Lesson 3: Linear differential equations of the first order
Solve each of the following differential equations by two methods.
Exercise 3.1.
y0 + y = 3
Solution.
Method 1. It is clear that
R
e dx = ex
is an integrating factor for this differential equation. Therefore,
ex [y 0 + y] = 3ex ,
or
[yex ]0 = 3ex .
The solution is therefore the following:
Z
yex = c + 3 ex dx,
or simply,
y = ce−x + 3.
Method 2. Consider a homogeneous equation
y 0 + y = 0.
The solution of this equation is
e−x .
To solve the initial equation, introduce a new variable v, to find the solution in the
form
y = e−x v.
(∗)
Then
y 0 = v 0 e−x − e−x v = e−x [v 0 − v].
Next,
y 0 + y = v 0 e−x = 3,
and
v 0 = 3ex ,
v = 3ex + c
Substitution it for (*) yields
y = 3 + ce−x .
Exercise 3.2.
y 0 − 2y = x
Solution.
Method 1. It is clear that
R
e (−2)dx = e−2x
is an integrating factor for this differential equation. Therefore,
e−2x [y 0 − 2y] = xe−2x ,
or
[ye−2x ]0 = xe−2x .
1
2
The solution is therefore the following:
Z
ye−2x = c + xe−2x dx = c − 0.5xe−2x − 0.25e−2x
or simply,
y = ce2x − 0.5x − 0.25
Method 2. Consider a homogeneous equation
y 0 − 2y = 0.
The solution of this equation is
e2x .
To solve the initial equation, introduce a new variable v, to find the solution in the
form
y = e2x v.
Then
y 0 = v 0 e2x + 2e2x v = e2x [v 0 + 2v].
Next,
y 0 − 2y = v 0 e2x = x,
and
v 0 = xe−2x ,
v = c − 0.5xe−2x − 0.25e−2x
Therefore,
y = e2x [c − 0.5xe−2x − 0.25e−2x ] = ce2x − 0.5x − 0.25
Exercise 3.3.
xy 0 + y = 2x + ex .
Solution.
Method 1. It is not difficult to see that the differential equation is exact. Therefore, rewriting it as
¡
¢
xdy + y − 2x − ex dx = 0,
we obtain the solution
Z x
Z y
¢
(0 − 2u − eu du +
xdv = c1
0
0
or
xy − x2 − ex + 1 = c1 ,
which in turn reduces to
xy − x2 − ex = c.
Method 2. Assuming that x 6= 0 let us divide the both sides of differential
equation to x. We obtain
ex
y
y0 + = 2 + .
x
x
Let us now consider the homogeneous differential equation
y
y 0 + = 0.
x
It is clear that
1
y=
x
3
is a solution. Introduce a new variable v, representing
v
y= .
x
Then,
xv 0 − v
y0 =
x2
and
ex
v0
=2+ ,
x
x
v 0 = 2x + ex ,
and
v = x2 + ex + c.
Finally,
xy = c + x2 + ex .
Exercise 3.4.
x2 y 0 − xy = x2 + 4.
Solution. Notice that x 6= 0. Then, one can divide the left and right sides of
differential equation to x2 . We have
y
4
y0 − = 1 + 2 .
x
x
Let us solve the last differential equation by two methods.
Method 1. Notice that,
e−
R
x−1 dx
= e− log x =
1
x
is an integrating factor. Therefore,
1³ 0 y´ 1³
4´
y −
=
1+ 2 ,
x
x
x
x
or
³ y ´0 ³ 1
´
4
=
+ 3
x
x x
By integrating we obtain
³y´
2
= log |x| − 2 + c,
x
x
or
yx = x2 log |x| − 2 + cx2 .
Method 2. The homogeneous equation
y
y 0 − = 0.
x
has the solution
y = x.
Introduce a new variable v, representing
y = xv.
Then,
y 0 = v + xv 0
4
and
4
,
x2
1
4
v0 = + 3 .
x x
xv 0 = 1 +
By integrating,
v = log |x| −
Finally,
Exercise 3.5.
2
+ c.
x2
vx2 = yx = x2 log |x| − 2 + cx2 .
y 0 + ay = b (a, b − constants).
Solution.
