrac{x}{y}
Note. On the interval (-2,2) the relation defines two explicit differentiable functions, y=\sqrt{4-x^{2}} and y=-\sqrt{4-x^{2}}. Each is an explicit solution of the DE.
4.4.2Number of solutions
A given differential equation usually has an infinite number of solutions, one for each value of the arbitrary constant(s).
- Differentiate.\dfrac{dy}{dx}=-\dfrac{c}{x^{2}}
- Substitute in the left side.x\left(-\dfrac{c}{x^{2}}\right)+\dfrac{c}{x}+1=-\dfrac{c}{x}+\dfrac{c}{x}+1=1
- This holds for every c, so there are infinitely many solutions. For c=0 we get the constant solution y=1.
- Differentiate the sum twice.y'=-4c_1\sin4x+4c_2\cos4x,\qquad y''=-16c_1\cos4x-16c_2\sin4x
- So y''=-16y, that is y''+16y=0. The sum is a solution for all c_1,c_2.y''+16y=-16y+16y=0
4.5Solution of first order, first degree equations
The solution of an algebraic equation such as x^{2}-5x+6=0 is a number: the values x=2 and x=3 satisfy it. The solution of a differential equation such as y\,dy=x\,dx is a function. To find it we remove the derivative by some technique, and every technique involves integration. After removing the derivative, the dependent variable y is given in terms of the independent variable x, either explicitly (y=f(x)) or implicitly (g(x,y)=0).
A DE of first order and first degree can be written as \dfrac{dy}{dx}=f(x,y) or as M(x,y)\,dx+N(x,y)\,dy=0. Some standard forms have ready methods:
- variable separable differential equations,
- equations reducible to variable separable form,
- homogeneous differential equations,
- equations reducible to homogeneous form.
4.5.1Variable separable equationsOften examined
A first order differential equation of the form
where f is a function of x only and g is a function of y only, is called separable, or is said to have separable variables.
For example, \dfrac{dy}{dx}=y^{3}xe^{2x+y} is separable, because it can be written as \dfrac{dy}{dx}=\left(xe^{2x}\right)\left(y^{3}e^{y}\right)=f(x)\,g(y). But \dfrac{dy}{dx}=y+\sin x is not separable: y+\sin x cannot be written as a product f(x)\,g(y).
Derivation Solving a separable equation
Aim. Find the general solution of \dfrac{dy}{dx}=f(x)\,g(y).
- g(y)\neq0, so that we may divide by it.
- Divide by g(y) and multiply by dx. Now every y is on the left with dy, and every x is on the right with dx.\dfrac{dy}{g(y)}=f(x)\,dx
- Integrate both sides, the left with respect to y and the right with respect to x.\int\dfrac{dy}{g(y)}=\int f(x)\,dx
- Each integral produces a constant; combine them into a single constant c written on one side.G(y)=F(x)+cG and F are anti-derivatives of \dfrac{1}{g(y)} and f(x).
- If possible, solve for y to get an explicit solution. Otherwise leave the implicit relation.
One integration gives one arbitrary constant, as expected for a first order DE. Any value y_0 with g(y_0)=0 gives a constant solution y=y_0, which may be lost by the division.
- Separate the variables: multiply both sides by y\,dx.y\,dy=x^{2}\,dx
- Integrate both sides.\int y\,dy=\int x^{2}\,dx\quad\Longrightarrow\quad \dfrac{y^{2}}{2}=\dfrac{x^{3}}{3}+c_1
- Multiply by 6 to clear the fractions.3y^{2}=2x^{3}+6c_1
- Write c=6c_1, which is still an arbitrary constant.3y^{2}=2x^{3}+c
- Move the dx term to the right.(1+x)\,dy=y\,dx
- Divide by y(1+x) to separate the variables.\dfrac{dy}{y}=\dfrac{dx}{1+x}
- Integrate both sides; write the constant as \ln c.\int\dfrac{dy}{y}=\int\dfrac{dx}{1+x}\quad\Longrightarrow\quad \ln y=\ln(x+1)+\ln c
- Combine the logarithms with \ln a+\ln b=\ln(ab).\ln y=\ln\left[c(x+1)\right]
- Take antilog of both sides.y=c(x+1)
- Separate the variables.\tan y\,dy=\dfrac{1}{x}\,dx
- Integrate. Use \int\tan y\,dy=-\ln(\cos y) and write the constant as \ln c.-\ln(\cos y)=\ln x+\ln c
- Bring everything to one side and combine.0=\ln(\cos y)+\ln x+\ln c=\ln(cx\cos y)
- Take antilog: e^{\ln(cx\cos y)}=e^{0}=1.cx\cos y=1
- Divide by c and rename \dfrac{1}{c}=C.x\cos y=C
4.5.1Initial condition and initial value problemOften examined
The general solution of a DE contains as many arbitrary constants as the order of the DE. Often we need the one solution that satisfies an extra (supplementary) condition, such as a known value of y at a given x.
Suppose we want the solution of \dfrac{dy}{dx}=f(x,y) with y=y_0 at x=x_0. Substituting x=x_0 and y=y_0 in the general solution gives a particular value of the constant, and so a particular solution. The condition y(x_0)=y_0 is called the initial condition, and the DE together with it is an initial value problem (IVP):
- Solve the DE to get the general solution, containing c.
- Substitute x=x_0 and y=y_0 to find c.
- Put this value of c back into the general solution.
- If you then solve for y, choose the sign or branch that agrees with the initial condition.
- Separate the variables.y\,dy=-x\,dx
- Integrate both sides.\int y\,dy=\int -x\,dx\quad\Longrightarrow\quad \dfrac{y^{2}}{2}=-\dfrac{x^{2}}{2}+c\quad\Longrightarrow\quad x^{2}+y^{2}=2c\qquad(1)
- Use the initial condition x=1, y=3 in (1).1^{2}+3^{2}=2c\quad\Longrightarrow\quad 2c=10\quad\Longrightarrow\quad c=5
- Substitute 2c=10 back into (1).x^{2}+y^{2}=10
This is the solution of the IVP. It is the circle with centre (0,0) and radius \sqrt{10}.
- Separate: dy=2x\,dx. Integrate both sides.\int dy=\int2x\,dx\quad\Longrightarrow\quad y=2\left(\dfrac{x^{2}}{2}\right)+c\quad\Longrightarrow\quad y-x^{2}=c\qquad(1)
- Use x=2, y=4 in (1).4-2^{2}=c\quad\Longrightarrow\quad c=0
- Substitute c=0 in (1).y-x^{2}=0\qquad\text{that is}\qquad y=x^{2}\qquad(2)
Note. The general solution (1) represents a family of parabolas for different values of c, whereas the particular solution (2) is the member of the family that passes through (2,4).
- Separate and integrate.dy=2\,dx\quad\Longrightarrow\quad y=2x+c
- For every c this is a straight line with slope 2 and y-intercept c. Equal slopes mean the lines are parallel.
- Use y(0)=1: 1=2(0)+c.c=1\quad\Longrightarrow\quad y=2x+1
4.6.1Homogeneous functions
Before homogeneous differential equations we need homogeneous functions. Roughly, a polynomial in x and y is homogeneous when every term has the same total degree (power of x plus power of y).
Definition. If a function f has the property
where t\in\mathbb{R}^{+} and n\in\mathbb{R}, then f is a **homogeneous function of degree n**. To test a function, replace x by tx and y by ty, simplify, and see whether a power t^{n} comes out as a common factor.
- (i) Replace x with tx and y with ty; take t^{3} out of the root.f(tx,ty)=\sqrt{(tx)^{3}+(ty)^{3}}=\sqrt{t^{3}\left(x^{3}+y^{3}\right)}=t^{3/2}\sqrt{x^{3}+y^{3}}=t^{3/2}f(x,y)
- So f is homogeneous of degree \dfrac{3}{2} (the degree need not be a whole number).
- (ii) The constant 2 does not get a factor t^{2}.f(tx,ty)=t^{2}x^{2}+t^{2}y^{2}+2\neq t^{2}f(x,y)
- So f is not homogeneous.
- (iii) The t cancels in the fraction.f(tx,ty)=\dfrac{tx}{2ty}+4=\dfrac{x}{2y}+4=t^{0}f(x,y)
- So f is homogeneous of degree 0.
- (a) Every term has total degree 2, so t^{2} factors out.f(tx,ty)=t^{2}x^{2}-3t^{2}xy+t^{2}y^{2}=t^{2}f(x,y)\quad(\text{degree }2)
- (b) Since t>0, \sqrt{t^{2}xy}=t\sqrt{xy}, so t factors out of every term.f(tx,ty)=tx-\sqrt{t^{2}xy}+5ty=t\left(x-\sqrt{xy}+5y\right)=t\,f(x,y)\quad(\text{degree }1)
4.6.2Homogeneous differential equationsOften examined
A differential equation of the form
is homogeneous if M and N are homogeneous functions of the same degree n, that is M(tx,ty)=t^{n}M(x,y) and N(tx,ty)=t^{n}N(x,y).
Equivalently, the equation \dfrac{dy}{dx}=f(x,y) is homogeneous when f is homogeneous of degree 0; then f depends only on the ratio \dfrac{y}{x}. Such an equation is reduced to a separable one by substituting y=ux (or x=vy), where u (or v) is a new dependent variable.
Derivation Why the substitution y=ux works
Aim. Show that y=ux turns the homogeneous equation (1) into an equation with separable variables x and u.
- M and N are homogeneous of the same degree n.
- x>0 (for x<0 the same steps work with |x|).
- Put y=ux, where u is a function of x. Differentiate with the product rule.\dfrac{dy}{dx}=u+x\dfrac{du}{dx}\qquad\text{or}\qquad dy=u\,dx+x\,du
- Substitute y=ux and dy=u\,dx+x\,du into (1).M(x,ux)\,dx+N(x,ux)\left[u\,dx+x\,du\right]=0
- Use homogeneity with t=x: M(x\cdot1,\,x\cdot u)=x^{n}M(1,u) and likewise for N.x^{n}M(1,u)\,dx+x^{n}N(1,u)\left[u\,dx+x\,du\right]=0
- Divide by x^{n} and collect the dx terms.\left[M(1,u)+u\,N(1,u)\right]dx+x\,N(1,u)\,du=0
- Divide by x\left[M(1,u)+uN(1,u)\right]. The first term contains only x and the second only u: the variables are separated.\dfrac{dx}{x}+\dfrac{N(1,u)\,du}{M(1,u)+u\,N(1,u)}=0\qquad(2)
After y=ux the variables separate. Integrate, then replace u by \dfrac{y}{x} to return to x and y.
- Check that the equation is homogeneous (M and N of the same degree, or \dfrac{dy}{dx} a function of \dfrac{y}{x}).
- Put y=ux and \dfrac{dy}{dx}=u+x\dfrac{du}{dx}.u+x\dfrac{du}{dx}=F(u)
- Cancel the powers of x and move u to the right.x\dfrac{du}{dx}=F(u)-u
- Separate the variables and integrate, with one constant.\int\dfrac{du}{F(u)-u}=\int\dfrac{dx}{x}
- Replace u by \dfrac{y}{x} and simplify. For an IVP, use the initial condition last.
