MATH/MTHE 225 - WINTER 2017
Problem Set 3
1. Solve the differential equation: 5y 0 + 30y = 25.
2. Solve the differential equation: y 0 + 2xy − 10x = 0.
3. Solve the IVP: xy 0 + (1 + 6x)y = 10xe−6x ;
y(1) = e−6 .
4. Solve the IVP: cos2 (x) sin (x)y 0 + cos3 (x)y = 2;
5. Solve the IVP: (x + 1)2 y 0 + 4(x + 1)y − 9 = 0;
6. Solve the differential equation: xy 0 + y = 6e6x .
1
y(π/4) = 0.
y(0) = 5.
Solution
1.
(a) Way 1:
Notice it is separable 1st order ODE.
We have 5y 0 + 30y = 25. Now divide both sides of the equation by 5 and get:
dy
dy
dy
+ 6y = 5 ⇒
= (5 − 6y) ⇒
= dx
dx
dx
(5 − 6y)
Z
Z
dy
ln (5 − 6y)
⇒
=
dx ⇒
= x + C (implicit solution)
(5 − 6y)
−6
y 0 + 6y = 5 ⇒
⇒ ln (5 − 6y) = −6x − 6C ⇒ 5 − 6y = e−6x−6C ⇒ y =
⇒y=
5 1 −6x−6C
− e
6 6
5 1 −6x −6C
1
5
− e e
⇒ y = + De−6x where D = − e−6C (explicit solution)
6 6
6
6
(b) Way 2:
We have 5y 0 + 30y = 25. Now divide both sides of the equation by 5 and get:
dy
+ 6 y = |{z}
5
dx |{z}
y 0 + 6y = 5 ⇒
P (x)
Q(x)
This is clearly a 1st order linear ODE.
R
So the integrating factor (I.F.) = e
P (x) dx
R
=e
6 dx
= e6x .
Now multiplying both sides of the differential equation by I.F. we get:
e
dy
+ 6y
dx
dy
d
+ 6e6x y = 5e6x ⇒
(ye6x ) = 5e6x
dx
dx
Z
Z
5
6x
6x
6x
⇒ d(ye ) = 5e dx ⇒
d(ye ) = 5e6x dx ⇒ ye6x = e6x + C (implicit solution)
6
6x
⇒ y=
= e6x (5) ⇒ e6x
5
+ Ce−6x (explicit solution)
6
2
2.
(a) Way 1:
Notice it is a separable 1st order ODE.
y 0 + 2xy − 10x = 0 ⇒ y 0 = −2xy + 10x ⇒
Z
⇒
dy
=−
y−5
⇒ y − 5 = e−x
Z
2 +C
dy
dy
= −2x(y − 5) ⇒
= −2x dx
dx
y−5
2x dx ⇒ ln (y − 5) = −x2 + C (implicit solution)
⇒ y = 5 + e−x
2 +C
2
⇒ y = 5 + e−x eC
2
⇒ y = 5 + De−x where D = eC (explicit solution)
(b) Way 2:
dy
+ 2x y = |{z}
10x
dx |{z}
We have y 0 + 2xy − 10x = 0 and we can rewrite is as:
P (x)
R
This is clearly a 1st order linear ODE. So the I.F. = e
P (x) dx
Q(x)
R
=e
2x dx
2
= ex .
Now multiplying both sides of the differential equation by I.F. we get:
2
d x2 dy
2
x2
x2
x2
+ 2xe y = 10xe ⇒
ye
= 10xe ⇒ d yex = 10xex dx
e
dx
dx
Z Z
2
2
2
2
⇒
d yex = 10xex dx ⇒ yex = 5ex + C (implicit solution)
x2
2
⇒ y = 5 + Ce−x (implicit solution)
Z
• For
Z
So
2
10xex dx we need to use u-substitution with u = x2 . That means du = 2x dx.
x2
10xe dx = 5
Z
x2
2xe dx = 5
Z
eu du = 5eu = 5ex
2
3.
We have xy 0 + (1 + 6x)y = 10xe−6x and so dividing both sides by x we get:
dy 1 + 6x
dy
1
−6x
+
y = 10e
⇒
+
+ 6 y = |10e{z−6x}
dx
x
dx
x
Q(x)
| {z }
P (x)
3
This is clearly a 1st order linear ODE.
