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Advanced Calculus Chapter 4 Differential Equations
4
65
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. When
n = 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 the
highest order of derivative appearing in the equation.
Example 1 The following are all examples of differential equations. 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
= 2x.
dx
(b)
ex
dy
=
.
dx
y+2
(c)
d2 y
dy
+3
+ 2y = 4 sin x.
2
dx
dx
(d)
d2 y
= −ω 2 y.
dx2
This is the equation for simple harmonic motion. It arises
naturally in the solution of many problems in mechanics.
(e)
d3 y
d2 y
dy
+
x
+ x2
+ x3 y = 0.
3
2
dx
dx
dx
(f)
∂z ∂z
+
= 3xy.
∂x ∂y
(g)
∂ 2z
= x2 + y 2 .
∂y∂x
(h)
∂U
∂ 2U
= α 2 . (The heat equation.)
∂t
∂x
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
dy
= 2x. However, there are other solutions.
differential equation
dx
Advanced Calculus Chapter 4 Differential Equations
66
For example, y = x2 + 5, y = x2 − 7 and y = x2 + 33.75 are
all solutions of this differential equation. In fact, y = x2 + c is a
solution of this differential equation for any constant c. This last
solution is called the general solution of the differential equation.
For most ordinary differential equations of order one, the general
solution involves an arbitrary constant. Given the value of y for a
particular value of x we can determine the value of c. Such values
of y and x are called boundary conditions for the differential
equation.
For an ordinary differential equation of order k, the general solution
normally involves k arbitrary constants. In this case, k independent
boundary conditions are required to find a unique solution.
The solutions of the differential equation in Example 2 gave y explicitly in terms of x. However, solutions of differential equations
often express y implicitly in terms of its variables, as the following
example illustrates.
Example 3 The general solution of the ordinary differential equady
ex
tion
=
is
dx
y+2
1 2
y + 2y = ex + c,
2
(1)
where c is a constant. We can check this by differentiating expression (1) implicitly with respect to x to get
y
dy
dy
+2
= ex .
dx
dx
Thus
dy
(y + 2) = ex ,
dx
and dividing throughout by y + 2 gives
ex
dy
=
,
dx
y+2
as required.
Note that starting from the given solution we get
y 2 + 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,
Advanced Calculus Chapter 4 Differential Equations
67
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
p
y = 2ex + f − 2.
So in this case it is also possible to give the solution for y explicitly
in terms x. However, this is not always possible (see the next
example), and it is debatable whether it is worth the effort to find
such a solution in this case.
Example 4 The general solution of the ordinary differential equady
y(x − 1)
tion
=
is
dx
x(y + 1)
y + ln y − x + ln x = c.
This can easily be verified by differentiating the solution with redy
spect to x and rearranging the result to get
. However, it is not
dx
possible to rearrange this solution to express y explicitly in terms
of x.
4.2
Ordinary differential equations of order one
In this section we consider the problem of finding the general solution of an ordinary differential equation of order one. This is not
always possible, but there are many methods for finding the general solution, depending on the nature of the differential equation,
and we consider some of these here. We also consider some special
methods for certain families of differential equations.
Variables separable differential equations
Let M and N be functions of one variable, and consider the differential equation
M (x)
dy
=
.
dx
N (y)
Then
dy
= M (x).
dx
Integrating both side with respect to x gives
Z
Z
dy
N (y) dx = M (x) dx,
dx
N (y)
Advanced Calculus Chapter 4 Differential Equations
68
and so, by the chain rule,
Z
Z
N (y) dy = M (x) dx.
Thus, if we can evaluate the two integrals, we can find the general
solution of the differential equation. Differential equations of this
nature are called variables separable, and the method outlined
for solving them is called the separation of variables.
Example 5 Consider the differential equation
9x
dy
x2 + 1
=
.
dx
2y
Separating the variables gives
Z
Z 2
x +1
2y dy =
dx.
9x
Thus
Z
1
2y dy =
9
¶
Z µ
1
x+
dx.
x
Evaluating the two integrals gives
µ
¶
1 x2
2
y =
+ ln x + c ,
9 2
where c is a constant, and so
r
1 x2
y=
+ ln x + c.
3 2
Example 6 Consider the differential equation
dy
= ay,
dx
where a is a constant.
Separating the variables gives
Z
Z
dy
= a dx.
y
Evaluating the two integrals gives
ln y = ax + c,
where c is a constant, and so
y = eax+c = Aeax ,
where A = ec .
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Advanced Calculus Chapter 4 Differential Equations
Example 7 Consider the differential equation
(3y + 2)3 cos2 x
dy
= 1.
dx
Separating the variables gives
Z
Z
3
(3y + 2) dy =
Thus
Z
dx
.
cos2 x
Z
3
(3y + 2) dy =
sec2 x dx.
Evaluating the two integrals gives
1
(3y
3
+ 2)4
= tan x + c,
4
where c is a constant, and so
(3y + 2)4 = 12 tan x + d,
where d = 12c.
Note that, with a bit more effort,
√ we can write2 the solution of this
1 4
differential equation as y = 3 12 tan x + d − 3 , but it is a matter
of mathematical taste as to whether this is better way of expressing
the solution than the one given.
Exact differential equations
Let M and N be functions of two variables. The differential equation
dy
M (x, y) + N (x, y)
=0
dx
is exact if
∂M
∂N
=
.
∂y
∂x
When a differential equation of the given form is exact then the
general solution has the form f (x, y) = c, for some constant c and
some function f , of two variables, satisfying
∂f
∂f
= M and
= N.
∂x
∂y
To see this, we differentiate f (x, y) = c with respect to x. This
gives
∂f dx ∂f dy
+
= 0.