- Numerator and denominator are both of degree 2, so the equation is homogeneous. Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}.u+x\dfrac{du}{dx}=\dfrac{x^{2}+u^{2}x^{2}}{2x\cdot ux}=\dfrac{1+u^{2}}{2u}
- Subtract u from both sides.x\dfrac{du}{dx}=\dfrac{1+u^{2}}{2u}-u=\dfrac{1+u^{2}-2u^{2}}{2u}=\dfrac{1-u^{2}}{2u}
- Separate the variables.\dfrac{2u}{1-u^{2}}\,du=\dfrac{dx}{x}
- Integrate. The numerator 2u is minus the derivative of 1-u^{2}, so the left side is -\ln(1-u^{2}).-\ln\left(1-u^{2}\right)=\ln x+\ln c\quad\Longrightarrow\quad \ln\left(1-u^{2}\right)+\ln x+\ln c=0
- Combine the logarithms and take antilog.\ln\left[cx\left(1-u^{2}\right)\right]=0\quad\Longrightarrow\quad cx\left(1-u^{2}\right)=1
- Replace u by \dfrac{y}{x}.cx\left(1-\dfrac{y^{2}}{x^{2}}\right)=1\quad\Longrightarrow\quad cx\left(\dfrac{x^{2}-y^{2}}{x^{2}}\right)=1\quad\Longrightarrow\quad x^{2}-y^{2}=\dfrac{x}{c}
- Rename \dfrac{1}{c}=C.x^{2}-y^{2}=Cx
- Divide by x. The right side depends only on \dfrac{y}{x}, so the equation is homogeneous of degree zero.\dfrac{dy}{dx}=\dfrac{y}{x}+e^{y/x},\qquad y(1)=1
- Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}. The u cancels.u+x\dfrac{du}{dx}=\dfrac{ux}{x}+e^{ux/x}=u+e^{u}\quad\Longrightarrow\quad x\dfrac{du}{dx}=e^{u}
- Separate and integrate.e^{-u}\,du=\dfrac{dx}{x}\quad\Longrightarrow\quad \int e^{-u}\,du=\int\dfrac{dx}{x}\quad\Longrightarrow\quad -e^{-u}=\ln x+c
- Replace u by \dfrac{y}{x}.-e^{-y/x}=\ln x+c\qquad(\text{i})
- Use x=1, y=1 in (i). Since \ln1=0:-e^{-1}=\ln1+c\quad\Longrightarrow\quad c=-e^{-1}
- Substitute back into (i) and rearrange.-e^{-y/x}=\ln x-e^{-1}\quad\Longrightarrow\quad e^{-1}-e^{-y/x}=\ln x
Solving for y gives the explicit form y=-x\ln\left(e^{-1}-\ln x\right), valid for 0<x<e^{1/e}\approx1.44 (where e^{-1}-\ln x>0).
4.7Applications of differential equationsOften examined
Differential equations are used to solve problems in engineering and the physical sciences, such as physics, chemistry, biology and economics. First order ODEs describe the flow of electricity, the motion of a pendulum and of a falling object, heat flow (thermodynamics) and population growth. In medicine they are used to follow the growth of diseases through graphs.
- Choose symbols and units: the independent variable (usually time t) and the unknown quantity.
- Translate the statement into a DE (see the table in the Introduction).
- Solve the DE, usually by separating the variables.
- Use the given data (an initial value and one more reading) to find the constants.
- Answer the question with units, and check that the answer is sensible.
- The DE. For downward motion the acceleration a=\dfrac{dv}{dt} equals g.\dfrac{dv}{dt}=g\qquad\text{or}\qquad dv=g\,dt
- Integrate both sides.v=gt+c_1
- At t=0, v=0 (released from rest), so c_1=0.v=gt\qquad(\text{i})
- Velocity after 1 s. Substitute g=9.8\,\mathrm{m\,s^{-2}} and t=1\,\mathrm{s}.v=9.8\times1=9.8\,\mathrm{m\,s^{-1}}
- Distance. Velocity is the rate of change of the distance s: \dfrac{ds}{dt}=gt, so ds=gt\,dt. Integrate.s=g\dfrac{t^{2}}{2}+c_2\qquad(\text{ii})
- At t=0, s=0, so c_2=0 and s=\dfrac{gt^{2}}{2}. The ball reaches the ground when s=20\,\mathrm{m}.20=9.8\times\dfrac{t^{2}}{2}\quad\Longrightarrow\quad t^{2}=\dfrac{40}{9.8}=4.08\quad\Longrightarrow\quad t=2.02\,\mathrm{s}
- Velocity at the ground. From (i):v=9.8\times2.02=19.8\,\mathrm{m\,s^{-1}}
4.7Population growth and decayOften examined
In 1798 Thomas Malthus proposed that the rate of increase of the population of a country is proportional to the population at that time. If P(t) is the population at time t,
where k is the constant of proportionality, called the growth constant. k>0 gives growth and k<0 gives decay. The same DE describes bacteria in a culture, money at continuous compound interest and radioactive decay.
Derivation Solution of the growth equation
Aim. Solve \dfrac{dP}{dt}=kP with P=P_0 at t=0.
- P>0, so \ln P is defined.
- k is constant.
- Separate the variables.\dfrac{dP}{P}=k\,dt
- Integrate both sides; write the constant as \ln c.\ln P=kt+\ln c
- Move \ln c to the left and take antilog.\ln\dfrac{P}{c}=kt\quad\Longrightarrow\quad P=c\,e^{kt}
- At t=0, P=P_0 and e^{0}=1, so c=P_0.P=P_0e^{kt}
In every time interval of the same length \tau, the quantity is multiplied by the same factor e^{k\tau}. This is exponential growth (k>0) or exponential decay (k<0).
**Finding k from two readings.** If P=P_1 at t=t_1, then P_1=P_0e^{kt_1}, so
Doubling time. The population doubles when e^{kt}=2, that is after t_d=\dfrac{\ln2}{k}, whatever the starting value.
- Model: \dfrac{dN}{dt}=kN, so N=400e^{kt} with t in hours.
- Use N(3)=1200.1200=400e^{3k}\quad\Longrightarrow\quad e^{3k}=3\quad\Longrightarrow\quad k=\dfrac{\ln3}{3}=0.366\ \text{per hour}
- Number after 5 hours: e^{5k}=\left(e^{3k}\right)^{5/3}=3^{5/3}.N(5)=400\times3^{5/3}=400\times6.240=2496
- Doubling time.t_d=\dfrac{\ln2}{k}=\dfrac{3\ln2}{\ln3}=1.89\ \text{hours}
4.7Radioactive decay and half-life
Radioactive substances disintegrate at a rate proportional to the amount present. If A(t) is the amount (mass or number of atoms) at time t,
The minus sign shows that A decreases, and \lambda is the decay constant. Solving exactly as in the growth equation (with k=-\lambda) gives A=A_0e^{-\lambda t}, where A_0 is the amount at t=0.
Derivation Half-life
Aim. Find the half-life T_{1/2}, the time in which half of the substance decays, and write A in terms of it.
- At t=T_{1/2} the amount is \dfrac{A_0}{2}.\dfrac{A_0}{2}=A_0e^{-\lambda T_{1/2}}\quad\Longrightarrow\quad e^{-\lambda T_{1/2}}=\dfrac12
- Take natural logarithms. \ln\tfrac12=-\ln2.-\lambda T_{1/2}=-\ln2\quad\Longrightarrow\quad T_{1/2}=\dfrac{\ln2}{\lambda}
- Write e^{-\lambda t}=\left(e^{-\lambda T_{1/2}}\right)^{t/T_{1/2}} and use step 1.A=A_0\left(\dfrac12\right)^{t/T_{1/2}}
The half-life does not depend on how much substance there is at the start.
- Use A=A_0\left(\tfrac12\right)^{t/T_{1/2}} with \dfrac{t}{T_{1/2}}=\dfrac{400}{1600}=\dfrac14.\dfrac{A}{A_0}=\left(\dfrac12\right)^{1/4}=0.841
- So about 84.1\% remains. Check: less than one half-life has passed, so more than 50\% must remain.
4.7Newton's law of cooling and warmingOften examined
Newton's law of cooling: the rate at which a body cools is proportional to the difference between its temperature T and the temperature T_s of the surrounding medium (air, water, ice). The same law describes the warming of a cold body placed in a warm room.
Here t is time and \dfrac{dT}{dt} is the rate of change of the temperature. For a cooling body T>T_s and T decreases, so \dfrac{dT}{dt}<0: the constant k is negative. (Many books write \dfrac{dT}{dt}=-k(T-T_s) with k>0; both forms give the same answers.)
Derivation Solution of Newton's law of cooling
Aim. Solve \dfrac{dT}{dt}=k(T-T_s) with T=T_0 at t=0.
- The surroundings stay at the constant temperature T_s.
- T>T_s for cooling (for warming, T_s-T>0 and the same result follows).
- Separate the variables.\dfrac{dT}{T-T_s}=k\,dt
- Integrate both sides; write the constant as \ln c.\ln\left(T-T_s\right)=kt+\ln c
- Take antilog.T-T_s=c\,e^{kt}
- At t=0, T=T_0, so c=T_0-T_s.T=T_s+\left(T_0-T_s\right)e^{kt}
The temperature difference T-T_s decays exponentially. The body approaches the temperature of the surroundings but never crosses it.
- Newton's law with a constant of proportionality k. In this example the book uses T_0 for the air temperature (called T_s in the derivation above), so T_0=30^\circ.\dfrac{dT}{dt}=k\left(T-T_0\right)=k\left(T-30\right)
- Separate the variables.\dfrac{dT}{T-30}=k\,dt
- Integrate both sides, with constant \ln c, and take antilog.\ln(T-30)=kt+\ln c\quad\Longrightarrow\quad T-30=c\,e^{kt}\qquad(3)
- Initial condition T(0)=90^\circ.90-30=c\quad\Longrightarrow\quad c=60\quad\Longrightarrow\quad T=60e^{kt}+30\qquad(4)
- Second condition T(5)=70^\circ gives k.70=60e^{5k}+30\quad\Longrightarrow\quad e^{5k}=\dfrac{40}{60}=\dfrac23\quad\Longrightarrow\quad k=\dfrac15\ln\dfrac23=-0.0811
- Substitute k in (4).T=60e^{-0.0811t}+30
- Let the outdoor temperature be T_s. The solution of Newton's law with T(0)=20 isT-T_s=\left(20-T_s\right)e^{kt}
- Use the two readings. The second time is double the first, so the second factor is the square of the first.25-T_s=\left(20-T_s\right)e^{8k},\qquad 30-T_s=\left(20-T_s\right)\left(e^{8k}\right)^{2}
- Eliminate e^{8k}: square the first equation and divide by (20-T_s).\left(25-T_s\right)^{2}=\left(20-T_s\right)\left(30-T_s\right)
- Expand both sides.625-50T_s+T_s^{2}=600-50T_s+T_s^{2}\quad\Longrightarrow\quad 625=600
- This is impossible, so no outdoor temperature fits these readings. Under Newton's law the temperature changes by smaller and smaller amounts in equal times, but here it rises by the same 5^\circ\mathrm{C} in each 8 minutes.
4.7Flow of electricity: series circuits
The SLOs of this unit also mention the flow of electricity in a series circuit. The book gives no worked example, so here is the standard model. It is solved by separating the variables.
