Notes 4 Differential Equations

7/27/2019 Notes 4 Differential Equations

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Advanced Calculus Chapter 4 Differential Equations 65

4 Differential Equations

4.1 Terminology

Let U ⊆ Rn, and let y : U → R. A differential equation in

y is an equation involving y and its (partial) derivatives. Whenn = 1 the differential equation is called an ordinary differential

equation; when n ≥ 2 the differential equation is called a partial

differential equation. The order of a differential equation is thehighest order of derivative appearing in the equation.

Example 1 The following are all examples of differential equa-tions. Equations (a)–(e) are ordinary differential equations and(f)–(h) are partial differential equations. Equations (a), (b) and (f)have order one, (c), (d), (g) and (h) have order 2, and (e) has order

three.

(a)dy

dx= 2x.

(b)dy

dx=

ex

y + 2.

(c)d2y

dx2+ 3

dy

dx+ 2y = 4 sin x.

(d)d2y

dx2

=−

ω2y.

This is the equation for simple harmonic motion. It arisesnaturally in the solution of many problems in mechanics.

(e)d3y

dx3+ x

d2y

dx2+ x2 dy

dx+ x3y = 0.

(f)∂z

∂x+

∂z

∂y= 3xy.

(g)∂ 2z

∂y∂x

= x2 + y2.

(h)∂U

∂t= α

∂ 2U

∂x2. (The heat equation.)

A solution of a differential equation in y is a relation between y

and its variables, free from the derivatives and higher derivatives of y, that is consistent with the equation.

Example 2 It is easy to check that y = x2 is a solution of the

differential equation

dy

dx = 2x. However, there are other solutions.




For example, y = x2 + 5, y = x2 − 7 and y = x2 + 33.75 areall solutions of this differential equation. In fact, y = x2 + c is asolution of this differential equation for any constant c. This lastsolution is called the general solution of the differential equation.

For most ordinary differential equations of order one, the generalsolution involves an arbitrary constant. Given the value of y for aparticular value of x we can determine the value of c. Such valuesof y and x are called boundary conditions for the differentialequation.

For an ordinary differential equation of order k, the general solutionnormally involves k arbitrary constants. In this case, k independentboundary conditions are required to find a unique solution.

The solutions of the differential equation in Example 2 gave y ex-

plicitly in terms of x. However, solutions of differential equations

often express y implicitly in terms of its variables, as the followingexample illustrates.

Example 3 The general solution of the ordinary differential equa-

tiondy

dx=

ex

y + 2is

1

2y2 + 2y = ex + c, (1)

where c is a constant. We can check this by differentiating expres-sion (1) implicitly with respect to x to get

ydy

dx+ 2

dy

dx= ex.

Thusdy

dx(y + 2) = ex,

and dividing throughout by y + 2 gives

dy

dx=

ex

y + 2,

as required.

Note that starting from the given solution we get

y2 + 4y = 2ex + d,

where d = 2c. Thus, completing the square of the left hand side of the above solution gives

(y + 2)2

−4 = 2ex + d,




and so(y + 2)2 = 2ex + f,

where f = d + 4. Taking square roots of both sides of this solution,and then subtracting 2 from both sides gives

y = 2ex + f

−2.

So in this case it is also possible to give the solution for y explicitlyin terms x. However, this is not always possible (see the nextexample), and it is debatable whether it is worth the effort to findsuch a solution in this case.

Example 4 The general solution of the ordinary differential equa-

tiondy

dx=

y(x − 1)

x(y + 1)is

y + ln y − x + ln x = c.

This can easily be verified by differentiating the solution with re-

spect to x and rearranging the result to getdy

dx. However, it is not

possible to rearrange this solution to express y explicitly in termsof x.

4.2 Ordinary differential equations of order one

In this section we consider the problem of finding the general solu-tion of an ordinary differential equation of order one. This is notalways possible, but there are many methods for finding the gen-eral solution, depending on the nature of the differential equation,and we consider some of these here. We also consider some specialmethods for certain families of differential equations.

Variables separable differential equations

Let M and N be functions of one variable, and consider the differ-ential equation

dy

dx=

M (x)

N (y).

Then

N (y)dy

dx= M (x).

Integrating both side with respect to x gives N (y)

dy

dxdx =

M (x) dx,




and so, by the chain rule, N (y) dy =

M (x) dx.

Thus, if we can evaluate the two integrals, we can find the generalsolution of the differential equation. Differential equations of this

nature are called variables separable, and the method outlinedfor solving them is called the separation of variables.

Example 5 Consider the differential equation

9xdy

dx=

x2 + 1

2y.

Separating the variables gives

2y dy =

x2 + 1

9x dx.

Thus 2y dy =

1

9

x +

1

x

dx.

Evaluating the two integrals gives

y2 =1

9

x2

2+ ln x + c

,

where c is a constant, and so

y =1

3

x2

2+ ln x + c.


dy

dx= ay,

where a is a constant.

Separating the variables gives dy

y=

a dx.


ln y = ax + c,


y = eax+c = Aeax,

where A = ec

.





(3y + 2)3 cos2 xdy

dx= 1.


(3y + 2)3 dy =

dx

cos2 x .

Thus (3y + 2)3 dy =

sec2 x dx.


13

(3y + 2)4

4= tan x + c,


(3y + 2)

4

= 12 tan x + d,where d = 12c.

Note that, with a bit more effort, we can write the solution of thisdifferential equation as y = 1

34√

12 tan x + d − 23

, but it is a matterof mathematical taste as to whether this is better way of expressingthe solution than the one given.

Exact differential equations

Let M and N be functions of two variables. The differential equa-tion

M (x, y) + N (x, y)dy

dx= 0

is exact if ∂M

∂y=

∂N

∂x.

When a differential equation of the given form is exact then thegeneral solution has the form f (x, y) = c, for some constant c andsome function f , of two variables, satisfying

∂f ∂x = M and ∂f

∂y = N.