1
So the I.F. = e ( x +6) dx = eln x+6x = eln x e6x = xe6x .
R
Now multiplying both sides of the differential equation by I.F. we get:
1
dy
6x
6x dy
+ xe
+ 6 y = 10x ⇒ xe6x
+ e6x + 6xe6x y = 10x
xe
dx
x
dx
Z
Z
⇒ d xe6x y = 10x dx ⇒
d xe6x y = 10x dx
⇒ xe6x y = 5x2 + C (implicit solution)
1
⇒ y = 5xe−6x + e−6x C (explicit solution)
x
But we need: y(1) = e−6 . That means e−6 = 5e−6 + e−6 C ⇒ 1 = 5 + C ⇒ C = −4.
4
Therefore the solution of the IVP is: y = 5xe−6x − e−6x .
x
4.
We have cos2 (x) sin (x)y 0 + cos3 (x)y = 2.
So dividing both sides by cos2 (x) sin (x) we get:
y0 +
2
2
dy cos (x)
cos (x)
y=
⇒
+
y=
2
sin (x)
sin (x)
dx sin (x)
cos (x) sin (x)
| {z }
|
{z
}
P (x)
Q(x)
So clearly this is a 1st order linear ODE.
R
So the I.F. = e
P (x) dx
R
=e
cos (x)
sin (x)
dx
= eln (sin (x)) = sin (x).
Now multiplying both sides of the differential equation by I.F. we get:
2
d
dy
⇒
[sin (x) y] = 2 sec2 (x) ⇒ d[sin (x) y] = 2 sec2 (x) dx
sin (x) +cos (x)y =
2
dx
cos (x)
dx
Z
Z
⇒
d[sin (x) y] = 2 sec2 (x) dx ⇒ sin (x) y = 2 tan (x) + C (implicit solution)
⇒ y=
2 tan (x) + C
(explicit solution)
sin (x)
But we need: y(π/4) = 0. That means 0 =
4
2 · tan (π/4) + C
2+C
⇒ 0 = √ ⇒ C = −2
sin (π/4)
1/ 2
2 tan (x) − 2
Therefore the solution of the IVP is: y =
sin (x)
Z
cos x
• For
dx we need to use u-substitution with u = sin x. That means du = cos x dx.
sin x
Z
Z
cos x
du
Therefore
dx =
= ln u = ln (sin x)
sin x
u
5.
We have (x + 1)2 y 0 + 4(x + 1)y − 9 = 0.
So dividing bot sides by (x + 1)2 we get:
y0 +
4
9
4
dy
y = 9(x + 1)−2
y=
+
⇒
2
| {z }
x+1
(x + 1)
dx |x {z
+ 1}
Q(x)
P (x)
R
So the I.F. = e
P (x) dx
R
=e
4
x+1
dx
= e4 ln (x+1) = (eln (x+1) )4 = (x + 1)4
Now multiplying both sides of the differential equation by I.F. we get:
d[(x + 1)4 y]
dy
+ 4(x + 1)3 y = 9(x + 1)2 ⇒
= 9(x + 1)2
dx
dx
Z
Z
4
2
4
⇒ d[(x + 1) y] = 9(x + 1) dx ⇒
d[(x + 1) y] = 9(x + 1)2 dx
(x + 1)4
⇒ (x + 1)4 y = 9
⇒ y=
(x + 1)3
+ C ⇒ (x + 1)4 y = 3(x + 1)3 + C (implicit solution)
3
3
C
+
(explicit solution)
x + 1 (x + 1)4
But we need y(0) = 5. That means 5 = 3 + C ⇒ C = 2.
Therefore the solution of the IVP is: y =
3
2
+
x + 1 (x + 1)4
6.
We have xy 0 + y = 6e6x
So dividing both sides by x we get:
1
6
dy
1
6
y 0 + y = e6x ⇒
+
y = e6x
x
x
dx |{z}
x
|x{z }
P (x)
R
R
1
Q(x)
So the I.F. = e P (x) dx = e x dx = eln x = x.
Now multiplying both sides of the differential equation by I.F. we get:
5
dy
d
x
+ y = 6e6x ⇒
(xy) = 6e6x ⇒ d(xy) = 6e6x dx ⇒
dx
dx
⇒ xy = e6x + C (implicit solution)
⇒ y=
e6x + C
(explicit solution)
x
6
Z
Z
d(xy) =
6e6x dx