By Kirchhoff's voltage law, the sum of the voltage drops round the loop equals the applied e.m.f. E. The drop across the resistance R is Ri, and across the inductance L it is L\dfrac{di}{dt}, where i(t) is the current. So
Derivation Growth of current in a series RL circuit
Aim. Find i(t) when the switch is closed at t=0, so i(0)=0.
- E, R and L are constants.
- E-Ri>0 while the current grows.
- Rearrange so that the variables can be separated.L\dfrac{di}{dt}=E-Ri\quad\Longrightarrow\quad \dfrac{L\,di}{E-Ri}=dt
- Integrate. The derivative of E-Ri is -R, so the left side is -\dfrac{L}{R}\ln(E-Ri).-\dfrac{L}{R}\ln\left(E-Ri\right)=t+c
- At t=0, i=0: c=-\dfrac{L}{R}\ln E. Substitute and multiply by -\dfrac{R}{L}.\ln\left(E-Ri\right)-\ln E=-\dfrac{R}{L}t\quad\Longrightarrow\quad \dfrac{E-Ri}{E}=e^{-Rt/L}
- Solve for i.i=\dfrac{E}{R}\left(1-e^{-Rt/L}\right)
The current does not jump to E/R at once: the inductor opposes the change. L/R is called the time constant.
- Steady current and exponent.\dfrac{E}{R}=\dfrac{12}{6}=2\,\mathrm{A},\qquad \dfrac{Rt}{L}=\dfrac{6(0.5)}{2}=1.5
- Substitute in the formula.i=2\left(1-e^{-1.5}\right)=2(1-0.2231)=1.55\,\mathrm{A}
RC circuit. A capacitor of capacitance C with charge q discharging through a resistance R obeys Ri+\dfrac{q}{C}=0 with i=\dfrac{dq}{dt}. So \dfrac{dq}{dt}=-\dfrac{q}{RC}, whose solution is q=q_0e^{-t/(RC)}: exponential decay, exactly like radioactive decay.
4.7Economics and finance
When interest is added continuously, the amount A in an account grows at a rate proportional to itself. If r is the annual interest rate (as a decimal) and t is the time in years,
where A_0 is the amount invested. This is the growth equation again. Similarly, a machine whose value V falls at a rate proportional to its value (depreciation) obeys \dfrac{dV}{dt}=-kV, so V=V_0e^{-kt}.
- Use A=A_0e^{rt} with A_0=50\,000, r=0.08, t=5.A=50\,000\,e^{0.08\times5}=50\,000\,e^{0.4}=50\,000\times1.4918=74\,591
- Doubling: e^{0.08t}=2.t=\dfrac{\ln2}{0.08}=8.66\ \text{years}
Summary: choosing the method
| Type | How to recognise it | Method | Book example |
|---|---|---|---|
| \dfrac{dy}{dx}=f(x) | no y on the right side | integrate directly | \dfrac{dy}{dx}=1+e^{2x} |
| Separable | right side factorises as f(x)\,g(y) | separate, integrate both sides | Example 8, 9, 10 |
| Homogeneous | M, N of the same degree; \dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right) | y=ux, then separate | Example 14, 15 |
| Growth or decay | rate proportional to the amount | P=P_0e^{kt} | Exercise 4.4 Q1, Q3, Q4 |
| Cooling or warming | rate proportional to T-T_s | T=T_s+(T_0-T_s)e^{kt} | Example 17 |
Formula sheet
Key terms
- Differential equation (DE)
- An equation containing one or more derivatives of an unknown function.
- Ordinary differential equation (ODE)
- A DE in which the unknown function depends on only one independent variable, so only ordinary derivatives appear.
- Partial differential equation (PDE)
- A DE whose unknown depends on two or more independent variables, so it contains partial derivatives.
- Dependent variable
- The unknown function whose derivatives appear in the DE, usually y.
- Independent variable
- The variable with respect to which we differentiate, usually x or t.
- Order
- The order of the highest order derivative present in the DE.
- Degree
- The power of the highest order derivative, after the DE is made free of radicals and fractional powers of derivatives. It is not defined if a derivative is inside a function such as \log or \sin.
- Linear DE
- A DE in which y and all its derivatives have power 1 and every coefficient depends only on x (or is constant).
- Non-linear DE
- A DE that is not linear, for example one containing y^{2}, (y')^{3}, y\,y'' or \sin y.
- Solution of a DE
- A function, defined on an interval, that turns the DE into an identity when substituted.
- Arbitrary constant
- A constant that may take any value, such as the constant of integration c.
- General solution
- A solution containing as many arbitrary constants as the order of the DE; it represents a family of curves.
- Particular solution
- A solution with no arbitrary constants, obtained by giving the constants particular values; it is one curve of the family.
- Trivial solution
- A solution that is identically zero on an interval, such as y=0.
- Family of curves
- The set of curves obtained from a general solution by giving its constant different values.
- Explicit solution
- A solution written as y=f(x).
- Implicit solution
- A solution given as a relation f(x,y)=0 that is not solved for y.
- Initial condition
- A condition y(x_0)=y_0 that fixes the value of the arbitrary constant.
- Initial value problem (IVP)
- A DE together with an initial condition, \dfrac{dy}{dx}=f(x,y), y(x_0)=y_0.
- Separable DE
- A first order DE that can be written \dfrac{dy}{dx}=f(x)\,g(y), so the variables can be separated and integrated.
- Homogeneous function
- A function with f(tx,ty)=t^{n}f(x,y) for all t>0; n is its degree.
- Homogeneous DE
- A DE M\,dx+N\,dy=0 in which M and N are homogeneous functions of the same degree; it is solved by y=ux.
- Slope field
- A picture of short line segments showing the slope \dfrac{dy}{dx} given by a DE at many points; solution curves follow the segments.
- Mathematical model
- A mathematical description, such as a DE, that imitates a real situation so that its behaviour can be predicted.
- Malthusian law
- The rate of increase of a population is proportional to the population: \dfrac{dP}{dt}=kP.
- Growth constant
- The constant k in \dfrac{dP}{dt}=kP; positive for growth, negative for decay.
- Decay constant
- The constant \lambda>0 in \dfrac{dA}{dt}=-\lambda A.
- Half-life
- The time in which half of a radioactive substance decays: T_{1/2}=\dfrac{\ln2}{\lambda}.
- Newton's law of cooling
- The rate of change of a body's temperature is proportional to the difference between its temperature and that of its surroundings.
- Superposition principle
- For a linear DE with right side 0, constant multiples and sums of solutions are again solutions.
Exercise 4.1p. 129–130
Order and degree, forming a DE by eliminating constants, and checking a solution by substitution.
- The derivatives present are y'' and y'. The highest is y'' (second order).
- y'' appears to the power 1.
- The only derivative is y' (first order).
- y' appears to the power 1. The product yy' makes the equation non-linear, but it does not change the degree.
- The highest derivative is y'' (second order).
- It appears cubed, so its power is 3.
- The highest derivative is y'' (second order).
- y'' appears to the power 1. The square belongs to y', which is not the highest derivative, so it does not decide the degree.
- The highest derivative is \dfrac{d^{2}y}{dx^{2}} (second order).
- It appears to the power 1; the cube is on \dfrac{dy}{dx}.
- A derivative is under a square root, so first square both sides to make the equation a polynomial in the derivatives.\left(\dfrac{dy}{dx}\right)^{2}=1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}
- The highest derivative is \dfrac{d^{2}y}{dx^{2}}, and it now appears to the power 2.
- The highest derivative is \dfrac{d^{4}y}{dx^{4}} (fourth order).
- It appears to the power 1. The power 5 on \dfrac{dy}{dx} does not count, because \dfrac{dy}{dx} is not the highest derivative.
- There are three arbitrary constants a, b, c, so differentiate three times. The first derivative removes c.y'=ae^{x}-be^{-x}
- Differentiate again.y''=ae^{x}+be^{-x}
- Differentiate a third time.y'''=ae^{x}-be^{-x}
- The right sides of y''' and y' are the same, so a and b are eliminated together.y'''=y'\quad\Longrightarrow\quad \dfrac{d^{3}y}{dx^{3}}-\dfrac{dy}{dx}=0
- One arbitrary constant b, so differentiate once.y'=-\sin(x+b)
- Square y and y' and add. Since \cos^{2}\theta+\sin^{2}\theta=1, the constant b disappears.y^{2}+(y')^{2}=\cos^{2}(x+b)+\sin^{2}(x+b)=1
- Two constants m and c, so differentiate twice. The first derivative removes c.\dfrac{dy}{dx}=m
- Differentiate again to remove m.\dfrac{d^{2}y}{dx^{2}}=0
So every straight line y=mx+c satisfies y''=0: a straight line has zero curvature.
- Two constants a and b, so differentiate twice.y'=2bx+2a\qquad(1)
- Differentiate again and solve for b.y''=2b\quad\Longrightarrow\quad b=\dfrac{y''}{2}
- Use b in (1) to find 2a.2a=y'-2bx=y'-xy''
- Substitute b and 2a into y=bx^{2}+(2a)x.y=\dfrac{y''}{2}x^{2}+x\left(y'-xy''\right)=xy'-\dfrac{x^{2}}{2}y''
- Multiply by 2 and rearrange.x^{2}y''-2xy'+2y=0
Check. With y=bx^{2}+2ax: x^{2}(2b)-2x(2bx+2a)+2(bx^{2}+2ax)=2bx^{2}-4bx^{2}-4ax+2bx^{2}+4ax=0.
- Differentiate the given function with the chain rule.y'=-\dfrac12e^{-x/2}
- Substitute in the left side.2y'+y=2\left(-\dfrac12e^{-x/2}\right)+e^{-x/2}=-e^{-x/2}+e^{-x/2}=0
- The left side equals the right side (0) for every x, so y=e^{-x/2} is a solution.
- Differentiate.\dfrac{dy}{dx}=3e^{3x}+20e^{2x}
- Substitute in the left side.\dfrac{dy}{dx}-2y=3e^{3x}+20e^{2x}-2e^{3x}-20e^{2x}=e^{3x}
- This equals the right side, so the function is a solution.
- Differentiate: \dfrac{d}{dx}\tan5x=5\sec^{2}5x.y'=25\sec^{2}5x
- Use \sec^{2}\theta=1+\tan^{2}\theta.y'=25\left(1+\tan^{2}5x\right)=25+25\tan^{2}5x=25+(5\tan5x)^{2}=25+y^{2}
- So y'=25+y^{2} on any interval where \tan5x is defined, for example -\dfrac{\pi}{10}<x<\dfrac{\pi}{10}.
- Differentiate term by term.y'=\dfrac12\cos x+\dfrac12\sin x-10e^{-x}
- Add y. The cosine terms and the exponential terms cancel.y'+y=\dfrac12\cos x+\dfrac12\sin x-10e^{-x}+\dfrac12\sin x-\dfrac12\cos x+10e^{-x}=\sin x
- This equals the right side, so the function is a solution.
- Write the DE with a derivative: divide by dx.x^{3}\dfrac{dy}{dx}-2=0
- Differentiate y=-x^{-2}+6.\dfrac{dy}{dx}=2x^{-3}=\dfrac{2}{x^{3}}
- Substitute in the left side.x^{3}\cdot\dfrac{2}{x^{3}}-2=2-2=0
- This holds for every x\neq0, so the function is a solution.