To see this, we differentiate f (x, y) = c with respect to x. Thisgives

∂f

∂x

dx

dx+

∂f

∂y

dy

dx= 0.

Since dxdx

= 1, ∂f ∂x

= M and ∂f ∂y

= N we get

M + N dy

dx= 0,

as required.





2xy + (x2 + y2)dy

dx= 0.

Here M = 2xy and N = x2+y2, and since ∂M ∂y

= 2x and ∂N ∂x

= 2x =∂M

∂y , the differential equation is exact. Thus we look for a functionf of two variables such that ∂f ∂x

= M and ∂f ∂y

= N . That is,

∂f

∂x= 2xy; (2)

∂f

∂y= x2 + y2. (3)

Integrating both sides of equation (2) with respect to x gives

f = x2y + g(y), (4)

where g is a function of y. To find g we differentiate equation (4)with respect to y to get

∂f

∂y= x2 + g(y).

Comparing this expression for ∂f ∂y

with the one in equation (3) gives

x2 + g(y) = x2 + y2.

Thus g

(y) = y2 and so g(y) = 13y3. Hence f (x, y) = x2y + 13y3, andso the solution of the original differential equation is

x2y +1

3y3 = c,

for some constant c.


1 + lnx

+ 2x

lny

+x2

y − 2y dy

dx = 0.

Here M = 1 + ln x + 2x ln y and N = x2

y− 2y, and since ∂M

∂y= 2x

y

and ∂N ∂x

= 2xy

= ∂M ∂y

, the differential equation is exact. Thus we look

for a function f of two variables such that ∂f ∂x

= M and ∂f ∂y

= N .That is,

∂f

∂x= 1 + ln x + 2x ln y; (5)

∂f

∂y =

x2

y − 2y.

(6)




Integrating both sides of equation (5) with respect to x gives To evaluate

lnx dx, useintegration by parts withu = lnx and dv

dx= 1.f = x ln x + x2 ln y + g(y), (7)

where g is a function of y. To find g we differentiate equation (7)with respect to y to get

∂f ∂y = x

2

y + g(y).

Comparing this expression for ∂f ∂y

with the one in equation (6) gives

x2

y+ g(y) =

x2

y− 2y.

Thus g(y) = −2y and so g(y) = −y2. Hencef (x, y) = x ln x + x2 ln y − y2, and so the solution of the originaldifferential equation is

x ln x + x2

ln y − y2

= c,


Integrating factors

Let M and N be functions of two variables. In general the differ-ential equation

M (x, y) + N (x, y)dy

dx= 0

will not be exact. However, suppose we can find a function µ of two variables such that the differential equation

µ(x, y)M (x, y) + µ(x, y)N (x, y)dy

dx= 0

is exact. Then we can solve the resulting differential equation usingthe method described, and hence find the general solution of theoriginal differential equation. The function µ is called an integrat-

ing factor for the original differential equation.

Unfortunately, in general, it is difficult to find an integrating factor

for most differential equations. However, there are certain classes of differential equations for which there is a method for finding one. Inthe next subsection we consider one important class of differentialequations that can be solved using an integrating factor.

Linear differential equations

A ordinary differential equation of order one is linear if it can bewritten in the form

dy

dx + P (x)y = Q(x),




where P and Q are functions of one variable.

Given a linear differential equation of this form, let

µ(x) = e P (x) dx.

We claim that µ is an integrating factor for the linear differential

equation.

First note that To differentiate µ(x), we use thechain rule and the fact thatd

dx

P (x) dx

= P (x).µ(x) = P (x)e

P (x) dx = P (x)µ(x).

Now rearrange the linear differential equation in the form

P (x)y − Q(x) +dy

dx= 0,

and then multiply both sides by µ(x) to get

µ(x)(P (x)y − Q(x)) + µ(x)dy

dx= 0.

We now need to show that this differential equation is exact. LetM = µ(x)(P (x)y −Q(x)) and N = µ(x). Then

∂M

∂y= µ(x)P (x);

∂N

∂x

= µ(x),

= µ(x)P (x).

Thus ∂M ∂y

= ∂N ∂x

, and so the differential equation is exact. Thus µ(x)is an integrating factor for the linear differential equation.

Note that in this case we do not have to reduce the linear differentialequation to the exact form since (using the Product Rule),

d

dx(µ(x)y) = µ(x)y + µ(x)

dy

dx,

= (µ(x)P (x))y + µ(x)dy

dx ,

= µ(x)

P (x)y +

dy

dx

,

= µ(x)Q(x).

Thusd

dx(µ(x)y) = µ(x)Q(x), and so integrating both sides with

respect to x gives

µ(x)y = µ(x)Q(x) dx.




Thus

y =1

µ(x)

µ(x)Q(x) dx. (8)

So to solve the linear differential equation

dy

dx+ P (x)y = Q(x),

first find the integrating factor µ(x) = e P (x) dx, and then use iden-

tity (8) to find y.

Example 10 Consider the linear differential equation

dy

dx+ ay = Aebx,

where A, a and b are constants.

Here P (x) = a and Q(x) = Aebx. Let

µ(x) = e adx = eax.

Then

y =1

µ(x)

µ(x)Q(x) dx,

= e−ax (eax)(Aebx) dx,

= e−ax

Ae(a+b)x dx,

= e−ax

Ae(a+b)x

a + b+ c

,

=A

a + bebx + ce−ax.

So the general solution of the differential equation is

y =A

a + bebx

+ ce−ax

,

where c is a constant.

Example 11 Consider the linear differential equation

(x2 + 1)dy

dx+ xy = x(x2 + 1).

Thendy

dx+

x

(x2 + 1)y = x.




Thus the differential equation is linear with P (x) = x(x2+1)

and

Q(x) = x. Let

µ(x) = e

x

(x2+1)dx

,

= e12

2x

(x2+1)dx

,

= e

1

2

ln(x2+1)

,

=

eln(x2+1)

12

,

= (x2 + 1)12 .