∂x dx ∂y dx
Since
dx
dx
= 1,
∂f
∂x
= M and
∂f
∂y
= N we get
M +N
as required.
dy
= 0,
dx
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Advanced Calculus Chapter 4 Differential Equations
Example 8 Consider the differential equation
2xy + (x2 + y 2 )
dy
= 0.
dx
Here M = 2xy and N = x2 +y 2 , and since ∂M
= 2x and ∂N
= 2x =
∂y
∂x
∂M
, the differential equation is exact. Thus we look for a function
∂y
f of two variables such that ∂f
= M and ∂f
= N . That is,
∂x
∂y
∂f
= 2xy;
∂x
∂f
= x2 + y 2 .
∂y
(2)
(3)
Integrating both sides of equation (2) with respect to x gives
f = x2 y + g(y),
(4)
where g is a function of y. To find g we differentiate equation (4)
with respect to y to get
∂f
= x2 + g 0 (y).
∂y
Comparing this expression for
∂f
∂y
with the one in equation (3) gives
x2 + g 0 (y) = x2 + y 2 .
Thus g 0 (y) = y 2 and so g(y) = 13 y 3 . Hence f (x, y) = x2 y + 13 y 3 , and
so the solution of the original differential equation is
1
x2 y + y 3 = c,
3
for some constant c.
Example 9 Consider the differential equation
µ 2
¶
x
dy
1 + ln x + 2x ln y +
− 2y
= 0.
y
dx
2
Here M = 1 + ln x + 2x ln y and N = xy − 2y, and since ∂M
= 2x
∂y
y
2x
∂M
and ∂N
=
=
,
the
differential
equation
is
exact.
Thus
we
look
∂x
y
∂y
for a function f of two variables such that ∂f
= M and ∂f
= N.
∂x
∂y
That is,
∂f
= 1 + ln x + 2x ln y;
∂x
∂f
x2
=
− 2y.
∂y
y
(5)
(6)
Advanced Calculus Chapter 4 Differential Equations
Integrating both sides of equation (5) with respect to x gives
f = x ln x + x2 ln y + g(y),
71
R
To evaluate ln x dx, use
integration by parts with
dv
(7) u = ln x and dx
= 1.
where g is a function of y. To find g we differentiate equation (7)
with respect to y to get
∂f
x2
=
+ g 0 (y).
∂y
y
Comparing this expression for
∂f
∂y
with the one in equation (6) gives
x2
x2
+ g 0 (y) =
− 2y.
y
y
Thus g 0 (y) = −2y and so g(y) = −y 2 . Hence
f (x, y) = x ln x + x2 ln y − y 2 , and so the solution of the original
differential equation is
x ln x + x2 ln y − y 2 = c,
for some constant c.
Integrating factors
Let M and N be functions of two variables. In general the differential equation
dy
M (x, y) + N (x, y)
=0
dx
will not be exact. However, suppose we can find a function µ of
two variables such that the differential equation
dy
=0
dx
is exact. Then we can solve the resulting differential equation using
the method described, and hence find the general solution of the
original differential equation. The function µ is called an integrating factor for the original differential equation.
µ(x, y)M (x, y) + µ(x, y)N (x, y)
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. In
the next subsection we consider one important class of differential
equations that can be solved using an integrating factor.
Linear differential equations
A ordinary differential equation of order one is linear if it can be
written in the form
dy
+ P (x)y = Q(x),
dx
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Advanced Calculus Chapter 4 Differential Equations
where P and Q are functions of one variable.
Given a linear differential equation of this form, let
R
µ(x) = e
P (x) dx
.
We claim that µ is an integrating factor for the linear differential
equation.
First note that
R
µ0 (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
= 0,
dx
and then multiply both sides by µ(x) to get
µ(x)(P (x)y − Q(x)) + µ(x)
dy
= 0.
dx
We now need to show that this differential equation is exact. Let
M = µ(x)(P (x)y − Q(x)) and N = µ(x). Then
∂M
∂y
∂N
∂x
= µ(x)P (x);
= µ0 (x),
= µ(x)P (x).
Thus ∂M
= ∂N
, and so the differential equation is exact. Thus µ(x)
∂y
∂x
is an integrating factor for the linear differential equation.
Note that in this case we do not have to reduce the linear differential
equation to the exact form since (using the Product Rule),
d
dy
(µ(x)y) = µ0 (x)y + µ(x) ,
dx
dx
dy
= (µ(x)P (x))y + µ(x) ,
µ
¶dx
dy
= µ(x) P (x)y +
,
dx
= µ(x)Q(x).
d
Thus
(µ(x)y) = µ(x)Q(x), and so integrating both sides with
dx
respect to x gives
Z
µ(x)y = µ(x)Q(x) dx.
To differentiate µ(x), we use the
chain
¡R rule and¢ the fact that
d
P (x) dx = P (x).
dx
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Advanced Calculus Chapter 4 Differential Equations
Thus
1
y =
µ(x)
Z
µ(x)Q(x) dx.
(8)
So to solve the linear differential equation
dy
+ P (x)y = Q(x),
dx
R
first find the integrating factor µ(x) = e
tity (8) to find y.
P (x) dx
, and then use iden-
Example 10 Consider the linear differential equation
dy
+ ay = Aebx ,
dx
where A, a and b are constants.
Here P (x) = a and Q(x) = Aebx . Let
µ(x) = e
R
a dx
= eax .
Then
y =
=
=
=
=
Z
1
µ(x)Q(x) dx,
µ(x)
Z
−ax
e
(eax )(Aebx ) dx,
Z
−ax
e
Ae(a+b)x dx,
µ (a+b)x
¶
Ae
−ax
e
+c ,
a+b
A bx
e + ce−ax .
a+b
So the general solution of the differential equation is
y=
A bx
e + ce−ax ,
a+b
where c is a constant.
Example 11 Consider the linear differential equation
(x2 + 1)
Then
dy
+ xy = x(x2 + 1).
dx
x
dy
+ 2
y = x.
dx (x + 1)
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Advanced Calculus Chapter 4 Differential Equations
Thus the differential equation is linear with P (x) =
Q(x) = x. Let
R
µ(x) = e
1
2
x
(x2 +1)
R
dx
2x
(x2 +1)
x
(x2 +1)
and
,
dx
= e
,
1
ln(x2 +1)
2
,
= e
1
³
´
ln(x2 +1) 2
= e
,
1
= (x2 + 1) 2 .