- Differentiate with the product rule.y'=x\cdot\dfrac1x+\ln x=1+\ln x
- Substitute in the left side.y'-\dfrac{y}{x}=1+\ln x-\dfrac{x\ln x}{x}=1+\ln x-\ln x=1
- This equals the right side for every x>0, so the function is a solution.
- Divide by dx to see the derivative.\dfrac{dy}{dx}-\sin x=0
- The only derivative is \dfrac{dy}{dx}, of first order, and it appears to the power 1. The equation is a polynomial in \dfrac{dy}{dx}, so the degree is defined.
- As a check, its general solution y=-\cos x+c has one arbitrary constant, as expected for a first order DE.
- Differentiate once.\dfrac{dy}{dx}=-a\sin x+b\cos x
- Differentiate again.\dfrac{d^{2}y}{dx^{2}}=-a\cos x-b\sin x=-\left(a\cos x+b\sin x\right)=-y
- Substitute in the left side of the DE.\dfrac{d^{2}y}{dx^{2}}+y=-y+y=0
- This is true for all x and for all real a, b. Since there are two arbitrary constants and the DE is second order, y=a\cos x+b\sin x is in fact its general solution.
- For y_1=x^{2}: y_1'=2x, y_1''=2. Substitute.x^{2}(2)-4x(2x)+6x^{2}=2x^{2}-8x^{2}+6x^{2}=0
- For y_2=x^{3}: y_2'=3x^{2}, y_2''=6x. Substitute.x^{2}(6x)-4x(3x^{2})+6x^{3}=6x^{3}-12x^{3}+6x^{3}=0
- Both left sides are 0, so y_1=x^{2} and y_2=x^{3} are solutions.
- For y=c_1x^{2}: y'=2c_1x, y''=2c_1. Every term contains the factor c_1.x^{2}(2c_1)-4x(2c_1x)+6c_1x^{2}=c_1\left(2x^{2}-8x^{2}+6x^{2}\right)=c_1\cdot0=0
- For y=c_2x^{3}: y'=3c_2x^{2}, y''=6c_2x.x^{2}(6c_2x)-4x(3c_2x^{2})+6c_2x^{3}=c_2\left(6x^{3}-12x^{3}+6x^{3}\right)=0
- So yes, both are solutions for any constants c_1, c_2. Reason: the DE is linear and its right side is 0, so a constant factor passes through every term.
- Let y=x^{2}+x^{3}. Then y'=2x+3x^{2} and y''=2+6x.x^{2}(2+6x)-4x\left(2x+3x^{2}\right)+6\left(x^{2}+x^{3}\right)
- Expand and collect.=2x^{2}+6x^{3}-8x^{2}-12x^{3}+6x^{2}+6x^{3}=0
- So y_1+y_2 is also a solution. In fact y=c_1x^{2}+c_2x^{3} is a solution for all c_1,c_2: with two arbitrary constants, it is the general solution of this second order DE.
Exercise 4.2p. 133
Separate the variables, integrate both sides with one constant, then (for Q9 to Q15) use the initial condition to find the constant.
- Write \dfrac{1}{e^{3x}}=e^{-3x} and separate: dy on the left, everything in x on the right.dy=-e^{-3x}\,dx
- Integrate both sides. Use \int e^{ax}\,dx=\dfrac{e^{ax}}{a} with a=-3.y=-\dfrac{e^{-3x}}{-3}+c=\dfrac13e^{-3x}+c
- Divide by xy and multiply by dx to separate.\dfrac{dy}{y}=\dfrac{4\,dx}{x}
- Integrate; write the constant as \ln c.\ln y=4\ln x+\ln c=\ln x^{4}+\ln c=\ln\left(cx^{4}\right)
- Take antilog.y=cx^{4}
Dividing by y assumed y\neq0; the solution y=0 is included by c=0.
- Separate: divide by y^{3} and multiply by dx.y^{-3}\,dy=x^{-2}\,dx
- Integrate with the power rule \int u^{n}\,du=\dfrac{u^{n+1}}{n+1}.\dfrac{y^{-2}}{-2}=\dfrac{x^{-1}}{-1}+c\quad\Longrightarrow\quad -\dfrac{1}{2y^{2}}=-\dfrac{1}{x}+c
- Multiply by -2 and rename C=-2c.\dfrac{1}{y^{2}}=\dfrac{2}{x}+C
Multiplying by xy^{2} gives the equivalent form x=y^{2}(2+Cx). The constant solution y=0 was lost when dividing by y^{3}, and it also satisfies the DE.
- Use e^{2x+3y}=e^{2x}\cdot e^{3y}; the equation is then separable.\dfrac{dy}{dx}=e^{2x}e^{3y}\quad\Longrightarrow\quad e^{-3y}\,dy=e^{2x}\,dx
- Integrate both sides.-\dfrac13e^{-3y}=\dfrac12e^{2x}+c
- Multiply by -6 to clear fractions, and rename C=-6c.2e^{-3y}=-3e^{2x}-6c\quad\Longrightarrow\quad 3e^{2x}+2e^{-3y}=C
- Separate: divide by y^{2} and multiply by dx.\dfrac{dy}{y^{2}}=\dfrac{x^{2}}{1+x}\,dx
- The right side is an improper fraction; divide x^{2} by x+1 first.\dfrac{x^{2}}{x+1}=x-1+\dfrac{1}{x+1}
- Integrate both sides.\int y^{-2}\,dy=\int\left(x-1+\dfrac{1}{x+1}\right)dx\quad\Longrightarrow\quad -\dfrac1y=\dfrac{x^{2}}{2}-x+\ln|1+x|+c
Explicitly, y=-\dfrac{1}{\tfrac12x^{2}-x+\ln|1+x|+c}. The constant solution y=0 also satisfies the DE.
- Divide by (x+1) to separate.2y\,dy=\dfrac{x}{x+1}\,dx
- Rewrite the fraction: \dfrac{x}{x+1}=\dfrac{(x+1)-1}{x+1}=1-\dfrac{1}{x+1}.2y\,dy=\left(1-\dfrac{1}{x+1}\right)dx
- Integrate both sides.y^{2}=x-\ln|x+1|+c
- Move the second term to the right and separate.\dfrac{dy}{dx}=-y^{2}\sin x\quad\Longrightarrow\quad \dfrac{dy}{y^{2}}=-\sin x\,dx
- Integrate: \int y^{-2}\,dy=-\dfrac1y and \int(-\sin x)\,dx=\cos x.-\dfrac1y=\cos x+c
- Solve for y and rename C=-c.y=-\dfrac{1}{\cos x+c}=\dfrac{1}{C-\cos x}
The constant solution y=0 (lost when dividing by y^{2}) also satisfies the DE.
- Divide both sides by \cos x to separate the variables.\dfrac{\sin x+\cos x}{\cos x}\,dx=\cot y\,dy\quad\Longrightarrow\quad(\tan x+1)\,dx=\cot y\,dy
- Integrate. Use \int\cot y\,dy=\ln|\sin y| and \int\tan x\,dx=-\ln|\cos x|.\ln|\sin y|=-\ln|\cos x|+x+c
- Collect the logarithms on the left.\ln|\sin y|+\ln|\cos x|=x+c\quad\Longrightarrow\quad \ln|\sin y\cos x|=x+c
- Take antilog and write C=\pm e^{c}.\sin y\cos x=Ce^{x}
- Separate and integrate.dy=\cos x\,dx\quad\Longrightarrow\quad y=\sin x+c
- Use y(0)=1: 1=\sin0+c.c=1
- Substitute c back.y=\sin x+1
- Divide by 2 and separate.dy=2xe^{-x}\,dx
- Integrate the right side by parts with u=x, dv=e^{-x}dx: \int xe^{-x}dx=-xe^{-x}-\int(-e^{-x})dx=-xe^{-x}-e^{-x}.y=2\left(-xe^{-x}-e^{-x}\right)+c=-2(x+1)e^{-x}+c
- Use y(0)=2: 2=-2(1)(1)+c.c=4
- Substitute c back.y=4-2(x+1)e^{-x}
- Move the y term to the right and separate.\dfrac{dy}{y}=-\dfrac{1+x}{x}\,dx=-\left(\dfrac1x+1\right)dx
- Integrate both sides.\ln|y|=-\ln|x|-x+c
- Take antilog; write A=\pm e^{c}.y=\dfrac{A\,e^{-x}}{x}
- Use y(1)=1: 1=Ae^{-1}.A=e
- Substitute A back.y=\dfrac{e\cdot e^{-x}}{x}=\dfrac{e^{1-x}}{x}
- Separate the variables.\dfrac{dy}{y}=-\tan2x\,dx
- Integrate. Since \int\tan2x\,dx=-\dfrac12\ln|\cos2x|, the right side is +\dfrac12\ln|\cos2x|.\ln|y|=\dfrac12\ln|\cos2x|+c
- Take antilog; write A=\pm e^{c}.y=A\sqrt{\cos2x}
- Use y(0)=2: 2=A\sqrt{\cos0}=A.A=2
- Substitute A back.y=2\sqrt{\cos2x}\qquad\left(\text{or } y^{2}=4\cos2x\right)
- Separate the variables.\dfrac{dy}{y^{2}+4}=dx
- Integrate with \int\dfrac{dy}{y^{2}+a^{2}}=\dfrac1a\tan^{-1}\dfrac{y}{a}, a=2.\dfrac12\tan^{-1}\dfrac{y}{2}=x+c
- Use y(0)=-2: \dfrac12\tan^{-1}(-1)=c, and \tan^{-1}(-1)=-\dfrac{\pi}{4}.c=-\dfrac{\pi}{8}
- Substitute c and multiply by 2.\tan^{-1}\dfrac{y}{2}=2x-\dfrac{\pi}{4}
- Take \tan of both sides.y=2\tan\left(2x-\dfrac{\pi}{4}\right)
- Move the dx term to the right, then multiply by y and divide by (1-x).(1-x)\,dy=-\dfrac{dx}{y}\quad\Longrightarrow\quad y\,dy=-\dfrac{dx}{1-x}
- Integrate. Since \dfrac{d}{dx}\ln|1-x|=-\dfrac{1}{1-x}, the right side integrates to \ln|1-x|.\dfrac{y^{2}}{2}=\ln|1-x|+c
- Use y(0)=2: \dfrac{4}{2}=\ln1+c.c=2
- Substitute c and multiply by 2.y^{2}=2\ln|1-x|+4
- Take the positive root, because y(0)=2>0.y=\sqrt{4+2\ln(1-x)},\qquad x<1
- The variables are already separated. Integrate both sides; \int2(y-1)\,dy=(y-1)^{2}.(y-1)^{2}=x^{3}+2x^{2}+2x+c
- Use y(0)=-1: (-1-1)^{2}=0+c.c=4
- Substitute c back.(y-1)^{2}=x^{3}+2x^{2}+2x+4
- Take the square root. At x=0, y-1=-2<0, so choose the negative root.y-1=-\sqrt{x^{3}+2x^{2}+2x+4}\quad\Longrightarrow\quad y=1-\sqrt{x^{3}+2x^{2}+2x+4}
Since x^{3}+2x^{2}+2x+4=(x+2)(x^{2}+2), the expression under the root is positive exactly when x>-2, so the solution is valid for x>-2.
Exercise 4.3p. 136
Q1 to Q3: test f(tx,ty)=t^{n}f(x,y). Q4 to Q15: substitute y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}, separate, integrate, and replace u by \dfrac{y}{x}.