Then

y =1

µ(x)

µ(x)Q(x) dx,

= (x2 + 1)−12 ((x2 + 1)

12 )x dx,

= (x2 + 1)−12

1

3(x2 + 1)

32 + c)

,

=1

3(x2 + 1) + c(x2 + 1)−

12 ,

=1

3(x2 + 1) +

c√ x2 + 1

.

So the general solution of the differential equation is

y =1

3

(x2 + 1) +c

√ x2 + 1

,

where c is a constant.

Homogeneous differential equations

Recall that a function f of two variables is homogeneous of degreen if

f (λx, λy) = λnf (x, y).

Let M and N be functions of two variables. Consider the differentialequation

M (x, y) + N (x, y)dy

dx= 0,

where M and N are both homogeneous of degree n. (Such a dif-ferential equation is called homogeneous.) Then we put y = vx,where v is a function of x.

By the product ruledy

dx= x

dv

dx+ v.




Also, since M is homogeneous of degree n,

M (x, y) = M (x,vx),

= M (x.1,x.v),

= xnM (1, v),

= xnM (v),

for some function M of one variable.

Similarly N (x, y) = xnN (v), for some function N of one variable.

So, with y = vx the original differential equation becomes

xnM (v) + xnN (v)

x

dv

dx+ v

= 0.

Since this is valid for all x, we can cancel the common factor of xn

to get

M (v) + N (v)

x

dv

dx+ v

= 0.

This final differential equation is variables separable. Thus, usingthe method given earlier, we can find v, and consequently y.

Note When solving a homogeneous differential equation, it is notnecessary to identify the functions M and N . You just have toremember to eliminate y by putting y = vx. If the resulting differ-ential equation is not variables separable then you have either made

a mistake, or the original differential equation is not homogeneous.


x3 + y3 + 2xy2 dy

dx= 0.

It is easy to verify that M = x3 + y3 and N = 2xy2 are both ho-mogeneous of degree 3. Putting y = vx in the differential equationgives

x3 + v3x3 + 2xv2x2

xdv

dx + v

= 0.

Cancelling the common factor of x3, and rearranging the resultingdifferential equation gives

2xv2 dv

dx= −(1 + 3v3).


2v2

1 + 3v3

dv =

− dx

x

.




Evaluating the two integrals we get

2

9ln(1 + 3v3) = − ln x + c,

where c is a constant. Since y = vx then v = yx

, and so substitutingfor v in the above solution, we get the general solution of the original

differential equation as follows:

2

9ln

1 +

3y3

x3

+ ln x = c.


y + x secy

x

− x

dy

dx= 0.

It is easy to verify that M = y + x sec yx and N =

−x are both ho-

mogeneous of degree 1. Putting y = vx in the differential equationgives

vx + x secvx

x

− x

x

dv

dx+ v

= 0.

Cancelling the common factor of x, and rearranging the resultingdifferential equation gives

xdv

dx= sec v.

Separating the variables gives dv

sec v=

dx

x,

and, since sec v = 1cos v

, we get cos v dv =

dx

x.

Evaluating the two integrals, we get

sin v = ln x + c,

where c is a constant. Since y = vx then v = yx

, and so substitutingfor v in the above solution, we get the general solution of the originaldifferential equation as follows:

siny

x

= ln x + c.




Change of variable

When solving a homogeneous differential equation we changed thevariables with the substitution y = vx. Often other first orderdifferential equations can be solved by changing the variables.

For some differential equations it is clear what the change of vari-ables should be, as illustrated in the following example.


dy

dx= (x + 4y + 3)2.

Here we put z = x + 4y + 3. Then

dz

dx= 1 + 4

dy

dx,

= 1 + 4(x + 4y + 3)2

,= 1 + 4z 2,

and so we get the variables separable equation

dz

dx= 1 + 4z 2.


dz

1 + 4z 2=

dx,

and evaluating the two integrals we get

1

2arctan(2z ) = x + c,


Substituting for z gives

1

2arctan(2(x + 4y + 3)) = x + c,

which is the general solution of the original equation.The general solution can be tidied up (with a little effort) to get

y =1

8(tan(2x + d) − 2x − 6),

where d = 2c.

In general, it is not always obvious what the change of variablesshould be. However, there are some classes of differential equationsthat can be solved with a standard change of variable. We give two

such special cases in the next two subsections.




Special case 1

Let a,b,c,d,e,f ∈ R with at least one of d, e or f nonzero. Considerthe differential equation

dy

dx

=ax + by + c

dx + ey + f

.

The differential equations in this class can be solved using a changeof variables, depending on whether the lines ax + by + c = 0 anddx + ey + f = 0 are parallel.

Suppose first that ax + by + c = 0 and dx + ey + f = 0 are notparallel. Then let (α, β ) be the point where these two lines intersect,and put

x = X + α,

y = Y + β.

We claim that dY dX

= dydx

. To see this, suppose that y = g(x) for

some function g. Then dydx

= g(x). Also Y + β = g(X + α) and so,by the chain rule,

dY

dX = g(X + α) = g(x) =

dy

dx.

Thus dY dX

= dydx

, as claimed.

Since (α, β ) is on the line ax+by+c = 0, it follows that aα+Bβ +c =0. Thus

ax + by + c = a(X + α) + b(Y + β ) + c,

= aX + bY + aα + Bβ + c,

= aX + bY.

Similarlydx + ey + f = dX + eY.

Using the above, after the change of variables, the differential equa-tion becomes

dY

dX

=aX + bY

dX + eY

.

This differential equation is homogeneous, and so can be solved us-ing the method given earlier. Thus we can find the general solutionof the original equation.

Suppose now that ax + by + c = 0 and dx + ey + f = 0 are parallel.Then there are λ, g ∈ R such that

dx + ey + f = λ(ax + by + g).