Method 1. Notice that,
e−
is an integrating factor. Therefore,
R
adx
= eax
eax [y 0 + ay] = beax
is an exact differential equation. Then,
¡ ax ¢0
e y = beax ,
or if a 6= 0,
eax y =
Finally, if a 6= 0
y=
otherwise
b ax
e + c.
a
b
+ ce−ax ,
a
y = bx + c.
Method 2. The homogeneous equation (a 6= 0)
y 0 + ay = 0.
has the solution
y = e−ax .
Introducing new variable v put
Then,
Therefore,
Finally,
y = e−ax v.
y 0 = −ave−ax + v 0 e−ax .
v 0 e−ax = b,
v 0 = beax
b
v = eax + c.
a
y=
Exercise 3.6.
y0 +
b
+ ce−ax .
a
2
1
y = ex .
x
5
Solution.
Method 1. Notice that,
R
−1
e x dx = elog x = x
is an integrating factor. Therefore,
³
2
1 ´
x y 0 + y = xex
x
is an exact differential equation. Then,
2
(xy)0 = xex .
By integrating we obtain
2
2xy = ex + c.
Method 2. The homogeneous equation
1
y0 + y = 0
x
has the solution
1
y= .
x
Introducing new variable v put
v
y= .
x
Then,
xv 0 − v
y0 =
x2
Next,
2
v0
= ex ,
x
2
2v = ex + c.
Finally,
2
2xy = ex + c.
Exercise 3.7.
y 00 + y 0 = 2.
Solution. Denoting y 0 = p we obtain
p0x + p = 2.
Let us now solve this linear differential equation by two methods.
Method 1. Notice that
R
e dx = ex
is an integrating factor. Therefore,
¡ x 0
e p) = 2ex ,
and by integrating
p = 2 + c1 e−x .
By integrating once again we obtain
y = c2 + 2x − c1 e−x .
6
Method 2. The homogeneous equation
p0x + p = 0
has the solution e−x . Therefore, introducing a new variable v we find the solution
as
p = e−x v.
Then,
p0 = e−x v 0 − e−x v.
Next,
e−x v 0 = 2,
and
v = c1 + 2ex .
The solution is
p = c1 e−x + 2.
Integrating once again we have
y = c2 + 2x − c1 e−x .
Exercise 3.8.
xy 00 + y 0 = x − 2.
Solution. Denoting y = p, we have
0
xp0x + p = x − 2.
Method 1. Notice that differential equation is exact. Therefore
Z x
Z p
x2
(0 − u + 2)du +
xdv = 2x −
+ xp = 0.5c1
2
0
0
or
4x − x2 + 2xp = c1
Rewriting it
p=
c1
x2 − 4x
+
2x
2x
and integrating, we obtain
y=
c1
x2
log |x| +
− 2x + 0.25c2 .
2
4
Finally,
4y = 2c1 log |x| + x2 − 2x + c2 .
Method 2. In the case x = 0 the solution is trivial y = 2x + c. Assuming that
x 6= 0 let us divide to x. We have
p
2
p0x + = 1 − .
x
x
Consider the homogeneous equation
p
p0x + = 0.
x
Its solution is p = x−1 . Therefore, the solution of general (nonhomogeneous equation) is p = x−1 v. Then
v0 x − v
.
p0 =
x2
7
That is
v0
2
=1− .
x
x
or
v0 = x − 2
By integrating we have
v = 0.25c1 +
and
p=
Integrating once again, we obtain
y=
x2
− 2x,
2
c1
x
+ −2
4x 2
c1
x2
log |x| +
− 2x + 0.25c2
4
4
or finally
4y = c1 log |x| + x2 − 2x + c2 .
Exercise 3.9.
y 2 dx + (y 2 x + 2xy − 1)dy = 0.
Solution. For convenience, rewrite this differential equation as follows:
dx
y2
+ (y 2 x + 2xy − 1) = 0.
dy
Taking into account that y 6= 0, let us divide the equation to y 2 . We have
³
dx
2´
1
+x 1+
= 2.
dy
y
y
Solve this equation by the two methods.
Method 1. Notice that
R
−1
e (1+2y )dy = y 2 ey
is an integrating factor. Therefore
³
´
dx
y 2 ey
+ x y 2 ey + 2yey = ey .
dy
By immediate integration, we obtain
Z x
Z y
2 y
y e du −
ev dv = c1 ,
0
or
0
xy 2 ey = c + ey
Finally,
´
1³
−y
.