Then
1
y =
µ(x)
=
=
=
=
Z
µ(x)Q(x) dx,
Z
1
2
− 21
(x + 1)
((x2 + 1) 2 )x dx,
µ
¶
3
1 2
2
− 12
(x + 1)
(x + 1) 2 + c) ,
3
1
1 2
(x + 1) + c(x2 + 1)− 2 ,
3
1 2
c
(x + 1) + √
.
3
x2 + 1
So the general solution of the differential equation is
1
c
y = (x2 + 1) + √
,
3
x2 + 1
where c is a constant.
Homogeneous differential equations
Recall that a function f of two variables is homogeneous of degree
n if
f (λx, λy) = λn f (x, y).
Let M and N be functions of two variables. Consider the differential
equation
dy
= 0,
M (x, y) + N (x, y)
dx
where M and N are both homogeneous of degree n. (Such a differential equation is called homogeneous.) Then we put y = vx,
where v is a function of x.
By the product rule
dy
dv
=x
+ v.
dx
dx
Advanced Calculus Chapter 4 Differential Equations
75
Also, since M is homogeneous of degree n,
M (x, y) =
=
=
=
M (x, vx),
M (x.1, x.v),
xn M (1, v),
xn M̂ (v),
for some function M̂ of one variable.
Similarly N (x, y) = xn N̂ (v), for some function N̂ of one variable.
So, with y = vx the original differential equation becomes
µ
¶
dv
n
n
x M̂ (v) + x N̂ (v) x
+ v = 0.
dx
Since this is valid for all x, we can cancel the common factor of xn
to get
¶
µ
dv
+ v = 0.
M̂ (v) + N̂ (v) x
dx
This final differential equation is variables separable. Thus, using
the method given earlier, we can find v, and consequently y.
Note When solving a homogeneous differential equation, it is not
necessary to identify the functions M̂ and N̂ . You just have to
remember to eliminate y by putting y = vx. If the resulting differential equation is not variables separable then you have either made
a mistake, or the original differential equation is not homogeneous.
Example 12 Consider the differential equation
x3 + y 3 + 2xy 2
dy
= 0.
dx
It is easy to verify that M = x3 + y 3 and N = 2xy 2 are both homogeneous of degree 3. Putting y = vx in the differential equation
gives
¶
µ
dv
3
3 3
2 2
x + v x + 2xv x x
+ v = 0.
dx
Cancelling the common factor of x3 , and rearranging the resulting
differential equation gives
2xv 2
dv
= −(1 + 3v 3 ).
dx
Separating the variables gives
Z
Z
2v 2
dx
dv
=
−
.
1 + 3v 3
x
Advanced Calculus Chapter 4 Differential Equations
76
Evaluating the two integrals we get
2
ln(1 + 3v 3 ) = − ln x + c,
9
where c is a constant. Since y = vx then v = xy , and so substituting
for v in the above solution, we get the general solution of the original
differential equation as follows:
µ
¶
3y 3
2
ln 1 + 3 + ln x = c.
9
x
Example 13 Consider the differential equation
³y´
dy
= 0.
x
dx
¡ ¢
It is easy to verify that M = y +x sec xy and N = −x are both homogeneous of degree 1. Putting y = vx in the differential equation
gives
µ
¶
³ vx ´
dv
vx + x sec
−x x
+ v = 0.
x
dx
y + x sec
−x
Cancelling the common factor of x, and rearranging the resulting
differential equation gives
x
dv
= sec v.
dx
Separating the variables gives
Z
Z
dv
dx
=
,
sec v
x
and, since sec v =
1
,
cos v
we get
Z
Z
dx
cos v dv =
.
x
Evaluating the two integrals, we get
sin v = ln x + c,
where c is a constant. Since y = vx then v = xy , and so substituting
for v in the above solution, we get the general solution of the original
differential equation as follows:
³y´
= ln x + c.
sin
x
Advanced Calculus Chapter 4 Differential Equations
77
Change of variable
When solving a homogeneous differential equation we changed the
variables with the substitution y = vx. Often other first order
differential equations can be solved by changing the variables.
For some differential equations it is clear what the change of variables should be, as illustrated in the following example.
Example 14 Consider the differential equation
dy
= (x + 4y + 3)2 .
dx
Here we put z = x + 4y + 3. Then
dz
dy
= 1+4 ,
dx
dx
= 1 + 4(x + 4y + 3)2 ,
= 1 + 4z 2 ,
and so we get the variables separable equation
dz
= 1 + 4z 2 .
dx
Separating the variables gives
Z
Z
dz
=
dx,
1 + 4z 2
and evaluating the two integrals we get
1
arctan(2z) = x + c,
2
for some constant c.
Substituting for z gives
1
arctan(2(x + 4y + 3)) = x + c,
2
which is the general solution of the original equation.
The general solution can be tidied up (with a little effort) to get
1
y = (tan(2x + d) − 2x − 6),
8
where d = 2c.
In general, it is not always obvious what the change of variables
should be. However, there are some classes of differential equations
that can be solved with a standard change of variable. We give two
such special cases in the next two subsections.
Advanced Calculus Chapter 4 Differential Equations
78
Special case 1
Let a, b, c, d, e, f ∈ R with at least one of d, e or f nonzero. Consider
the differential equation
dy
ax + by + c
=
.
dx
dx + ey + f
The differential equations in this class can be solved using a change
of variables, depending on whether the lines ax + by + c = 0 and
dx + ey + f = 0 are parallel.