Thusdy

dx=

ax + by + c

λ(ax + by + g).




Put z = ax + by. Then dzdx

= a + bdydx

, and so If |λ| < 1 it is slightly easier toput z = dx + ey.

dy

dx=

dz

dx− a

b.

Thus, changing the variables in the differential equation we getdz

dx− a

b =

z + c

λ(z + g),

which can be rearranged to get

dz

dx=

b(z + c)

λ(z + g)+ a.

This differential equation is variables separable, and so can besolved using the method given earlier. Thus we can find the general

solution of the original equation.


dy

dx=

x + y − 6

x − y − 2.

The lines x + y − 6 = 0 and x − y − 2 = 0 are not parallel and To find the point where the linesintersect solve the equations

x + y = 6,

x− y = 2.

intersect at the point (4, 2). Putting x = X +4 and y = Y + 2 intothe differential equation gives

dY

dX =

X + 4 + Y + 2

−6

X + 4 − Y − 2− 2 =

X + Y

X − Y .

This differential equation is homogeneous, so putting Y = XV weget dY

dX = X dV

dX + V , and the differential equation becomes

X dV

dX + V =

X + XV

X −XV =

1 + V

1− V .

Thus

X dV

dX =

1 + V

1

−V − V =

1 + V − V (1 − V )

1

−V

=1 + V 2

1

−V

.

The last differential equation is variables separable and so 1 − V

1 + V 2dV =

dX

X .

Hence 1

1 + V 2− V

1 + V 2

dV =

dX

X .


arctan V −1

2 ln(1 + V 2

) = ln X + c,




where c is a constant. Multiplying throughout by 2, and rearrangingthe solution gives

2 arctan V − (ln(1 + V 2) + 2ln X ) = d,

where d = 2c. Now 2ln X = ln X 2, and so

ln(1 + V 2) + 2 ln X = ln(1 + V 2) + ln X 2 = ln(X 2 + X 2V 2).

Thus2 arctan V − ln(X 2 + X 2V 2) = d.

Since V = Y X

the solution becomes

2 arctan

Y

X

− ln(X 2 + Y 2) = d.

Finally, putting X = x

−4 and Y = y

−2 we get

2 arctan

y − 2

x − 4

− ln((x − 4)2 + (y − 2)2) = d,

which is the general solution of the original differential equation.


dy

dx=

x + 2y + 5

2x + 4y − 3.

The lines x + 2y + 5 = 0 and 2x + 4y− 3 = 0 are parallel so we putz = x + 2y. Then

dz

dx= 1 + 2

dy

dx,

and sody

dx=

1

2

dz

dx− 1

.

Thus, changing the variable, the differential equation becomes

1

2dz

dx −1 =

z + 5

2z − 3

.

Hencedz

dx=

2z + 10

2z − 3+ 1 =

4z + 7

2z − 3.

This differential equation is variables separable, and so separatingthe variables gives

2z − 3

4z + 7dz =

dx. (9)




Now 2z − 3

4z + 7dz =

1

2

4z − 6

4z + 7dz,

=1

2

4z + 7 − 13

4z + 7dz,

= 12

1 − 13

4z + 7dz,

=1

2

z − 13

4ln(4z + 7)

.

Thus evaluating the two integrals in equality (9) gives

1

2

z − 13

4ln(4z + 7)

= x + c,

where c is a constant. Multiplying both sides of the last equality

by 8 we get4z − 13ln(4z + 7) = 8x + d,

where d = 8c. Substituting z = x + 2y in the last expression andsimplifying the result gives

8y − 4x − 13ln(4x + 8y + 7) = d,

which is the general solution of the original differential equation.

Special case 2

Let P and Q be functions of one variable, and consider the differ-ential equation

dy

dx+ P (x)y = Q(x)yn.

To solve this differential equation first divide both sides by yn toget

1

yndy

dx+

P (x)

yn−1= Q(x).

Now put u =

1

yn−1 = y

1−n

. For this change of variable

du

dx= (1 − n)y−n

dy

dx=

1 − n

yndy

dx,

and sody

dx=

yn

1− n

du

dx.

After the change of variable, the differential equation becomes

1

yn yn

1 − n

du

dx+ uP (x) = Q(x),




and so1

1 − n

du

dx+ uP (x) = Q(x).

Multiplying both sides by 1 − n yields the differential equation

du

dx+ (1 − n)uP (x) = (1 − n)Q(x),

which is linear, so it can be solved by finding an integrating factorand using the method given earlier.


dy

dx+

y

x= x3y6.

Dividing both sides by y6 gives

1

y6

dy

dx+

1

xy5= x3.

Putting u =1

y6 , we getdu

dx = −6

y6dy

dx , and sody

dx = −y6

5

du

dx . Changingthe variable in the differential equation gives

1

y6

−y6

5

du

dx

+

u

x= x3.

Simplifying this differential equation, and then multiplying bothsides by −5 gives

du

dx− 5

xu = −5x3.

The preceding differential equation is linear so we first find theintegrating factor

µ(x) = e − 5x dx,

= e−5 ln x,

= elnx−5

,

= x−5 =1

x5.

Then

u =1

µ(x)

µ(x)(−5x3) dx,

= x5 −5

x2

dx,

= x5

5

x+ c

,

where c is a constant. Thus

u = x4(5 + cx).

Since u = 1y5

, the general solution of the original differential equa-tion is

1

y5= x4(5 + cx),

or x

4

y

5

(5 + cx) = 1.




4.3 Ordinary differential equations of order two

To find the general solution of a differential equation of order twois more difficult than for order one. In this section we only considerlinear differential equations. That is, differential equations of theform

P (x) d2

ydx2

+ Q(x) dydx

+ R(x)y = S (x),

where P , Q, R and S are functions of one variable.