1
+
ce
y2
Method 2. The homogeneous differential equation
³
dx
2´
+x 1+
=0
dy
y
x=
has the solution x = e−y y −2 . Therefore, the solution of nonhomogeneous differential equation is
1
x = e−y 2 v,
y
8
x0 =
³h
− ey
´
2 −y i
1
0 −y 1
−
e
v
+
v
e
.
y2
y
y2
Then,
v 0 e−y
1
1
= 2,
y2
y
and
v 0 = ey ,
v = c + ey
Finally,
x=c
1 −y
1
e + 2.
y2
y
Exercise 3.10.
y + (x − y 3 − 2)y 0 = 0.
Solution. Rewrite the equation as follows
yx0 + x − y 3 − 2 = 0.
Method 1. Notice that the differential equation is exact. Therefore,
Z x
Z y
y4
− 2y = 0.25c
0du +
(x − v 3 − 2)dv = xy −
4
0
0
Finally,
4xy − y 4 − 8y = c.
Method 2. Assuming that y 6= 0 let us represent the differential equation as
follows
1
2
x0 + x = y 2 − .
y
y
Consider first the homogeneous equation
1
x0 + x = 0.
y
Its solution is x = y −1 . Therefore, the general solution is
v
x= .
y
Then,
v0 y − v
x0 =
.
y2
Therefore
v0
2
= y2 − ,
y
y
v 0 = y 3 − 2.
By integrating we have
y4
− 2y + 0.25c
v=
4
Finally,
4xy = y 4 − 84 + c.
Exercise 3.11.
y 0 − y = 1, y(0) = 0.
9
Solution. For the homogeneous differential equation
y0 − y = 0
we have the solution y = ex . Therefore, for nonhomogeneous equation we have
y = ex v,
y 0 = vex + ex v 0 .
Then,
ex v 0 = 1,
v 0 = e−x ,
v = c − e−x ,
and finally
y = cex − 1.
Next,
c = (y + 1)e−x .
Substituting x = 0 and y = 0, we obtain c = 1. Thus,
y = ex − 1
is the solution of the problem.
Exercise 3.12.
xy 0 + y = 2x, y(1) = 1.
Solution. Notice that the differential equation is exact. Therefore
Z x
Z y
(0 − 2u)du +
xdv = xy − x2 = c.
0
0
Putting x = 1 and y = 1 we find c = 0. Thus, finally
xy − x2 = 0
or
y = x.
Exercise 3.13.
1 1
+ .
2 e
Solution. Multiplying the left and right sides to ex we obtain
e−x y 0 + 2ex y = ex , y(0) =
y 0 + 2e2x y = e2x .
This is a linear differential equation. The homogeneous differential equation
y 0 + 2e2x y = 0
has the solution y = exp(− exp(2x)). Therefore, the solution of the nonhomogeneous differential equation has solution
2x
y = e−e v.
y 0 = −2ve−e e2x + e−e v 0 .
2x
Therefore
or
2x
e−e v 0 = e2x ,
2x
2x
2x
v 0 = e2x ee = e2x+e .
10
By integrating we have
Z
v =c+
e2x+e dx = c +
2x
1
2
Z
e+e de2x
2x
1 2x
= c + ee .
2
and
1
2
we obtain c = 1. Therefore,
2x
1
y = e−e + .
2
y = ce−e
2x
Putting x = 0, y = 0.5 + e−1
+
Exercise 3.14.
³π ´
1
= .
2
2
Solution. If x = 0 the solution of the differential equation is trivial as y = −1.
Let us assume that x 6= 0 dividing both sides of the differential equation to sin x.
Then we obtain
cos 2x
y 0 + (cot x)y =
.
sin x
The homogeneous differential equation
(sin x)y 0 + (cos x)y = − cos 2x, y
y 0 + (cot x)y = 0
has a solution (sin x)−1 . Therefore the solution of the nonhomogeneous differential
equation is represented
v
y=
,
sin x
0
(sin x)v − (cos x)v
y0 =
.
sin2 x
Next,
v0
cos 2x
=
,
sin x
sin x
or
v 0 = cos 2x.
By integrating we obtain
2v = c + sin 2x,
and
2y sin x = c + sin 2x,
c
2y =
+ 2 cos x
sin x
Now, putting y = 0.5 and x = 0.5π we obtain c = 1 and the final solution is
1
+ 2 cos x.
2y =
sin x
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