- Replace x by tx and y by ty.f(tx,ty)=6(tx)(ty)^{3}-(tx)^{2}(ty)^{2}=6t^{4}xy^{3}-t^{4}x^{2}y^{2}
- Take out the common factor t^{4}.f(tx,ty)=t^{4}\left(6xy^{3}-x^{2}y^{2}\right)=t^{4}f(x,y)
- So f is homogeneous of degree 4 (each term has total degree 1+3=2+2=4).
- Replace x by tx and y by ty.f(tx,ty)=t^{2}x^{2}-ty
- The two terms carry different powers of t (t^{2} and t^{1}), so no single power t^{n} can be taken out.f(tx,ty)\neq t^{n}f(x,y)
- So f is not homogeneous: its terms have total degrees 2 and 1.
- Replace x by tx and y by ty.f(tx,ty)=\dfrac{2t^{3}y^{3}}{t^{2}x^{2}\cdot ty}-7=\dfrac{2t^{3}y^{3}}{t^{3}x^{2}y}-7
- The t^{3} cancels, so the function is unchanged.f(tx,ty)=\dfrac{2y^{3}}{x^{2}y}-7=t^{0}f(x,y)
- So f is homogeneous of degree 0. The constant -7 is allowed here because the fraction also has degree 3-3=0.
- M=x-y and N=x are both of degree 1, so the equation is homogeneous. Write it as \dfrac{dy}{dx}.\dfrac{dy}{dx}=-\dfrac{x-y}{x}=\dfrac{y-x}{x}
- Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}.u+x\dfrac{du}{dx}=\dfrac{ux-x}{x}=u-1
- The u cancels.x\dfrac{du}{dx}=-1\quad\Longrightarrow\quad du=-\dfrac{dx}{x}
- Integrate.u=-\ln|x|+c
- Replace u by \dfrac{y}{x} and multiply by x.\dfrac{y}{x}=c-\ln|x|\quad\Longrightarrow\quad y=x\left(c-\ln|x|\right)
- M=y and N=-(y-x) are both of degree 1, so the equation is homogeneous.\dfrac{dy}{dx}=\dfrac{y}{y-x}
- Put y=ux.u+x\dfrac{du}{dx}=\dfrac{ux}{ux-x}=\dfrac{u}{u-1}
- Subtract u.x\dfrac{du}{dx}=\dfrac{u}{u-1}-u=\dfrac{u-u^{2}+u}{u-1}=\dfrac{u(2-u)}{u-1}
- Separate the variables.\dfrac{u-1}{u(2-u)}\,du=\dfrac{dx}{x}
- Partial fractions: u-1=A(2-u)+Bu. Put u=0: A=-\tfrac12. Put u=2: B=\tfrac12.\dfrac{u-1}{u(2-u)}=-\dfrac{1}{2u}+\dfrac{1}{2(2-u)}
- Integrate. Note \int\dfrac{du}{2-u}=-\ln|2-u|.-\dfrac12\ln|u|-\dfrac12\ln|2-u|=\ln|x|+c
- Multiply by -2 and combine the logarithms.\ln\left|u(2-u)\right|+\ln x^{2}=-2c\quad\Longrightarrow\quad u(2-u)\,x^{2}=C_1
- Replace u by \dfrac{y}{x}: \dfrac{y}{x}\left(2-\dfrac{y}{x}\right)x^{2}=y(2x-y).2xy-y^{2}=C_1\quad\Longrightarrow\quad y^{2}-2xy=C
- Numerator and denominator are both of degree 1, so the equation is homogeneous. Put y=ux.u+x\dfrac{du}{dx}=\dfrac{ux-x}{x+ux}=\dfrac{u-1}{1+u}
- Subtract u.x\dfrac{du}{dx}=\dfrac{u-1-u-u^{2}}{1+u}=-\dfrac{1+u^{2}}{1+u}
- Separate the variables.\dfrac{1+u}{1+u^{2}}\,du=-\dfrac{dx}{x}
- Split the fraction and integrate: \int\dfrac{du}{1+u^{2}}=\tan^{-1}u and \int\dfrac{u\,du}{1+u^{2}}=\dfrac12\ln(1+u^{2}).\tan^{-1}u+\dfrac12\ln\left(1+u^{2}\right)=-\ln|x|+c
- Replace u by \dfrac{y}{x}. Note \dfrac12\ln\dfrac{x^{2}+y^{2}}{x^{2}}=\dfrac12\ln(x^{2}+y^{2})-\ln|x|, so the \ln|x| terms cancel.\tan^{-1}\dfrac{y}{x}+\dfrac12\ln\left(x^{2}+y^{2}\right)=c
- Multiply by 2 and rename C=2c.\ln\left(x^{2}+y^{2}\right)+2\tan^{-1}\dfrac{y}{x}=C
- Numerator and denominator are of degree 2, so the equation is homogeneous. Put y=ux.u+x\dfrac{du}{dx}=\dfrac{u^{2}x^{2}+ux^{2}}{x^{2}}=u^{2}+u
- Subtract u and separate.x\dfrac{du}{dx}=u^{2}\quad\Longrightarrow\quad \dfrac{du}{u^{2}}=\dfrac{dx}{x}
- Integrate.-\dfrac1u=\ln|x|+c
- Replace u by \dfrac{y}{x} and solve for y.-\dfrac{x}{y}=\ln|x|+c\quad\Longrightarrow\quad y=-\dfrac{x}{\ln|x|+c}
The solution y=0 (u=0, lost when dividing by u^{2}) also satisfies the DE.
- Take x>0, y>0 so that \sqrt{xy} is real. Numerator and denominator are of degree 1. Put y=ux; then \sqrt{xy}=\sqrt{ux^{2}}=x\sqrt u.u+x\dfrac{du}{dx}=\dfrac{ux}{x+x\sqrt u}=\dfrac{u}{1+\sqrt u}
- Subtract u.x\dfrac{du}{dx}=\dfrac{u-u-u\sqrt u}{1+\sqrt u}=-\dfrac{u^{3/2}}{1+\sqrt u}
- Separate the variables and split the fraction.\dfrac{1+\sqrt u}{u^{3/2}}\,du=-\dfrac{dx}{x}\quad\Longrightarrow\quad\left(u^{-3/2}+u^{-1}\right)du=-\dfrac{dx}{x}
- Integrate.-2u^{-1/2}+\ln u=-\ln x+c
- Replace u by \dfrac{y}{x}: u^{-1/2}=\sqrt{\dfrac{x}{y}} and \ln u=\ln y-\ln x. The \ln x terms cancel.-2\sqrt{\dfrac{x}{y}}+\ln y-\ln x=-\ln x+c\quad\Longrightarrow\quad \ln y=2\sqrt{\dfrac{x}{y}}+c
Taking antilog gives the equivalent form y=Ce^{2\sqrt{x/y}}.
- Numerator and denominator are of degree 3, so the equation is homogeneous. Put y=ux.u+x\dfrac{du}{dx}=\dfrac{3x^{3}+u^{3}x^{3}}{x\cdot u^{2}x^{2}}=\dfrac{3+u^{3}}{u^{2}}=\dfrac{3}{u^{2}}+u
- Subtract u and separate.x\dfrac{du}{dx}=\dfrac{3}{u^{2}}\quad\Longrightarrow\quad u^{2}\,du=\dfrac{3\,dx}{x}
- Integrate.\dfrac{u^{3}}{3}=3\ln|x|+c
- Replace u by \dfrac{y}{x}, multiply by 3x^{3} and rename C=3c.\dfrac{y^{3}}{x^{3}}=9\ln|x|+C\quad\Longrightarrow\quad y^{3}=9x^{3}\ln|x|+Cx^{3}
- Numerator and denominator are of degree 1. Put y=ux.u+x\dfrac{du}{dx}=\dfrac{x+3ux}{3x+ux}=\dfrac{1+3u}{3+u}
- Subtract u.x\dfrac{du}{dx}=\dfrac{1+3u-3u-u^{2}}{3+u}=\dfrac{1-u^{2}}{3+u}
- Separate the variables.\dfrac{3+u}{(1-u)(1+u)}\,du=\dfrac{dx}{x}
- Partial fractions: 3+u=A(1+u)+B(1-u). Put u=1: A=2. Put u=-1: B=1.\dfrac{3+u}{(1-u)(1+u)}=\dfrac{2}{1-u}+\dfrac{1}{1+u}
- Integrate; \int\dfrac{2\,du}{1-u}=-2\ln|1-u|. Write the constant as \ln c.-2\ln|1-u|+\ln|1+u|=\ln|x|+\ln c\quad\Longrightarrow\quad \dfrac{1+u}{(1-u)^{2}}=cx
- Replace u by \dfrac{y}{x}: \dfrac{(x+y)/x}{(x-y)^{2}/x^{2}}=\dfrac{x(x+y)}{(x-y)^{2}}.\dfrac{x(x+y)}{(x-y)^{2}}=cx\quad\Longrightarrow\quad x+y=c(x-y)^{2}
- Divide by x\,dx. The right side depends only on \dfrac{y}{x}, so the equation is homogeneous.\dfrac{dy}{dx}=\dfrac{y}{x}+\cot\dfrac{y}{x}
- Put y=ux. The u cancels.u+x\dfrac{du}{dx}=u+\cot u\quad\Longrightarrow\quad x\dfrac{du}{dx}=\cot u
- Separate: \dfrac{1}{\cot u}=\tan u.\tan u\,du=\dfrac{dx}{x}
- Integrate; \int\tan u\,du=-\ln|\cos u|. Write the constant as -\ln C.-\ln|\cos u|=\ln|x|-\ln C\quad\Longrightarrow\quad \ln\left|x\cos u\right|=\ln C
- Take antilog and replace u by \dfrac{y}{x}.x\cos\dfrac{y}{x}=C
- Divide by xy^{2}. Numerator and denominator are of degree 3, so the equation is homogeneous.\dfrac{dy}{dx}=\dfrac{y^{3}-x^{3}}{xy^{2}}
- Put y=ux.u+x\dfrac{du}{dx}=\dfrac{u^{3}x^{3}-x^{3}}{x\cdot u^{2}x^{2}}=\dfrac{u^{3}-1}{u^{2}}=u-\dfrac{1}{u^{2}}
- Subtract u and separate.x\dfrac{du}{dx}=-\dfrac{1}{u^{2}}\quad\Longrightarrow\quad u^{2}\,du=-\dfrac{dx}{x}
- Integrate.\dfrac{u^{3}}{3}=-\ln|x|+c
- Replace u by \dfrac{y}{x}, multiply by 3x^{3}, and rename C=3c.y^{3}=x^{3}\left(C-3\ln|x|\right)
- Use y(1)=2: 8=1\cdot(C-0).C=8
- Substitute C back.y^{3}=x^{3}\left(8-3\ln x\right)
- Both sides have degree 2, so the equation is homogeneous.\dfrac{dy}{dx}=\dfrac{x^{2}+2y^{2}}{xy}
- Put y=ux.u+x\dfrac{du}{dx}=\dfrac{x^{2}+2u^{2}x^{2}}{x\cdot ux}=\dfrac{1+2u^{2}}{u}
- Subtract u.x\dfrac{du}{dx}=\dfrac{1+2u^{2}-u^{2}}{u}=\dfrac{1+u^{2}}{u}
- Separate the variables.\dfrac{u\,du}{1+u^{2}}=\dfrac{dx}{x}
- Integrate: the numerator is half the derivative of 1+u^{2}. Write the constant as \ln\sqrt{C}.\dfrac12\ln\left(1+u^{2}\right)=\ln|x|+\dfrac12\ln C\quad\Longrightarrow\quad 1+u^{2}=Cx^{2}
- Replace u by \dfrac{y}{x} and multiply by x^{2}.x^{2}+y^{2}=Cx^{4}
- Use y(1)=1: 1+1=C.C=2
- Substitute C back.x^{2}+y^{2}=2x^{4}