The homogeneous case with constant coefficients

We start with the simplest case where P , Q and R are constantfunctions and S is identically zero. That is, the differential equationhas the form

P d

2y

dx2 + Q dy

dx + Ry = 0, (10)

where P,Q,R ∈ R. Such a differential equation is called homoge-

neous with constant coefficients.

In looking for the general solution of equation (10) we note firstthat if y is a solution of the differential equation then so is Ay, fora constant A. Also, since the differential equation is of order two,we expect the general solution to involve two arbitrary constants.Finally, in Example 6 we found that the general solution of a homo-geneous linear differential equation of order one was an exponential

function. Inspired by this, we look for solutions of equation (10)having the form y = etx, for some constant t. For this y, dy

dx= tetx

and d2ydx2

= t2etx. Substituting for y, dydx

and d2ydx2

in equation (10)gives

P t2etx + Qtetx + Retx = 0.

Since etx = 0, we can cancel this factor from the previous equationto get

P t2 + Qt + R = 0.

This is a quadratic equation in t, and is called the auxiliary equa-

tion of the differential equation given in (10).Let α and β be the roots of the auxiliary equation. Then the generalsolution of (10) depends on the nature of α and β . There are threecases to consider.

Case 1: α, β ∈ R with α = β . This is the easiest case; since weknow that eαx and eβx are distinct solutions of equation (10), thegeneral solution is

y = Aeαx + Beβx,

where A and B are (real) constants.




Case 2: α, β ∈ R with α = β . In this case, since α = β , eαx = eβx

and so we only have one solution of equation (10). However, itis straightforward to check that xeαx is also a solution of equa-tion (10), and so the general solution is

y = (A + Bx)eαx,

where A and B are (real) constants.

Case 3: α, β ∈ R. We first recall that it is not possible to haveone of α or β in R and the other not in R, since the non-real rootsof a polynomial equation with coefficients in R occur in conjugatepairs. Thus, in this case, α = a + bi and β = a− ib where a, b ∈ R, Recall that i =

√ −1.

with b = 0. Doing the same as in case 1 we could write the generalsolution of equation (10) in the form

y = Ae(a+bi)x + Be(a−ib)x,

where A and B are complex constants. However, it is unsatisfac-tory to have the solution of a differential equation involving a realfunction expressed in terms of complex functions, so we aim to finda better way of expressing the solution. Now Here we use the standard rules

for exponentials together withthe result that

eiθ = cos θ + i sin θ,

for all θ ∈ R.

e(a+bi)x = eax+bxi = eaxebxi = eax(cos(bx) + i sin(bx)).

Similarly,

e(a−bi)x = eax−bxi = eaxe−bxi = eax(cos(bx) − i sin(bx)).

Using these equivalent expressions for e(a+bi)x and e(a−bi)x we get

ˆAe

(a+bi)x

+ˆ

Be(a−ib)x

=ˆ

Aeax

(cos(bx) + i sin(bx)) +Beax(cos(bx) − i sin(bx)),

= eax((A + B) cos(bx) + i(A− B) sin(bx)),

= eax(A cos(bx) + B sin(bx)),

where A = A + B and B = i(A − B). Note, since A,B ∈ R, it followsthat B is the complex conjugateof A.In summary, we have the following method for finding the general

solution of equation (10). First find the solutions α and β of theauxiliary equation

P t2 + Qt + R = 0.

Then the following table gives the general solution of equation (10),where A and B are constants.

Nature of the roots of General solutionthe auxiliary equation of equation (10)

α, β ∈ R, α = β y = Aeαx + Beβx

α, β ∈ R, α = β y = (A + Bx)eαx

α = a + ib, β = a − ib, y = eax(A cos(bx) + B sin(bx))a, b

∈R, b

= 0





d2y

dx2− 5

dy

dx+ 6y = 0.

The auxiliary equation of this differential equation is

t2

− 5t + 6 = 0.

Thus (t− 2)(t− 3) = 0, and so t = 2, 3. Hence the general solutionof the differential equation is

y = Ae2x + Be3x,

where A and B are constants.


4 d2

ydx2 − 4 dy

dx+ y = 0.


4t2 − 4t + 1 = 0.

Thus (2t−1)2 = 0, and so t = 12

(twice). Hence the general solutionof the differential equation is

y = (Ax + B)e12x,



d2y

dx2− 10

dy

dx+ 29y = 0.


t2 − 10t + 29 = 0.

Thus

t =10 ±√ 100 − 4 × 29

2,

=10 ±√ −16

2,

= 5 ± 2i.

Hence the general solution of the differential equation is

y = e5x(A cos(2x) + B sin(2x)),





The nonhomogeneous case with constant coefficients

We now consider the case where P , Q and R are constant functionsand S is a function of x. That is, the differential equation has theform

P

d2y

dx2 + Q

dy

dx + Ry = S (x), (11)where P,Q,R ∈ R. Such a differential equation is called nonho-

mogeneous with constant coefficients.

To find the general solution of equation (11) we have to do twothings. First we find the general solution C (x)of the homogeneous

part

P d2y

dx2+ Q

dy

dx+ Ry = 0.

This is called the complementary function of equation (11).Next we find any solution p(x) of equation (11). This is called aparticular integral of the differential equation. Then the generalsolution of equation (11) is

y = C (x) + p(x).

Finally, if there are any boundary conditions we use them to deter-mine the arbitrary constants in the general solution.

Finding a particular integral

In the above method, we know how to find the complementaryfunction. However, finding a particular integral is very hard formost functions S (x), but there are some families of functions forwhich it is possible to find a particular integral. Here we give threesuch families of functions.

Family 1: S (x) = M eγx, where M, γ ∈ R. In this case we lookfor a particular integral of the form

p(x) = Gx jeγx,

where j is the number of times γ is a root of the auxiliary equation,and G is a constant.