Suppose first that ax + by + c = 0 and dx + ey + f = 0 are not
parallel. Then let (α, β) be the point where these two lines intersect,
and put
x = X + α,
y = Y + β.
dy
dY
We claim that dX
= dx
. To see this, suppose that y = g(x) for
dy
some function g. Then dx = g 0 (x). Also Y + β = g(X + α) and so,
by the chain rule,
dY
dy
= g 0 (X + α) = g 0 (x) =
.
dX
dx
Thus
dY
dX
=
dy
,
dx
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.
Similarly
dx + ey + f = dX + eY.
Using the above, after the change of variables, the differential equation becomes
dY
aX + bY
=
.
dX
dX + eY
This differential equation is homogeneous, and so can be solved using the method given earlier. Thus we can find the general solution
of 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).
Thus
ax + by + c
dy
=
.
dx
λ(ax + by + g)
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Advanced Calculus Chapter 4 Differential Equations
dy
= a + b dx
, and so
µ
¶,
dy
dz
=
−a
b.
dx
dx
Put z = ax + by. Then
dz
dx
If |λ| < 1 it is slightly easier to
put z = dx + ey.
Thus, changing the variables in the differential equation we get
µ
¶,
dz
z+c
−a
b=
,
dx
λ(z + g)
which can be rearranged to get
dz
b(z + c)
=
+ a.
dx
λ(z + g)
This differential equation is variables separable, and so can be
solved using the method given earlier. Thus we can find the general
solution of the original equation.
Example 15 Consider the differential equation
dy
x+y−6
=
.
dx
x−y−2
The lines x + y − 6 = 0 and x − y − 2 = 0 are not parallel and To find the point where the lines
intersect at the point (4, 2). Putting x = X + 4 and y = Y + 2 into intersect solve the equations
the differential equation gives
x + y = 6,
x − y = 2.
dY
X +4+Y +2−6
X +Y
=
=
.
dX
X +4−Y −2−2
X −Y
This differential equation is homogeneous, so putting Y = XV we
dY
dV
get dX
= X dX
+ V , and the differential equation becomes
X
dV
X + XV
1+V
+V =
=
.
dX
X − XV
1−V
Thus
X
dV
1+V
1 + V − V (1 − V )
1+V2
=
−V =
=
.
dX
1−V
1−V
1−V
The last differential equation is variables separable and so
Z
Z
1−V
dX
dV =
.
2
1+V
X
Hence
Z µ
1
V
−
2
1+V
1+V2
¶
Z
dV =
dX
.
X
Evaluating the two integrals gives
arctan V −
1
ln(1 + V 2 ) = ln X + c,
2
Advanced Calculus Chapter 4 Differential Equations
80
where c is a constant. Multiplying throughout by 2, and rearranging
the solution gives
2 arctan V − (ln(1 + V 2 ) + 2 ln X) = d,
where d = 2c. Now 2 ln X = ln X 2 , and so
ln(1 + V 2 ) + 2 ln X = ln(1 + V 2 ) + ln X 2 = ln(X 2 + X 2 V 2 ).
Thus
2 arctan V − ln(X 2 + X 2 V 2 ) = d.
Since V =
Y
X
the solution becomes
µ ¶
Y
− ln(X 2 + Y 2 ) = d.
2 arctan
X
Finally, putting X = x − 4 and Y = y − 2 we get
µ
¶
y−2
− ln((x − 4)2 + (y − 2)2 ) = d,
2 arctan
x−4
which is the general solution of the original differential equation.
Example 16 Consider the differential equation
dy
x + 2y + 5
=
.
dx
2x + 4y − 3
The lines x + 2y + 5 = 0 and 2x + 4y − 3 = 0 are parallel so we put
z = x + 2y. Then
dz
dy
=1+2 ,
dx
dx
and so
µ
¶
dy
1 dz
=
−1 .
dx
2 dx
Thus, changing the variable, the differential equation becomes
¶
µ
1 dz
z+5
−1 =
.
2 dx
2z − 3
Hence
dz
2z + 10
4z + 7
=
+1=
.
dx
2z − 3
2z − 3
This differential equation is variables separable, and so separating
the variables gives
Z
Z
2z − 3
dz =
dx.
(9)
4z + 7
Advanced Calculus Chapter 4 Differential Equations
81
Now
Z
Z
2z − 3
1
4z − 6
dz =
dz,
4z + 7
2
4z + 7
Z
4z + 7 − 13
1
=
dz,
2
4z + 7
Z
1
13
=
1−
dz,
2
4z + 7
µ
¶
1
13
=
z−
ln(4z + 7) .
2
4
Thus evaluating the two integrals in equality (9) gives
µ
¶
13
1
z−
ln(4z + 7) = x + c,
2
4
where c is a constant. Multiplying both sides of the last equality
by 8 we get
4z − 13 ln(4z + 7) = 8x + d,
where d = 8c. Substituting z = x + 2y in the last expression and
simplifying the result gives
8y − 4x − 13 ln(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 differential equation
dy
+ P (x)y = Q(x)y n .
dx
To solve this differential equation first divide both sides by y n to
get
1 dy P (x)
+ n−1 = Q(x).
y n dx
y
Now put u =
1
y n−1
= y 1−n . For this change of variable
du
dy
1 − n dy
= (1 − n)y −n
=
,
dx
dx
y n dx
and so
dy
y n du
=
.
dx
1 − n dx
After the change of variable, the differential equation becomes
µ n
¶
1
y du
+ uP (x) = Q(x),
y n 1 − n dx
82
Advanced Calculus Chapter 4 Differential Equations
and so
1 du
+ uP (x) = Q(x).
1 − n dx
Multiplying both sides by 1 − n yields the differential equation
du
+ (1 − n)uP (x) = (1 − n)Q(x),
dx
which is linear, so it can be solved by finding an integrating factor
and using the method given earlier.
Example 17 Consider the differential equation
dy y
+ = x3 y 6 .
dx x
Dividing both sides by y 6 gives
1 dy
1
+ 5 = x3 .
6
y dx xy
6
dy
dy
= − y66 dx
, and so dx
= − y5 du
. Changing
Putting u = y16 , we get du
dx
dx
the variable in the differential equation gives
µ 6 ¶
1
y du
u
−
+ = x3 .