- Divide by 2x^{2}. Every term has degree 2, so the equation is homogeneous.\dfrac{dy}{dx}=\dfrac{3xy+y^{2}}{2x^{2}}
- Put y=ux.u+x\dfrac{du}{dx}=\dfrac{3ux^{2}+u^{2}x^{2}}{2x^{2}}=\dfrac{3u+u^{2}}{2}
- Subtract u.x\dfrac{du}{dx}=\dfrac{3u+u^{2}-2u}{2}=\dfrac{u(1+u)}{2}
- Separate the variables.\dfrac{2\,du}{u(1+u)}=\dfrac{dx}{x}
- Partial fractions: \dfrac{1}{u(1+u)}=\dfrac1u-\dfrac{1}{1+u}. Integrate.2\ln|u|-2\ln|1+u|=\ln|x|+c\quad\Longrightarrow\quad \ln\left(\dfrac{u}{1+u}\right)^{2}=\ln|x|+c
- Take antilog (C=e^{c}) and replace u by \dfrac{y}{x}: \dfrac{u}{1+u}=\dfrac{y}{x+y}.\left(\dfrac{y}{x+y}\right)^{2}=Cx
- Use y(1)=-2: \left(\dfrac{-2}{1-2}\right)^{2}=C.C=4\quad\Longrightarrow\quad y^{2}=4x\,(x+y)^{2}
- Solve for y. At x=1, \dfrac{y}{x+y}=\dfrac{-2}{-1}=2>0, so take \dfrac{y}{x+y}=+2\sqrt x.y=2\sqrt x\,(x+y)\quad\Longrightarrow\quad y\left(1-2\sqrt x\right)=2x\sqrt x\quad\Longrightarrow\quad y=\dfrac{2x^{3/2}}{1-2\sqrt x}
- Write the equation as \dfrac{dy}{dx}. Dividing numerator and denominator by x shows it depends only on \dfrac{y}{x}, so it is homogeneous.\dfrac{dy}{dx}=\dfrac{x+ye^{y/x}}{xe^{y/x}}
- Put y=ux.u+x\dfrac{du}{dx}=\dfrac{x+uxe^{u}}{xe^{u}}=\dfrac{1+ue^{u}}{e^{u}}=e^{-u}+u
- The u cancels. Separate the variables.x\dfrac{du}{dx}=e^{-u}\quad\Longrightarrow\quad e^{u}\,du=\dfrac{dx}{x}
- Integrate.e^{u}=\ln|x|+c\quad\Longrightarrow\quad e^{y/x}=\ln|x|+c
- Use y(1)=0: e^{0}=\ln1+c.c=1
- Substitute c and take \ln of both sides.e^{y/x}=1+\ln x\quad\Longrightarrow\quad y=x\ln\left(1+\ln x\right)
Exercise 4.4p. 138
Write the DE from the words, solve it by separating the variables, then use the two pieces of data to find the constants.
- P(0) = 20\ \text{million} present population
- P(4) = 25\ \text{million}
- P(12)
- Model. Let P(t) be the population in millions after t years. Introduce the constant of proportionality k.\dfrac{dP}{dt}=kP
- Separate the variables and integrate.\dfrac{dP}{P}=k\,dt\quad\Longrightarrow\quad \ln P=kt+\ln c\quad\Longrightarrow\quad P=c\,e^{kt}
- Use P(0)=20: c=20.P=20e^{kt}
- Use P(4)=25 to find k.25=20e^{4k}\quad\Longrightarrow\quad e^{4k}=\dfrac54\quad\Longrightarrow\quad k=\dfrac14\ln1.25=0.0558\ \text{per year}
- Population after 12 years: e^{12k}=\left(e^{4k}\right)^{3}.P(12)=20\left(\dfrac54\right)^{3}=20\times\dfrac{125}{64}=39.06\ \text{million}
- T(0) = 150^\circ\mathrm{C}
- T(4) = 90^\circ\mathrm{C}
- T_s = 20^\circ\mathrm{C} room temperature
- t when T=40^\circ\mathrm{C}
- Model. Newton's law of cooling, with t in minutes.\dfrac{dT}{dt}=k\left(T-20\right)
- Separate, integrate and take antilog.\dfrac{dT}{T-20}=k\,dt\quad\Longrightarrow\quad \ln(T-20)=kt+\ln c\quad\Longrightarrow\quad T-20=c\,e^{kt}
- Use T(0)=150: c=130.T=20+130e^{kt}
- Use T(4)=90 to find k.90=20+130e^{4k}\quad\Longrightarrow\quad e^{4k}=\dfrac{70}{130}=\dfrac{7}{13}\quad\Longrightarrow\quad k=\dfrac14\ln\dfrac{7}{13}=-0.1548
- Put T=40 and solve for t.40=20+130e^{kt}\quad\Longrightarrow\quad e^{kt}=\dfrac{2}{13}\quad\Longrightarrow\quad t=\dfrac{\ln(2/13)}{k}=\dfrac{4\ln(2/13)}{\ln(7/13)}
- Evaluate.t=\dfrac{4(-1.8718)}{-0.6190}=12.1\ \text{min}
So the pizza reaches 40^\circ\mathrm{C} about 12.1 minutes after it is taken out of the oven, that is about 8.1 minutes after the 90^\circ\mathrm{C} reading.
- P(0) = 20
- P(2) = 4\times20=80 four times in two days
- P(10)
- Model. Let P(t) be the number of bacteria after t days.\dfrac{dP}{dt}=kP\quad\Longrightarrow\quad P=P(0)\,e^{kt}=20e^{kt}
- Use P(2)=80.80=20e^{2k}\quad\Longrightarrow\quad e^{2k}=4\quad\Longrightarrow\quad k=\dfrac12\ln4=\ln2=0.693\ \text{per day}
- After 10 days: e^{10k}=\left(e^{2k}\right)^{5}=4^{5}.P(10)=20\times4^{5}=20\times1024=20\,480
- A(0) = 50\,\mathrm{g}
- T_{1/2} = 1000\ \text{years} so A(1000)=25\,\mathrm{g}
- A(800)
- Model. Let A(t) be the amount in grams after t years.\dfrac{dA}{dt}=kA\quad\Longrightarrow\quad A=50e^{kt}
- Half-life: A(1000)=25.\begin{aligned}25&=50e^{1000k}\\ \Longrightarrow\ e^{1000k}&=\dfrac12\\ \Longrightarrow\ k&=-\dfrac{\ln2}{1000}\\ &=-0.000693\ \text{per year}\end{aligned}
- After 800 years: e^{800k}=\left(e^{1000k}\right)^{0.8}=\left(\tfrac12\right)^{0.8}.A(800)=50\left(\dfrac12\right)^{0.8}=50\times0.5743=28.72\,\mathrm{g}
- T(0) = 80^\circ\mathrm{F}
- T_s = 30^\circ\mathrm{F} block of ice
- T(1) = 40^\circ\mathrm{F}
- t when T=70^\circ\mathrm{F}
- Model. Newton's law of cooling, t in minutes. As in Q2 the solution isT=30+(80-30)e^{kt}=30+50e^{kt}
- Use T(1)=40.40=30+50e^{k}\quad\Longrightarrow\quad e^{k}=\dfrac15\quad\Longrightarrow\quad k=-\ln5=-1.609\ \text{per minute}
- Put T=70.70=30+50e^{kt}\quad\Longrightarrow\quad e^{kt}=\dfrac45
- Take logarithms and solve for t.t=\dfrac{\ln(4/5)}{-\ln5}=\dfrac{\ln1.25}{\ln5}=\dfrac{0.2231}{1.6094}=0.139\ \text{min}
- Convert to seconds.t=0.139\times60\approx8.3\,\mathrm{s}
Note. 70^\circ\mathrm{F} lies between the starting reading 80^\circ\mathrm{F} and the one-minute reading 40^\circ\mathrm{F}, so the thermometer passes 70^\circ\mathrm{F} during the first minute. The small answer is therefore expected.
- v(0) = 40\,\mathrm{m\,s^{-1}} upward
- g = 9.8\,\mathrm{m\,s^{-2}} downward
- v(1)
- maximum height H
- DE. Take upward as positive. Gravity reduces the velocity at the constant rate g.\dfrac{dv}{dt}=-g\qquad\text{or}\qquad dv=-g\,dt
- Integrate and use v(0)=40.v=-gt+c_1,\quad c_1=40\quad\Longrightarrow\quad v=40-9.8t\qquad(\text{i})
- Velocity after 1 s.v(1)=40-9.8=30.2\,\mathrm{m\,s^{-1}}\ \ (\text{upward})
- Height. \dfrac{ds}{dt}=40-9.8t. Integrate, with s(0)=0.s=40t-4.9t^{2}\qquad(\text{ii})
- At the maximum height the ball stops for an instant: v=0 in (i).0=40-9.8t\quad\Longrightarrow\quad t=\dfrac{40}{9.8}=4.08\,\mathrm{s}
- Substitute t=4.08\,\mathrm{s} in (ii).H=40(4.082)-4.9(4.082)^{2}=163.3-81.6=81.6\,\mathrm{m}
Review exercisep. 139
Mixed revision of the whole unit: MCQs, order and degree, verifying solutions, and solving separable and homogeneous equations.
- A1
- B2
- C3Answer
- D4
The highest derivative present is \dfrac{d^{3}y}{dx^{3}}, so the order is 3.
Option (d) 4 is the power of \dfrac{dy}{dx}. That power decides neither the order nor the degree (the degree here is 1, the power of \dfrac{d^{3}y}{dx^{3}}).
- A0
- B1Answer
- C2
- D3
The highest derivative \dfrac{d^{2}y}{dx^{2}} appears to the power 1, so the degree is 1.
The cube in 9y^{3} is on y, not on a derivative, so (d) is wrong; (c) 2 is the order, not the degree.
- A\dfrac{dy}{dx}+8y=32
- B\dfrac{dy}{dx}+6y=32
- C\dfrac{dy}{dx}+5y=32
- D\dfrac{dy}{dx}+4y=32Answer
For the constant function y=8, \dfrac{dy}{dx}=0. The left side becomes k(8), which must equal 32, so k=4.
Check (d): 0+4(8)=32. The other options give 64, 48 and 40.
- A1
- B2Answer
- C3
- D4
f(tx,ty)=\dfrac{t^{3}\left(x^{3}-y^{3}\right)}{t(x-y)}=t^{2}f(x,y), so the degree is 2.
Quick rule: degree of numerator minus degree of denominator, 3-1=2. Indeed f=x^{2}+xy+y^{2} for x\neq y. Option (c) forgets the denominator.