Family 2: S (x) = M cos(cx) + N sin(cx), where M , N , c ∈ R. Inthis case we look for a particular integral of the form

p(x) = G cos(cx) + H sin(cx),

where G and H are constants, unless the roots of the auxiliaryequation are ±ci, in which case we look for a particular integral of the form

p(x) = x(G cos(cx) + H sin(cx)).




Family 3: S (x) = M kxk + M k−1xk−1 + · · · + M 1x + M 0, whereM k, M k−1, . . . , M 1, M 0 ∈ R. In this case we look for a particularintegral of the form

p(x) = x j(mkxk + mk−1xk−1 + · · · + m1x + m0),

where j is the number of times 0 is a root of the auxiliary equation,and mk, mk−1, . . . , m1, m0 are constants.

Observe that, if 0 is a root of the auxiliary equation then the dif-ferential equation

P d2y

dx2+ Q

dy

dx+ Ry = S (x)

must have R = 0. That is the differential equation is

P d2y

dx2+ Q

dy

dx= S (x).

In this case, it is easier to change the variable by putting u = dydx

(so dudx

= d2ydx2

) to get

P du

dx+ Qu = S (x),

and then solve the resulting first order differential equation. If we dothis, then whenever S (x) is a polynomial of degree k the particularintegral is also a polynomial of degree k.

In general, we note that for all three families of functions considered,

the particular integral has the same form as S (x), except when theroots of the auxiliary equation take certain values. In the lattercase we multiple the form of the particular integral by a suitablepower of x; this is required because, in this case, S (x) is part of thecomplementary function and so cannot be a particular integral of the nonhomogeneous differential equation.

Finally, we observe that if S (x) is a sum of functions from the threefamilies considered then the particular integral will be a sum of thecorresponding particular integrals.


d2y

dx2− 5

dy

dx+ 6y = 20e−3x,

with boundary conditions y = 5, dydx

= 3 when x = 0.

From Example 18, the auxiliary equation of the homogeneous partof this differential equation is t2 − 5t + 6 = 0, with roots t = 2, 3.Hence the complementary function of the differential equation is

C = Ae2x

+ Be3x

,





Here S = 20e−3x, so we look for a particular integral of the form p = Ge−3x. Then d p

dx= −3Ge−3x and d2 p

dx2= 9Ge−3x. Now, since p is

a particular integral of the differential equation it must satisfy thedifferential equation. So putting y = p in the differential equationgives

9Ge−3x − 5(−3Ge−3x) + 6Ge−3x = 20e−3x.

Cancelling the common factor of e−3x from this expression, andsimplifying the result gives 30G = 20, and so G = 2

3. Thus the

particular integral is

p =2

3e−3x.

From the above, the general solution of the differential equation is

y = C + p = Ae2x + Be3x +2

3

e−3x.

We now use the boundary conditions to determine A and B. Dif-ferentiating the general solution with respect to x gives

dy

dx= 2Ae2x + 3Be3x − 2e−3x.

Now, when x = 0, y = 5 and dydx

= 3. Thus, putting x = 0 in thegeneral solution and its derivative, and then comparing them withthe corresponding boundary condition gives

A + B +2

3= 5,

2A + 3B − 2 = 3.

Simplifying these equations we get

3A + 3B = 13, (12)

2A + 3B = 5. (13)

Subtracting equation (13) from equation (12) gives A = 8; sub-

stituting A = 8 in equation (13) gives 16 + 3B = 5, which givesB = −11

3. Thus the solution of the differential equation with the

given boundary conditions is

y = 16e2x − 11

3e3x +

2

3e−3x.


d2y

dx2− 5

dy

dx+ 6y = 3e2x.




As in the previous example, the roots of the auxiliary equation aret = 2, 3 and the complementary function of the differential equationis

C = Ae2x + Be3x,


Here S = 3e2x. Since 2 is a single root of the auxiliary equation,we look for a particular integral of the form p = Gxe2x. Then

d p

dx= Ge2x + 2Gxe2x,

= G(1 + 2x)e2x;

d2 p

dx2= 2Ge2x + 2G(1 + 2x)e2x,

= 4G(1 + x)e2x.

So putting y = p in the differential equation gives

4G(1 + x)e2x − 5(G(1 + 2x)e2x) + 6Gxe2x = 3e2x.

Cancelling the common factor of e2x from this expression gives

G(4 + 4x − 5− 10x + 6x) = 3.

Thus −G = 3, and so G = −3. Thus the particular integral is Note that all of the x termscancel. If this does not happenthen we have done somethingwrong.

p = −3xe2x.


y = C + p = Ae2x + Be3x − 3xe2x.


d2y

dx2− 10

dy

dx+ 29y = 29x2 + 38x − 18.

From Example 20, the auxiliary equation of the homogeneous part

of this differential equation is t2−10t+29 = 0, with roots t = 5±2i.Hence the complementary function of the differential equation is

C = e5x(A cos(2x) + B sin(2x)),


Here S = 29x2 + 38x − 18, so we look for a particular integral of the form p = ax2 + bx + c. Then d p

dx= 2ax + b and d2 p

dx2= 2a. So

putting y = p in the differential equation gives

2a − 10(2ax + b) + 29(ax

2

+ bx + c) = 29x

2

+ 38x − 18.




Comparing the coefficients of x2 and x, and the constant term,respectively, for both sides of the previous expression gives

29a = 29,

−20a + 29b = 38,

2a

−10b + 29c =

−18.

Solving the system of linear equations gives a = 1, b = 2 and c = 0.Thus the particular integral is

p = x2 + 2x.


y = C + p = e5x(A cos(2x) + B sin(2x)) + x2 + 2x.


d2y

dx2+ y = 4 sin x + 3 cos x.