6
y
5 dx
x
Simplifying this differential equation, and then multiplying both
sides by −5 gives
du 5
− u = −5x3 .
dx x
The preceding differential equation is linear so we first find the
integrating factor
R
5
µ(x) = e − x dx ,
= e−5 ln x ,
−5
= eln x ,
1
= x−5 = 5 .
x
Then
Z
1
u =
µ(x)(−5x3 ) dx,
µ(x)
Z
5
5
= x
− 2 dx,
x
µ
¶
5
5
= x
+c ,
x
where c is a constant. Thus
u = x4 (5 + cx).
Since u =
tion is
1
,
y5
the general solution of the original differential equa-
or x4 y 5 (5 + cx) = 1.
1
= x4 (5 + cx),
5
y
Advanced Calculus Chapter 4 Differential Equations
4.3
83
Ordinary differential equations of order two
To find the general solution of a differential equation of order two
is more difficult than for order one. In this section we only consider
linear differential equations. That is, differential equations of the
form
d2 y
dy
P (x) 2 + Q(x)
+ R(x)y = S(x),
dx
dx
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 constant
functions and S is identically zero. That is, the differential equation
has the form
d2 y
dy
+Q
+ Ry = 0,
(10)
2
dx
dx
where P, Q, R ∈ R. Such a differential equation is called homogeneous with constant coefficients.
P
In looking for the general solution of equation (10) we note first
that if y is a solution of the differential equation then so is Ay, for
a 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 homogeneous linear differential equation of order one was an exponential
function. Inspired by this, we look for solutions of equation (10)
dy
having the form y = etx , for some constant t. For this y, dx
= tetx
2
2
d y
dy
d y
2 tx
and dx
2 = t e . Substituting for y, dx and dx2 in equation (10)
gives
P t2 etx + Qtetx + Retx = 0.
Since etx 6= 0, we can cancel this factor from the previous equation
to get
P t2 + Qt + R = 0.
This is a quadratic equation in t, and is called the auxiliary equation of the differential equation given in (10).
Let α and β be the roots of the auxiliary equation. Then the general
solution of (10) depends on the nature of α and β. There are three
cases to consider.
Case 1: α, β ∈ R with α 6= β. This is the easiest case; since we
know that eαx and eβx are distinct solutions of equation (10), the
general solution is
y = Aeαx + Beβx ,
where A and B are (real) constants.
84
Advanced Calculus Chapter 4 Differential Equations
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, it
is straightforward to check that xeαx is also a solution of equation (10), and so the general solution is
y = (A + Bx)eαx ,
where A and B are (real) constants.
Case 3: α, β 6∈ R. We first recall that it is not possible to have
one of α or β in R and the other not in R, since the non-real roots
of a polynomial equation with coefficients in R occur in conjugate
√
pairs. Thus, in this case, α = a + bi and β = a − ib where a, b ∈ R, Recall that i = −1.
with b 6= 0. Doing the same as in case 1 we could write the general
solution of equation (10) in the form
y = Âe(a+bi)x + B̂e(a−ib)x ,
where  and B̂ are complex constants. However, it is unsatisfactory to have the solution of a differential equation involving a real
function expressed in terms of complex functions, so we aim to find
a better way of expressing the solution. Now
Here we use the standard rules
e(a+bi)x = eax+bxi = eax ebxi = eax (cos(bx) + i sin(bx)).
Similarly,
(a−bi)x
e
for exponentials together with
the result that
eiθ = cos θ + i sin θ,
ax−bxi
=e
ax −bxi
=e e
ax
= e (cos(bx) − i sin(bx)).
(a+bi)x
Using these equivalent expressions for e
(a−bi)x
and e
for all θ ∈ R.
we get
Âe(a+bi)x + B̂e(a−ib)x = Âeax (cos(bx) + i sin(bx)) +
B̂eax (cos(bx) − i sin(bx)),
= eax ((Â + B̂) cos(bx) + i(Â − B̂) sin(bx)),
= eax (A cos(bx) + B sin(bx)),
where A = Â + B̂ and B = i(Â − B̂).
In summary, we have the following method for finding the general
solution of equation (10). First find the solutions α and β of the
auxiliary 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
the auxiliary equation
General solution
of equation (10)
α, β ∈ R, α 6= β
y = Aeαx + Beβx
α, β ∈ R, α = β
y = (A + Bx)eαx
α = a + ib, β = a − ib,
a, b ∈ R, b 6= 0
y = eax (A cos(bx) + B sin(bx))
Note, since A, B ∈ R, it follows
that B̂ is the complex conjugate
of Â.
85
Advanced Calculus Chapter 4 Differential Equations
Example 18 Consider the differential equation
d2 y
dy
−5
+ 6y = 0.
2
dx
dx
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 solution
of the differential equation is
y = Ae2x + Be3x ,
where A and B are constants.
Example 19 Consider the differential equation
d2 y
dy
4 2 −4
+ y = 0.
dx
dx
The auxiliary equation of this differential equation is
4t2 − 4t + 1 = 0.
Thus (2t−1)2 = 0, and so t =
of the differential equation is
1
2
(twice). Hence the general solution
1
y = (Ax + B)e 2 x ,
where A and B are constants.
Example 20 Consider the differential equation
d2 y
dy
− 10
+ 29y = 0.
2
dx
dx
The auxiliary equation of this differential equation is
t2 − 10t + 29 = 0.
Thus
√
100 − 4 × 29
,
√ 2
10 ± −16
,
=
2
= 5 ± 2i.
t =
10 ±
Hence the general solution of the differential equation is
y = e5x (A cos(2x) + B sin(2x)),
where A and B are constants.
Advanced Calculus Chapter 4 Differential Equations
86
The nonhomogeneous case with constant coefficients
We now consider the case where P , Q and R are constant functions
and S is a function of x. That is, the differential equation has the
form
dy
d2 y
+ Ry = S(x),
(11)
P 2 +Q
dx
dx
where P, Q, R ∈ R. Such a differential equation is called nonhomogeneous with constant coefficients.