- Ay=x+cAnswer
- By=x^{2}+c
- Cy^{2}=x^{2}+c
- Dy^{2}=x+c
Integrate both sides: \int dy=\int dx gives y=x+c.
Check: \dfrac{d}{dx}(x+c)=1, so dy=dx.
- A1Answer
- B2
- C3
- D0
The general solution contains as many arbitrary constants as the order of the DE. A first order DE is solved by one integration, which gives exactly one constant.
- Ae^{-x-y}=c
- Be^{-x}+e^{y}=c
- Ce^{x}+e^{y}=c
- De^{x}+e^{-y}=cAnswer
Write e^{x+y}=e^{x}e^{y} and separate: e^{-y}\,dy=e^{x}\,dx.
Integrate: -e^{-y}=e^{x}+c_1, so e^{x}+e^{-y}=c with c=-c_1.
- Ax^{2}-y^{2}=c
- Bx^{3}+y^{3}=c
- Cx^{3}-y^{3}=cAnswer
- Dx^{2}+y^{2}=c
The variables are already separated: y^{2}\,dy=x^{2}\,dx. Integrate: \dfrac{y^{3}}{3}=\dfrac{x^{3}}{3}+c_1.
Multiply by 3: x^{3}-y^{3}=c with c=-3c_1. Options (a) and (d) forget that \int x^{2}dx raises the power to 3.
- A\sin x\cos y=c
- B\cos x\cos y=c
- C\cos x\sin y=c
- D\sin x\sin y=cAnswer
Divide by \sin x\sin y: \cot x\,dx+\cot y\,dy=0. Integrate: \ln\sin x+\ln\sin y=\ln c, so \sin x\sin y=c.
Check: the left side of the DE is exactly d(\sin x\sin y) by the product rule.
- A-1Answer
- B1
- C10
- D100
The order is the order of a derivative that appears, so it is always a positive integer (1, 2, 3, ...). -1 is impossible; 10 and 100 are unusual but possible.
- The only derivative is \dfrac{dy}{dx} (first order).
- The equation is already a polynomial in \dfrac{dy}{dx}. The root \sqrt{xy} contains no derivative, so it is just a coefficient and need not be removed.
- The highest power of \dfrac{dy}{dx} is 2.
- A derivative is under the square root, so square both sides first.\left(\dfrac{dy}{dx}\right)^{2}=1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{4}
- The highest derivative is \dfrac{d^{2}y}{dx^{2}} (second order), and its power is 4.
Note. In part (ii) the book prints the second derivative as \dfrac{d^{2}y}{d^{2}x}; it should be \dfrac{d^{2}y}{dx^{2}}.
- Differentiate.\dfrac{dy}{dx}=-\dfrac12e^{-x/2}
- Substitute in the left side.2\left(-\dfrac12e^{-x/2}\right)+e^{-x/2}=-e^{-x/2}+e^{-x/2}=0
- The left side equals the right side, so y=e^{-x/2} is a solution.
- Differentiate: \dfrac{d}{dx}e^{-20x}=-20e^{-20x}.\dfrac{dy}{dx}=-\dfrac65\left(-20e^{-20x}\right)=24e^{-20x}
- Substitute in the left side.24e^{-20x}+20\left(\dfrac65-\dfrac65e^{-20x}\right)=24e^{-20x}+24-24e^{-20x}=24
- The left side equals 24, so the function is a solution.
- Differentiate with the chain rule.\dfrac{dy}{dx}=25\sec^{2}5x
- Use \sec^{2}\theta=1+\tan^{2}\theta.\dfrac{dy}{dx}=25+25\tan^{2}5x=25+(5\tan5x)^{2}=25+y^{2}
- So the function satisfies the DE wherever \tan5x is defined.
- Divide by dx.x^{2}\dfrac{dy}{dx}+2xy=0
- Differentiate y=x^{-2}.\dfrac{dy}{dx}=-2x^{-3}=-\dfrac{2}{x^{3}}
- Substitute in the left side.x^{2}\left(-\dfrac{2}{x^{3}}\right)+2x\cdot\dfrac{1}{x^{2}}=-\dfrac2x+\dfrac2x=0
- True for every x\neq0, so the function is a solution.
- Separate: dy=x\ln x\,dx, then integrate by parts with u=\ln x, dv=x\,dx (so du=\dfrac{dx}{x}, v=\dfrac{x^{2}}{2}).y=\dfrac{x^{2}}{2}\ln x-\int\dfrac{x^{2}}{2}\cdot\dfrac1x\,dx
- Finish the integral.y=\dfrac{x^{2}}{2}\ln x-\dfrac{x^{2}}{4}+c
- Take the equation as (y+1)\dfrac{dy}{dx}+x\sin x=0 and separate the variables.(y+1)\,dy=-x\sin x\,dx
- Integrate the right side by parts: \int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x.\int(y+1)\,dy=-\left(-x\cos x+\sin x\right)+c
- Write the result.\dfrac{(y+1)^{2}}{2}=x\cos x-\sin x+c
- Numerator and denominator have degree 2, so the equation is homogeneous. Put y=ux.u+x\dfrac{du}{dx}=\dfrac{x^{2}+u^{2}x^{2}}{x\cdot ux}=\dfrac{1+u^{2}}{u}=\dfrac1u+u
- Subtract u and separate.x\dfrac{du}{dx}=\dfrac1u\quad\Longrightarrow\quad u\,du=\dfrac{dx}{x}
- Integrate.\dfrac{u^{2}}{2}=\ln|x|+c
- Replace u by \dfrac{y}{x} and multiply by 2x^{2}.y^{2}=2x^{2}\left(\ln|x|+c\right)
- Use y(1)=2: 4=2(0+c).c=2
- Substitute c. Since y(1)=2>0, take the positive root for the explicit form.y^{2}=2x^{2}\left(\ln x+2\right)\quad\Longrightarrow\quad y=x\sqrt{2\ln x+4}
FBISE 2026 questions
Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.
- Ay^{2}=x^{2}+CAnswer
- By^{2}=-x^{2}+C
- Cy^{2}=2x^{2}+C
- Dy=x+C
Separate the variables: y\,dy=x\,dx. Integrate: \dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}+c, so y^{2}=x^{2}+C with C=2c.
Option (B) comes from a sign slip (it solves \dfrac{dy}{dx}=-\dfrac{x}{y}), and (C) from integrating x as x^{2} without the factor \tfrac12.
- P(0) = 50\,000 in 2020, so t=0
- P(5) = 60\,000 in 2025
- (a) k
- (b) P(10), the population in 2030
Model and general solution
- "Rate proportional to the current population": with t in years after 2020,\dfrac{dP}{dt}=kP
- Separate the variables and integrate.\dfrac{dP}{P}=k\,dt\quad\Longrightarrow\quad \ln P=kt+\ln c\quad\Longrightarrow\quad P=c\,e^{kt}
- At t=0, P=50\,000, so c=50\,000.P=50\,000\,e^{kt}
(a) Growth constant
- Use P(5)=60\,000.60\,000=50\,000\,e^{5k}\quad\Longrightarrow\quad e^{5k}=\dfrac{6}{5}=1.2
- Take natural logarithms.k=\dfrac{1}{5}\ln1.2=\dfrac{0.18232}{5}=0.0365\ \text{per year}
(b) Population in 2030
- In 2030, t=10. Use e^{10k}=\left(e^{5k}\right)^{2}=1.2^{2}.P(10)=50\,000\times(1.2)^{2}=50\,000\times1.44=72\,000
Note. Same method as Exercise 4.4 Q1 (Malthus model) with different data.
- Separate the variables: divide by (y+1) and multiply by dx.\dfrac{dy}{y+1}=\dfrac{dx}{x-2}
- Integrate both sides; write the constant as \ln|c|.\ln|y+1|=\ln|x-2|+\ln|c|
- Combine the logarithms and take antilog.\ln|y+1|=\ln\left|c(x-2)\right|\quad\Longrightarrow\quad y+1=c(x-2)
- Check. From y=c(x-2)-1: \dfrac{dy}{dx}=c, and \dfrac{y+1}{x-2}=\dfrac{c(x-2)}{x-2}=c. Both sides agree.
Dividing by y+1 assumed y\neq-1; the constant solution y=-1 is included by c=0.
Step 1: Show that the equation is homogeneous
Let f(x,y)=\dfrac{x^{2}+y^{2}}{xy}. Then f(tx,ty)=\dfrac{t^{2}\left(x^{2}+y^{2}\right)}{t^{2}xy}=t^{0}f(x,y). Numerator and denominator are both of degree 2, so f is homogeneous of degree 0 and the DE is homogeneous. It can be solved with y=ux.
Step 2: Substitute y = ux
- Put y=ux, where u is a function of x. By the product rule:\dfrac{dy}{dx}=u+x\dfrac{du}{dx}
- Substitute in the DE and cancel x^{2}.u+x\dfrac{du}{dx}=\dfrac{x^{2}+u^{2}x^{2}}{x\cdot ux}=\dfrac{1+u^{2}}{u}=\dfrac{1}{u}+u
Step 3: Separate the variables
- Subtract u from both sides. The u terms cancel.x\dfrac{du}{dx}=\dfrac{1}{u}
- Multiply by u\,dx and divide by x.u\,du=\dfrac{dx}{x}
Step 4: Integrate
- Integrate both sides, with one constant c.\int u\,du=\int\dfrac{dx}{x}\quad\Longrightarrow\quad \dfrac{u^{2}}{2}=\ln|x|+c
Step 5: Return to x and y
- Replace u by \dfrac{y}{x}.\dfrac{y^{2}}{2x^{2}}=\ln|x|+c
- Multiply by 2x^{2} and write C=2c.y^{2}=2x^{2}\ln|x|+Cx^{2}
Verification
- Differentiate y^{2}=2x^{2}\ln|x|+Cx^{2} implicitly.2y\dfrac{dy}{dx}=4x\ln|x|+2x+2Cx\quad\Longrightarrow\quad \dfrac{dy}{dx}=\dfrac{2x\ln|x|+x+Cx}{y}
- The right side of the DE gives the same value, because y^{2}=2x^{2}\ln|x|+Cx^{2}.\dfrac{x^{2}+y^{2}}{xy}=\dfrac{x^{2}+2x^{2}\ln|x|+Cx^{2}}{xy}=\dfrac{x+2x\ln|x|+Cx}{y}\ \checkmark
Note. Nearly the same as Review Exercise Q4 (iii), which adds the initial condition y(1) = 2.
- Ay^{2}=2x^{3}+cAnswer
- By=6x^{3}+c
- Cy^{2}=x^{3}+c
- Dy^{2}=3x^{3}+c
Separate: y\,dy=3x^{2}\,dx. Integrate: \dfrac{y^{2}}{2}=x^{3}+c. Multiply by 2: y^{2}=2x^{3}+C.
Option (C) forgets to multiply x^{3} by 2, and (D) integrates 3x^{2} wrongly.
- A2000
- B1000
- C1350Answer
- D1500
The intended model is P(t)=1000e^{0.03t}. At t=10: P=1000e^{0.3}=1000\times1.3499\approx1350.
Option (A) 2000 would need e^{0.3}=2, and (B) 1000 is the starting population.
Note. The paper prints the exponent as e^{0.03} without t. Read literally, P=1000e^{0.03}\approx1030 for all t, which would make (B) closest; the intended answer is clearly (C) for P=1000e^{0.03t}.