The auxiliary equation of the homogeneous part of this differentialequation is t2 + 1 = 0, and so t = ±i. Hence the complementaryfunction of the differential equation is

C = A cos x + B sin x,


Here S = 4sin x + 3 cos x. Since±

i are roots of the auxiliaryequation, we look for a particular integral of the form

p = x(G sin x + H cos x). Then

d p

dx= G sin x + H cos x + x(G cos x −H sin x);

d2 p

dx2= G cos x − H sin x + G cos x − H sin x + x(−G sin x − H cos x),

= 2(G cos x − H sin x) − x(G sin x + H cos x).

Putting y = p in the differential equation gives

2(G cos x−H sin x)−x(G sin x+H cos x)+x(G sin x+H cos x) = 4 sin x+3 cos x.

Thus 2(G cos x − H sin x) = 4sin x + 3cos x. Thus G = 32

andH = −2, and so the particular integral is

p = x

3

2sin x − 2cos x

.


y = C + p = A cos x + B sin x + x3

2

sin x

−2cos x .





d2y

dx2+

dy

dx+ y = 2ex + 3x + 4.

The auxiliary equation of the homogeneous part of this differentialequation is t2 + t + 1 = 0, and so

t =−1±√ 1 − 4

2,

= −1

2±√

3

2i.

Hence the complementary function of the differential equation is

C = e−12x

A cos

√ 3

2x

+ B sin

√ 3

2x

,


Here S = 2ex + 3x + 4. This is the sum of functions from two of the families considered. Hence we look for a particular integral of the form p = Gex + Hx + K . Then

d p

dx= Gex + H ;

d2 p

dx2= Gex.

Putting y = p in the differential equation gives

Gex + Gex + H + Gex + Hx + K = 2ex + 3x + 4.

Thus 3Gex + Hx + H + K = 2ex + 3x +4. Thus G = 23

, H = 3 andH + K = 4, so K = 1. Thus the particular integral is

p =2

3ex + 3x + 1.


y = C + p = e−12x

A cos

√ 3

2x

+ B sin

√ 3

2x

+

2

3ex+3x+1.

Order two linear differential equations

We now consider general order two linear differential equations.That is, differential equations of the form

P (x)

d2y

dx2 + Q(x)

dy

dx + R(x)y = S (x), (14)




where P , Q, R and S are functions of one variable. For such adifferential equation, the homogeneous part is the differentialequation

P (x)d2y

dx2+ Q(x)

dy

dx+ R(x)y = 0.

As in the case with constant coefficients, finding the general solution

of equation (14) depends upon being able to find a solution of thehomogeneous part. To see this, suppose y = g is a solution of thehomogeneous part, for some function g of x. Then

P (x)d2g

dx2+ Q(x)

dg

dx+ R(x)g = 0.

Now we change the variable in equation (14) by putting y = gv forsome function v of x. Then, using the product rule, we get

dy

dx

= gdv

dx

+ vdg

dx

;

d2y

dx2= g

d2v

dx2+ 2

dg

dx

dv

dx+ v

d2g

dx2.

Thus, with the change the variable, equation (14) becomes

P (x)

g

d2v

dx2+ 2

dg

dx

dv

dx+ v

d2g

dx2

+Q(x)

g

dv

dx+ v

dg

dx

+R(x)gv = S (x),

which can be rewritten as

v P (x) d2g

dx2

+ Q(x)dg

dx

+ R(x)g+

P (x)

g d2vdx2

+ 2 dgdx

dvdx

+ Q(x)g dv

dx= S (x).

(15)

Now, since g is a solution of the homogeneous part of equation (14),the first term of equation (15) is zero. Thus equation (15) becomes

P (x)

g

d2v

dx2+ 2

dg

dx

dv

dx

+ Q(x)g

dv

dx= S (x),

which can be rewritten as

P (x)g d2

vdx2 +

2P (x) dg

dx+ Q(x)g

dvdx

= S (x). (16)

Now we put w = dvdx

in equation (16) to get

P (x)gdw

dx+

2P (x)

dg

dx+ Q(x)g

w = S (x),

which is a first order differential equation for w. Solving this dif-ferential equation gives w. We can then find the general solutionof equation (14) by integrating w with respect to x to get v and

finally getting y = vg.




Note You do not have to remember the first order differentialequation for w in terms of P,Q,S and g to solve equation (14); justremember the change of variable y = gv, and derive the first orderdifferential equation from scratch.


x2(x + 1)d2y

dx2+ x(3x − 2)

dy

dx− (3x − 2)y = 10x3.

It is easy to verify that y = x is a solution of the homogeneous partof this equation. So putting y = vx we get

dy

dx= x

dv

dx+ v;

d2y

dx2= x

d2v

dx2+ 2

dv

dx.

Thus the differential equation becomes

x2(x+1)

x

d2v

dx2+ 2

dv

dx

+x(3x−2)

x

dv

dx+ v

−(3x−2)vx = 10x3.

Collecting terms in the previous equation involving d2vdx2

and dvdx

weget

x3(x + 1)d2v

dx2+

2x2(x + 1) + x2(3x − 2)

dv

dx= 10x3,

and simplifying gives

x3(x + 1)d2v

dx2+ 5x3 dv

dx= 10x3.

Dividing the resulting equation throughout by x3(x + 1), and thenputting w = dv

dxwe get

dw

dx+

5w

x + 1=

10

x + 1,

which is a first order linear differential equation. To solve thisdifferential equation we first find the integrating factor

µ(x) = e

5x+1

dx,

= e5ln(x+1),

= eln(x+1)5,

= (x + 1)5.




Then

w =1

µ(x)

µ(x)

10

x + 1dx,

=1

(x + 1)5

(x + 1)5

10

x + 1dx,

= 1(x + 1)5

10(x + 1)4 dx,

=1

(x + 1)5

2(x + 1)5 + C

,

= 2 +C

(x + 1)5,

where C is a constant.

Now, dvdx

= w, and so

v =

w dx,

=

2 +

C

(x + 1)5

dx,

= 2x +D

(x + 1)4+ E,

where D = −C 4

and E is a constant.