To find the general solution of equation (11) we have to do two
things. First we find the general solution C(x)of the homogeneous
part
dy
d2 y
P 2 +Q
+ Ry = 0.
dx
dx
This is called the complementary function of equation (11).
Next we find any solution p(x) of equation (11). This is called a
particular integral of the differential equation. Then the general
solution of equation (11) is
y = C(x) + p(x).
Finally, if there are any boundary conditions we use them to determine the arbitrary constants in the general solution.
Finding a particular integral
In the above method, we know how to find the complementary
function. However, finding a particular integral is very hard for
most functions S(x), but there are some families of functions for
which it is possible to find a particular integral. Here we give three
such families of functions.
Family 1: S(x) = M eγx , where M, γ ∈ R. In this case we look
for a particular integral of the form
p(x) = Gxj eγx ,
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. In
this 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 auxiliary
equation are ±ci, in which case we look for a particular integral of
the form
p(x) = x(G cos(cx) + H sin(cx)).
Advanced Calculus Chapter 4 Differential Equations
87
Family 3: S(x) = Mk xk + Mk−1 xk−1 + · · · + M1 x + M0 , where
Mk , Mk−1 , . . . , M1 , M0 ∈ R. In this case we look for a particular
integral of the form
p(x) = xj (mk xk + mk−1 xk−1 + · · · + m1 x + 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 differential equation
P
d2 y
dy
+Q
+ Ry = S(x)
2
dx
dx
must have R = 0. That is the differential equation is
P
d2 y
dy
+Q
= S(x).
2
dx
dx
In this case, it is easier to change the variable by putting u =
d2 y
(so du
= dx
2 ) to get
dx
P
dy
dx
du
+ Qu = S(x),
dx
and then solve the resulting first order differential equation. If we do
this, then whenever S(x) is a polynomial of degree k the particular
integral 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 the
roots of the auxiliary equation take certain values. In the latter
case we multiple the form of the particular integral by a suitable
power of x; this is required because, in this case, S(x) is part of the
complementary 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 three
families considered then the particular integral will be a sum of the
corresponding particular integrals.
Example 21 Consider the differential equation
dy
d2 y
−
5
+ 6y = 20e−3x ,
dx2
dx
with boundary conditions y = 5,
dy
dx
= 3 when x = 0.
From Example 18, the auxiliary equation of the homogeneous part
of 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 ,
Advanced Calculus Chapter 4 Differential Equations
88
where A and B are constants.
Here S = 20e−3x , so we look for a particular integral of the form
dp
d2 p
−3x
p = Ge−3x . Then dx
= −3Ge−3x and dx
. Now, since p is
2 = 9Ge
a particular integral of the differential equation it must satisfy the
differential equation. So putting y = p in the differential equation
gives
9Ge−3x − 5(−3Ge−3x ) + 6Ge−3x = 20e−3x .
Cancelling the common factor of e−3x from this expression, and
simplifying the result gives 30G = 20, and so G = 23 . Thus the
particular integral is
2
p = e−3x .
3
From the above, the general solution of the differential equation is
2
y = C + p = Ae2x + Be3x + e−3x .
3
We now use the boundary conditions to determine A and B. Differentiating the general solution with respect to x gives
dy
= 2Ae2x + 3Be3x − 2e−3x .
dx
dy
Now, when x = 0, y = 5 and dx
= 3. Thus, putting x = 0 in the
general solution and its derivative, and then comparing them with
the corresponding boundary condition gives
2
= 5,
3
2A + 3B − 2 = 3.
A+B+
Simplifying these equations we get
3A + 3B = 13,
2A + 3B = 5.
(12)
(13)
Subtracting equation (13) from equation (12) gives A = 8; substituting A = 8 in equation (13) gives 16 + 3B = 5, which gives
. Thus the solution of the differential equation with the
B = − 11
3
given boundary conditions is
y = 16e2x −
11 3x 2 −3x
e + e .
3
3
Example 22 Consider the differential equation
d2 y
dy
−
5
+ 6y = 3e2x .
dx2
dx
Advanced Calculus Chapter 4 Differential Equations
89
As in the previous example, the roots of the auxiliary equation are
t = 2, 3 and the complementary function of the differential equation
is
C = Ae2x + Be3x ,
where A and B are constants.
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
dp
=
dx
=
2
dp
=
dx2
=
Ge2x + 2Gxe2x ,
G(1 + 2x)e2x ;
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
p = −3xe2x .
From the above, the general solution of the differential equation is
y = C + p = Ae2x + Be3x − 3xe2x .
Example 23 Consider the differential equation
d2 y
dy
− 10
+ 29y = 29x2 + 38x − 18.
2
dx
dx
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)),
where A and B are constants.
Here S = 29x2 + 38x − 18, so we look for a particular integral of
d2 p
dp
= 2ax + b and dx
the form p = ax2 + bx + c. Then dx
2 = 2a. So
putting y = p in the differential equation gives
2a − 10(2ax + b) + 29(ax2 + bx + c) = 29x2 + 38x − 18.
Note that all of the x terms
cancel. If this does not happen
then we have done something
wrong.
90
Advanced Calculus Chapter 4 Differential Equations
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.
From the above, the general solution of the differential equation is
y = C + p = e5x (A cos(2x) + B sin(2x)) + x2 + 2x.
Example 24 Consider the differential equation
d2 y
+ y = 4 sin x + 3 cos x.
dx2
The auxiliary equation of the homogeneous part of this differential
equation is t2 + 1 = 0, and so t = ±i. Hence the complementary
function of the differential equation is
C = A cos x + B sin x,
where A and B are constants.