- P(0) = 50\,000 in 2020, so t=0
- P(5) = 60\,000 in 2025
- (a) k
- (b) P(10), the population in 2030
Model and general solution
- "Rate proportional to the current population": with t in years after 2020,\dfrac{dP}{dt}=kP
- Separate the variables and integrate.\dfrac{dP}{P}=k\,dt\quad\Longrightarrow\quad \ln P=kt+\ln c\quad\Longrightarrow\quad P=c\,e^{kt}
- At t=0, P=50\,000, so c=50\,000.P=50\,000\,e^{kt}
(a) Growth constant
- Use P(5)=60\,000.60\,000=50\,000\,e^{5k}\quad\Longrightarrow\quad e^{5k}=\dfrac{6}{5}=1.2
- Take natural logarithms.k=\dfrac{1}{5}\ln1.2=\dfrac{0.18232}{5}=0.0365\ \text{per year}
(b) Population in 2030
- In 2030, t=10. Use e^{10k}=\left(e^{5k}\right)^{2}=1.2^{2}.P(10)=50\,000\times(1.2)^{2}=50\,000\times1.44=72\,000
Note. The same question appeared in Set B. Same method as Exercise 4.4 Q1 with different data.
- Separate the variables: divide by (y-2) and multiply by dx.\dfrac{dy}{y-2}=\dfrac{dx}{x+6}
- Integrate both sides; write the constant as \ln|c|.\ln|y-2|=\ln|x+6|+\ln|c|
- Combine and take antilog.y-2=c\,(x+6)
- Check. From y=c(x+6)+2: \dfrac{dy}{dx}=c and \dfrac{y-2}{x+6}=\dfrac{c(x+6)}{x+6}=c. Both sides agree.
The solutions are straight lines through the point (-6,2). The constant solution y=2 (lost when dividing by y-2) is included by c=0.
Step 1: Show that the equation is homogeneous
f(x,y)=\dfrac{x^{2}+3y^{2}}{xy} gives f(tx,ty)=\dfrac{t^{2}\left(x^{2}+3y^{2}\right)}{t^{2}xy}=t^{0}f(x,y). Numerator and denominator are both of degree 2, so the DE is homogeneous.
Step 2: Substitute y = ux
- Put y=ux and \dfrac{dy}{dx}=u+x\dfrac{du}{dx}. Cancel x^{2}.u+x\dfrac{du}{dx}=\dfrac{x^{2}+3u^{2}x^{2}}{x\cdot ux}=\dfrac{1+3u^{2}}{u}
Step 3: Separate the variables
- Subtract u.x\dfrac{du}{dx}=\dfrac{1+3u^{2}-u^{2}}{u}=\dfrac{1+2u^{2}}{u}
- Separate.\dfrac{u\,du}{1+2u^{2}}=\dfrac{dx}{x}
Step 4: Integrate
- The derivative of 1+2u^{2} is 4u, so the left side is \dfrac14\ln\left(1+2u^{2}\right). Write the constant as \dfrac14\ln C.\dfrac14\ln\left(1+2u^{2}\right)=\ln|x|+\dfrac14\ln C
- Multiply by 4 and take antilog.\ln\left(1+2u^{2}\right)=\ln\left(Cx^{4}\right)\quad\Longrightarrow\quad 1+2u^{2}=Cx^{4}
Step 5: Return to x and y
- Replace u by \dfrac{y}{x} and multiply by x^{2}.1+\dfrac{2y^{2}}{x^{2}}=Cx^{4}\quad\Longrightarrow\quad x^{2}+2y^{2}=Cx^{6}
Verification
- Differentiate x^{2}+2y^{2}=Cx^{6} implicitly and use C=\dfrac{x^{2}+2y^{2}}{x^{6}}.2x+4y\dfrac{dy}{dx}=6Cx^{5}=\dfrac{6\left(x^{2}+2y^{2}\right)}{x}
- Solve for \dfrac{dy}{dx}: this is the given DE.4y\dfrac{dy}{dx}=\dfrac{6x^{2}+12y^{2}-2x^{2}}{x}=\dfrac{4\left(x^{2}+3y^{2}\right)}{x}\quad\Longrightarrow\quad\dfrac{dy}{dx}=\dfrac{x^{2}+3y^{2}}{xy}\ \checkmark
Note. Nearly the same as Exercise 4.3 Q13, (x^{2}+2y^{2})dx=xy\,dy, which has coefficient 2 instead of 3 and an initial condition.
- Ay=e^{x^{2}/2}+C
- By=Ce^{x}
- Cy=\dfrac{x^{2}}{2}+C
- Dy=Ce^{x^{2}/2}Answer
Separate: \dfrac{dy}{y}=x\,dx. Integrate: \ln y=\dfrac{x^{2}}{2}+c. Take antilog: y=e^{c}e^{x^{2}/2}=Ce^{x^{2}/2}.
Option (A) is tempting, but the constant multiplies the exponential; it is not added to it.
- Separate the variables: divide by y and multiply by dx.\dfrac{dy}{y}=\dfrac{dx}{x}
- Integrate both sides; write the constant as \ln|c|.\ln|y|=\ln|x|+\ln|c|=\ln|cx|
- Take antilog.y=cx
- Check. \dfrac{dy}{dx}=c=\dfrac{cx}{x}=\dfrac{y}{x}.
The general solution is the family of straight lines through the origin (for x\neq0). The equation is also homogeneous, but separation is quicker.
Note. Nearly the same as Exercise 4.2 Q2, x\dfrac{dy}{dx}=4y (answer y=cx^{4}).
- A1
- B2
- C3Answer
- D4
The highest derivative is \dfrac{d^{3}y}{dx^{3}}, so the order is 3.
Option (D) 4 is the power of \dfrac{dy}{dx}, which is not the highest derivative. (The degree of this DE is 1.)
- A\dfrac{dM}{dt}=-kMAnswer
- B\dfrac{dM}{dt}=kM
- C\dfrac{dM}{dt}=kt
- D\dfrac{dM}{dt}=-kt
"Rate" is \dfrac{dM}{dt}, "proportional to M" means k\times M, and "decays" means the rate is negative: \dfrac{dM}{dt}=-kM with k>0.
Option (B) describes growth, and (C), (D) make the rate depend on time instead of on the mass.
- Separate the variables: divide by y and multiply by dx.\dfrac{dy}{y}=\dfrac{dx}{x+1}
- Integrate both sides; write the constant as \ln|c|.\ln|y|=\ln|x+1|+\ln|c|
- Combine and take antilog.\ln|y|=\ln\left|c(x+1)\right|\quad\Longrightarrow\quad y=c(x+1)
- Check. \dfrac{dy}{dx}=c and \dfrac{y}{x+1}=\dfrac{c(x+1)}{x+1}=c.
This is the book's Example 9, (1+x)\,dy-y\,dx=0, written in another form. The solutions are straight lines through (-1,0).
Step 1: Show that the equation is homogeneous
f(x,y)=\dfrac{x+y}{x-y} gives f(tx,ty)=\dfrac{t(x+y)}{t(x-y)}=t^{0}f(x,y): numerator and denominator are both of degree 1, so the DE is homogeneous.
Step 2: Substitute y = ux
- Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}, and cancel x.u+x\dfrac{du}{dx}=\dfrac{x+ux}{x-ux}=\dfrac{1+u}{1-u}
Step 3: Separate the variables
- Subtract u.x\dfrac{du}{dx}=\dfrac{1+u-u(1-u)}{1-u}=\dfrac{1+u^{2}}{1-u}
- Separate.\dfrac{1-u}{1+u^{2}}\,du=\dfrac{dx}{x}
Step 4: Integrate
- Split the fraction: \dfrac{1-u}{1+u^{2}}=\dfrac{1}{1+u^{2}}-\dfrac{u}{1+u^{2}}. Use \int\dfrac{du}{1+u^{2}}=\tan^{-1}u and \int\dfrac{u\,du}{1+u^{2}}=\dfrac12\ln\left(1+u^{2}\right).\tan^{-1}u-\dfrac12\ln\left(1+u^{2}\right)=\ln|x|+c
Step 5: Return to x and y
- Replace u by \dfrac{y}{x}. Since \dfrac12\ln\dfrac{x^{2}+y^{2}}{x^{2}}=\dfrac12\ln\left(x^{2}+y^{2}\right)-\ln|x|, the \ln|x| terms cancel.\tan^{-1}\dfrac{y}{x}-\dfrac12\ln\left(x^{2}+y^{2}\right)+\ln|x|=\ln|x|+c
- Simplify.\tan^{-1}\dfrac{y}{x}=\ln\sqrt{x^{2}+y^{2}}+c
Verification
- Differentiate both sides implicitly: \dfrac{d}{dx}\tan^{-1}\dfrac{y}{x}=\dfrac{xy'-y}{x^{2}+y^{2}} and \dfrac{d}{dx}\ln\sqrt{x^{2}+y^{2}}=\dfrac{x+yy'}{x^{2}+y^{2}}.xy'-y=x+yy'\quad\Longrightarrow\quad y'(x-y)=x+y\quad\Longrightarrow\quad y'=\dfrac{x+y}{x-y}\ \checkmark
Exam focus
- Find the order and degree of a DE, including one with a derivative under a root such as \dfrac{dy}{dx}=\sqrt{1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}}. (short)A regular MCQ and short question; the 2026 Old syllabus paper asked for the order in MCQ 16.Go to answer
- Form the DE by eliminating the arbitrary constants from y=a\cos x+b\sin x or y=bx^{2}+2ax. (problem)Tests differentiation and elimination; Examples 6 and Exercise 4.1 Q2.Go to answer
- Verify that a given function is a solution of a DE, for example y=5\tan5x for y'=25+y^{2}. (problem)Quick marks by substitution; Exercise 4.1 Q3 and Review Q3.Go to answer
- Solve a separable DE such as \dfrac{dy}{dx}=\dfrac{y+1}{x-2}. (problem)Asked for 4 marks in both 2026 new-syllabus sets (Q.2 (iv)) and in both old-syllabus papers (Q.2 (iv) OR, Q.2 (xi)).Go to answer
- Solve an initial value problem such as \dfrac{dy}{dx}=y^{2}+4, y(0)=-2. (problem)Combines separation with finding the constant and choosing the correct branch; Exercise 4.2 Q9 to Q15.Go to answer
- Test whether f(x,y) is homogeneous and find its degree. (short)Exercise 4.3 Q1 to Q3, Example 13 and Review MCQ (iv).Go to answer
- Solve a homogeneous DE by y=ux, for example \dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}. (problem)The 8 mark Section C question in 2026 Set B, Set D and the Old syllabus paper.Go to answer
- Population growth: find k from two readings and predict a future value. (problem)2026 Sets B and D Q.2 (i) OR; Exercise 4.4 Q1 and Q3; Set D MCQ 9.Go to answer
- Newton's law of cooling: find the temperature as a function of time, or the time to reach a given temperature. (problem)Example 17 and Exercise 4.4 Q2 and Q5.Go to answer
- Radioactive decay: write the DE and use the half-life. (problem)Exercise 4.4 Q4; the 2026 Old syllabus MCQ 17 asked for \dfrac{dM}{dt}=-kM.Go to answer
- Define general and particular solutions and state how many arbitrary constants a general solution has. (short)Definitions for short questions and MCQs (Review MCQ (vi)).Go to answer