Finally, y = vx, and so the general solution of the original differen-tial equation is

y = 2x2 +Dx

(x + 1)4 + Ex.

Euler equations

In general, it is difficult to find a solution of the homogeneous partof an order two linear differential equation. However, there are somefamilies of equations where this is possible. Here we consider onesuch family, the Euler equations. These are differential equationsof the form

ax2d2y

dx2 + bxdy

dx + cy = S (x),

where a, b and c are constants, and S is a function of one variable.

There are two possible ways to look for the general solution of anEuler equation.

• Look for a solution of the homogeneous part of the equationof the form y = xk. If a suitable value of k is found then applythe method described above to find the general solution of thenonhomogeneous equation. (This method does not always

work as the values of k found could be non-real.)




• Change the variable by putting x = et. This reduces thedifferential equation to one with constant coefficients. (Thismethod always works, but is probably more difficult.)

We now give some examples to illustrate the two methods.


x2 d2y

dx2− 7x

dy

dx+ 16y = 16x4.

We look for a solution of

x2 d2y

dx2− 7x

dy

dx+ 16y = 0

of the form y = xk. Then dydx

= kxk−1 and d2ydx2

= k(k

−1)xk−2.

Substituting for y, dydx and d2y

dx2 in the homogeneous part gives

x2k(k − 1)xk−2 − 7xkxk−1 + 16xk = 0,

and sok(k − 1)xk − 7kxk + 16xk = 0.

The last equation is valid for all x and so we can cancel the xk term.Thus

k(k − 1) − 7k + 16 = 0 ⇔ k2 − 8k + 16 = 0,

⇔ (k − 4)2 = 0,

⇔ k = 4.

So y = x4 is a solution of the homogeneous part of the differentialequation. We now change the variable in the differential equationby putting y = vx4. Then

dy

dx= x4 dv

dx+ 4x3v,

= x3xdv

dx

+ 4v ;

d2y

dx2= x3

x

d2v

dx2+

dv

dx+ 4

dv

dx

+ 3x2

x

dv

dx+ 4v

,

= x2

x2 d2v

dx2+ 8x

dv

dx+ 12v

.

Substituting the expressions found for y, dvdx

and d2vdx2

in the differ-ential equation we get

x4x2 d2v

dx2

+ 8xdv

dx

+ 12v− 7x4xdv

dx

+ 4v+ 16vx4 = 16x4.




Cancelling the common factor of x4 in the previous equation, andsimplifying the result gives

x2 d2v

dx2+ x

dv

dx= 16.

Putting w = dv

dx

in this equation, and dividing throughout by x2 weget

dw

dx+

w

x=

16

x2,

which is a first order linear differential equation. To solve thisdifferential equation we first find the integrating factor

µ(x) = e

1xdx,

= elnx,

= x.

Then

w =1

µ(x)

µ(x)

16

x2dx,

=1

x

x

16

x2dx,

=1

x

16

xdx,

=1

x(16ln x + C ) ,

=16ln x

x +C

x

where C is a constant.

Now, dvdx

= w, and so

v =

w dx,

=

16ln x

x+

C

x

dx,

= 8(ln x)2 + C ln x + D,

where D is a constant.

To evaluate

16lnx

xdx, put

u = lnx.

Finally, y = vx4, and so the general solution of the original differ-ential equation is

y = (8(ln x)2 + C ln x + D)x4.

Example 28 Consider again the differential equation

x2 d2y

dx2 − 7x

dy

dx + 16y = 16x4

.




We now consider the second way of solving this equation using thechange of variable x = et. The

dy

dt=

dy

dx

dx

dt,

=dy

dx

et,

= xdy

dx;

d2y

dt2=

d

dt

dy

dt

,

=d

dt

x

dy

dx

,

=dx

dt

d

dx

x

dy

dx

,

= etdy

dx + x

d2y

dx2

,

= x

dy

dx+ x

d2y

dx2

,

= xdy

dx+ x2 d2y

dx2.

Using the final expressions for dydt

and d2ydt2

we get

xdy

dx=

dy

dt;

x2 d2y

dx2=

d2y

dt2− x

dy

dx,

=d2y

dt2− dy

dt.

Replacing there terms xdydx

and x2 d2ydx2

of the original equation withthe above equivalent terms gives

d2y

dt2− dy

dt− 7

dy

dt+ 16y = 16x4.

Simplifying the previous equation, and noting that x4 = (et)4 = e4t,we get

d2y

dt2− 8

dy

dt+ 16y = 16e4t.

This is a linear differential equation with constant coefficients, so wecan solve it using the method given earlier. The auxiliary equationof this equation is s2 − 8s + 16 = 0. Thus (s − 4)2 = 0, and sos = 4 (twice). Hence the complementary function of the differentialequation is

C (t) = (At + B)e4t

,





Since 4 is a double root of the auxiliary equation, we look for aparticular integral of the form p = Gt2e4t. Then

d p

dt= 2Gte4t + 4Gt2e4t,

= 2Gt(1 + 2t)e4t;d2 p

dt2= 2G

(1 + 4t)e4t + 4t(1 + 2t)e4t

,

= 2G(1 + 8t + 8t2)e4t.

So putting y = p in the differential equation gives

2G(1 + 8t + 8t2)e4t − 16Gt(1 + 2t)e4t + 16Gt2e4t = 16e4t.

Cancelling the common factor of e4t from this expression gives

G(2 + 16t + 16t2 − 16t − 32t2 + 16t2) = 16.

Thus 2G = 16, and so G = 8. Thus the particular integral is

p = 8t2e4t.


y = C + p,

= (At + B)e4t + 8t2e4t,

= (At + B + 8t2)e4t.

Now et = x, and so e4t = x4 and t = ln x. Thus

y = (A ln x + B + 8(ln x)2)x4.

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Notes 4 Differential Equations

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