Here S = 4 sin x + 3 cos x. Since ±i are roots of the auxiliary
equation, we look for a particular integral of the form
p = x(G sin x + H cos x). Then
dp
= G sin x + H cos x + x(G cos x − H sin x);
dx
d2 p
= G cos x − H sin x + G cos x − H sin x + x(−G sin x − H cos x),
dx2
= 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) = 4 sin x + 3 cos x. Thus G =
H = −2, and so the particular integral is
¶
µ
3
sin x − 2 cos x .
p=x
2
3
2
and
From the above, the general solution of the differential equation is
µ
¶
3
y = C + p = A cos x + B sin x + x
sin x − 2 cos x .
2
Advanced Calculus Chapter 4 Differential Equations
91
Example 25 Consider the differential equation
d2 y dy
+
+ y = 2ex + 3x + 4.
dx2 dx
The auxiliary equation of the homogeneous part of this differential
equation is t2 + t + 1 = 0, and so
√
−1 ± 1 − 4
t =
,
2√
3
1
= − ±
i.
2
2
Hence the complementary function of the differential equation is
Ã
Ã√ !
à √ !!
1
3
3
C = e− 2 x A cos
x + B sin
x
,
2
2
where A and B are constants.
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
dp
= Gex + H;
dx
d2 p
= Gex .
dx2
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 = 32 , H = 3 and
H + K = 4, so K = 1. Thus the particular integral is
2
p = ex + 3x + 1.
3
From the above, the general solution of the differential equation is
Ã
Ã√ !
à √ !!
1
3
3
2
y = C +p = e− 2 x A cos
x + B sin
x
+ ex +3x+1.
2
2
3
Order two linear differential equations
We now consider general order two linear differential equations.
That is, differential equations of the form
P (x)
dy
d2 y
+ Q(x)
+ R(x)y = S(x),
2
dx
dx
(14)
Advanced Calculus Chapter 4 Differential Equations
92
where P , Q, R and S are functions of one variable. For such a
differential equation, the homogeneous part is the differential
equation
d2 y
dy
P (x) 2 + Q(x)
+ R(x)y = 0.
dx
dx
As in the case with constant coefficients, finding the general solution
of equation (14) depends upon being able to find a solution of the
homogeneous part. To see this, suppose y = g is a solution of the
homogeneous part, for some function g of x. Then
P (x)
d2 g
dg
+ Q(x)
+ R(x)g = 0.
2
dx
dx
Now we change the variable in equation (14) by putting y = gv for
some function v of x. Then, using the product rule, we get
dy
dv
dg
= g
+v ;
dx
dx
dx
d2 y
d2 v
dg dv
d2 g
=
g
+
2
.
+
v
dx2
dx2
dx dx
dx2
Thus, with the change the variable, equation (14) becomes
µ 2
¶
µ
¶
dv
dg dv
d2 g
dv
dg
P (x) g 2 + 2
+ v 2 +Q(x) g
+v
+R(x)gv = S(x),
dx
dx dx
dx
dx
dx
which can be rewritten as
³
´
d2 g
dg
v P (x) dx
+
Q(x)
+
R(x)g
+
2
dx
³ 2
´
dg dv
d v
dv
P (x) g dx
+ Q(x)g dx
= S(x).
2 + 2 dx dx
(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
µ 2
¶
dv
dg dv
dv
P (x) g 2 + 2
+ Q(x)g
= S(x),
dx
dx dx
dx
which can be rewritten as
µ
¶
dg
dv
d2 v
+ Q(x)g
= S(x).
P (x)g 2 + 2P (x)
dx
dx
dx
Now we put w =
(16)
dv
dx
in equation (16) to get
µ
¶
dg
dw
+ 2P (x)
+ Q(x)g w = S(x),
P (x)g
dx
dx
which is a first order differential equation for w. Solving this differential equation gives w. We can then find the general solution
of equation (14) by integrating w with respect to x to get v and
finally getting y = vg.
93
Advanced Calculus Chapter 4 Differential Equations
Note You do not have to remember the first order differential
equation for w in terms of P, Q, S and g to solve equation (14); just
remember the change of variable y = gv, and derive the first order
differential equation from scratch.
Example 26 Consider the differential equation
x2 (x + 1)
d2 y
dy
+ x(3x − 2)
− (3x − 2)y = 10x3 .
2
dx
dx
It is easy to verify that y = x is a solution of the homogeneous part
of this equation. So putting y = vx we get
dy
dv
= x
+ v;
dx
dx
d2 y
d2 v
dv
=
x
+2 .
2
2
dx
dx
dx
Thus the differential equation becomes
µ 2
¶
µ
¶
dv
dv
dv
2
x (x+1) x 2 + 2
+x(3x−2) x
+ v −(3x−2)vx = 10x3 .
dx
dx
dx
Collecting terms in the previous equation involving
get
x3 (x + 1)
d2 v
dx2
and
dv
dx
we
¢ dv
d2 v ¡ 2
2
+
2x
(x
+
1)
+
x
(3x
−
2)
= 10x3 ,
dx2
dx
and simplifying gives
x3 (x + 1)
d2 v
dv
+ 5x3
= 10x3 .
2
dx
dx
Dividing the resulting equation throughout by x3 (x + 1), and then
dv
putting w = dx
we get
dw
5w
10
+
=
,
dx x + 1
x+1
which is a first order linear differential equation. To solve this
differential equation we first find the integrating factor
µ(x) =
=
=
=
R
5
e x+1 dx ,
e5 ln(x+1) ,
5
eln(x+1) ,
(x + 1)5 .
Advanced Calculus Chapter 4 Differential Equations
Then
94
Z
1
10
µ(x)
dx,
µ(x)
x+1
Z
1
5 10
(x
+
1)
dx,
(x + 1)5
x+1
Z
1
10(x + 1)4 dx,
(x + 1)5
¡
¢
1
2(x + 1)5 + C ,
5
(x + 1)
C
2+
,
(x + 1)5
w =
=
=
=
=
where C is a constant.
Now,
dv
dx
= w, and so
Z
v =
w dx,
Z µ
=
2+
¶
C
dx,
(x + 1)5
D
= 2x +
+ E,
(x + 1)4
where D = − C4 and E is a constant.
Finally, y = vx, and so the general solution of the original differential equation is
Dx
y = 2x2 +
+ Ex.
(x + 1)4
Euler equations
In general, it is difficult to find a solution of the homogeneous part
of an order two linear differential equation. However, there are some
families of equations where this is possible. Here we consider one
such family, the Euler equations. These are differential equations
of the form
dy
d2 y
ax2 2 + bx
+ cy = S(x),
dx
dx
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 an
Euler equation.
• Look for a solution of the homogeneous part of the equation
of the form y = xk . If a suitable value of k is found then apply
the method described above to find the general solution of the
nonhomogeneous equation. (This method does not always
work as the values of k found could be non-real.)
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Advanced Calculus Chapter 4 Differential Equations
• Change the variable by putting x = et . This reduces the
differential equation to one with constant coefficients. (This
method always works, but is probably more difficult.)
We now give some examples to illustrate the two methods.
Example 27 Consider the differential equation
x2
d2 y
dy
− 7x
+ 16y = 16x4 .
2
dx
dx
We look for a solution of
x2
d2 y
dy
− 7x
+ 16y = 0
2
dx
dx
2
dy
d y
k−2
of the form y = xk . Then dx
= kxk−1 and dx
.
2 = k(k − 1)x
2
dy
d y
Substituting for y, dx and dx2 in the homogeneous part gives
x2 k(k − 1)xk−2 − 7xkxk−1 + 16xk = 0,
and so
k(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 ⇔ k 2 − 8k + 16 = 0,
⇔ (k − 4)2 = 0,
⇔ k = 4.
So y = x4 is a solution of the homogeneous part of the differential
equation. We now change the variable in the differential equation
by putting y = vx4 . Then
dy
dv
= x4
+ 4x3 v,
dx
dx
¶
µ
dv
3
+ 4v ;
= x x
dx
¶
µ 2
¶
µ
d2 y
dv
d v dv
dv
3
2
= x x 2+
+4
+ 4v ,
+ 3x x
dx2
dx
dx
dx
dx
µ
¶
2
dv
2
2d v
= x x
+ 8x
+ 12v .
dx2
dx
2
dv
d v
Substituting the expressions found for y, dx
and dx
2 in the differential equation we get
¶
¶
µ
µ
2
dv
dv
4
4
2d v
+ 8x
+ 12v − 7x x
+ 4v + 16vx4 = 16x4 .
x x
dx2
dx
dx
Advanced Calculus Chapter 4 Differential Equations
96
Cancelling the common factor of x4 in the previous equation, and
simplifying the result gives
2
2d v
x
dx2
Putting w =
get
dv
dx
+x
dv
= 16.
dx
in this equation, and dividing throughout by x2 we
dw w
16
+ = 2,
dx
x
x
which is a first order linear differential equation. To solve this
differential equation we first find the integrating factor
R
1
µ(x) = e x dx ,
= eln x ,
= x.
Then
Z
1
16
µ(x) 2 dx,
µ(x)
x
Z
1
16
x 2 dx,
x
x
Z
1
16
dx,
x
x
1
(16 ln x + C) ,
x
16 ln x C
+
x
x
w =
=
=
=
=
where C is a constant.
Now,
dv
dx
= w, and so
Z
v =
w dx,
¶
Z µ
16 ln x C
=
+
dx,
x
x
= 8(ln x)2 + C ln x + D,
where D is a constant.
Finally, y = vx4 , and so the general solution of the original differential equation is
y = (8(ln x)2 + C ln x + D)x4 .
Example 28 Consider again the differential equation
x2
dy
d2 y
− 7x
+ 16y = 16x4 .
2
dx
dx
To evaluate
u = ln x.
R
16 ln x
x
dx, put
97
Advanced Calculus Chapter 4 Differential Equations
We now consider the second way of solving this equation using the
change of variable x = et . The
dy
dy dx
=
,
dt
dx dt
dy t
=
e,
dx
dy
= x ;
dxµ ¶
d2 y
d dy
=
,
dt2
dt dt
µ
¶
dy
d
x
,
=
dt
dx
µ
¶
dx d
dy
=
x
,
dt dx
dx
µ
¶
dy
d2 y
t
= e
+x 2 ,
dx
dx
µ
¶
dy
d2 y
= x
+x 2 ,
dx
dx
dy
d2 y
= x
+ x2 2 .
dx
dx
Using the final expressions for
dy
dt
and
d2 y
dt2
we get
dy
dy
=
;
dx
dt
d2 y
d2 y
dy
x2 2 =
−x ,
2
dx
dt
dx
d2 y dy
=
− .
dt2
dt
x
2
dy
d y
Replacing there terms x dx
and x2 dx
2 of the original equation with
the above equivalent terms gives
d2 y dy
dy
−
− 7 + 16y = 16x4 .
2
dt
dt
dt
4
Simplifying the previous equation, and noting that x4 = (et ) = e4t ,
we get
dy
d2 y
−
8
+ 16y = 16e4t .
2
dt
dt
This is a linear differential equation with constant coefficients, so we
can solve it using the method given earlier. The auxiliary equation
of this equation is s2 − 8s + 16 = 0. Thus (s − 4)2 = 0, and so
s = 4 (twice). Hence the complementary function of the differential
equation is
C(t) = (At + B)e4t ,
Advanced Calculus Chapter 4 Differential Equations
98
where A and B are constants.
Since 4 is a double root of the auxiliary equation, we look for a
particular integral of the form p = Gt2 e4t . Then
dp
=
dt
=
d2 p
=
dt2
=
2Gte4t + 4Gt2 e4t ,
2Gt(1 + 2t)e4t ;
¡
¢
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 + 16Gt2 e4t = 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 = 8t2 e4t .
From the above, the general solution of the differential equation is
y = C + p,
= (At + B)e4t + 8t2 e4t ,
= (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 .