Download Homework 7 Solutions

Math 3213 - Differential Equations
Homework #7 - 2007.03.05
Due Date - 2007.03.12
Solutions
Consider the differential equation
xy 00 − (x + N )y 0 + N y = 0.
1. Verify that one solution is given by y1 (x) = ex .
This one is much too obvious...
2. By setting y2 (x) = c2 v(x)y1 (x), derive a new differential equation for v(x).
So we have
y2 (x) = c2 v(x)ex ,
y20 (x) = c2 (v(x)ex + v 0 (x)ex ) ,
y200 (x) = c2 (v(x)ex + 2v 0 (x)ex + v 00 (x)ex ) ,
and plugging this into xy 00 − (x + N )y 0 + N y = 0 gives
xv 00 + (x − N )v 0 = 0.
3. Solve the equation found in problem 2 and show that
Z
v(x) = xN e−x dx.
By setting w = v 0 , we get
xw0 + (x − N )w = 0,
which is separable or can be solved by an integrating factor. We will use the
separable approach:
N −x N
dw
=
=
− 1.
w
x
x
Solving this yields
Z
N −x
w(x) = x e ⇒ v(x) = xN e−x dx.
1
4. Find a closed form for the v(x) found in problem 3.
We can use tabular integration here to get
v(x) = − xN e−x − N xN −1 e−x − N (N − 1)xN −2 e−x
− N (N − 1)(N − 2)xN −3 e−x − . . . − N (N − 1)(N − 2) · · · 2xe−x − N !e−x .
5. If c2 = − N1 ! , what function is y2 (x) the N th Maclaurin series the approximation of?
Setting y2 (x) = − N1 ! ex v(x) gives
xN
x2 x3
+
+ ... +
.
y2 (x) = 1 + x +
2
3!
N!
Notice that this is the N th Maclaurin series of y1 (x).
Solve the following differential equations:
6. y 00 + 9y = x cos(x)
The homogeneous solution is given by yh (x) = c1 cos(3x) + c2 sin(3x). To find the
particular solution, we guess
yp (x) = (A + Bx) cos(x) + (C + Dx) sin(x),
which gives
yp0 (x) = (B + C + Dx) cos(x) + (D − A − Bx) sin(x),
yp00 (x) = (2D − A − Bx) cos(x) − (2B + C + Dx) sin(x).
Plugging this into y 00 +9y = x cos(x) gives the system of equations for the coefficients
to be


D
=0



2D + 8A = 0

8B
=1



8C − 2B = 0
which means that B =
1
8
and C =
1
32
so
1
1
yp (x) = x cos(x) +
sin(x).
8
32
So our final solution is given by
1
1
y(x) = c1 cos(3x) + c2 sin(3x) + x cos(x) +
sin(x).
8
32
2
7. y 00 − 4y 0 + 3y = 1
The homogeneous solution is given by yh (x) = c1 ex + c2 e3x . To find the particular
solution, we guess
yp (x) = ax + b,
which gives
yp0 (x) = a
yp00 (x) = 0.
Plugging this into y 00 −4y 0 +3y = 1 gives the system of equations for the coefficients
to be
(
−4a + 3b = 1
3a
=0
which means that a = 0 and b =
1
3
so
1
yp (x) = .
3
So our final solution is given by
1
y(x) = c1 ex + c2 e3x + .
3
8. y 00 − y = 4xex
The homogeneous solution is given by yh (x) = c1 ex + c2 e−x . To find the particular
solution, we guess
yp (x) = ax2 + bx + c ex ,
which gives
yp0 (x) = ax2 + (2a + b)x + (b + c) ex
yp00 (x) = ax2 + (4a + b)x + (2a + 2b + c) ex .
Plugging this into y 00 − y = 4xex gives the system of equations for the coefficients
to be
(
2a + 2b = 0
4a
=4
which means that a = 1, b = −1 and c = c so
yp (x) = x2 − x + c ex .
3
So our final solution is given by
y(x) = c1 ex + c2 e−x + x2 − x + c ex = c1 ex + c2 e−x + x2 − x ex .
9. y 00 + 4y = cos(2x) + cos(4x)
The homogeneous solution is given by yh (x) = c1 cos(2x) + c2 sin(2x). To find the
particular solution, we guess
yp (x) = (A + Bx) cos(2x) + (C + Dx) sin(2x) + E cos(4x),
which gives
yp0 (x) = (2(C + Dx) + B) cos(2x) + (−2(A + Bx) + D) sin(2x) − 4E sin(4x)
yp00 (x) = (2D − 4A − 4Bx + 2D) cos(2x) + (−4B − 4C − 4Dx) sin(2x) − 16 cos(4x).
Plugging this into y 00 + 4y = cos(2x) + cos(4x) gives the system of equations for
the coefficients to be


4D
=1



−4B
=0

−12A = 1



−12E = 1
1
1
which means that E = − 12
, D = 14 , B = 0, C = C and A = − 12
so
1
1
yp (x) = − cos(2x) + C + x sin(2x).
12
4
So our final solution is given by
1
1
1
y(x) =c1 cos(2x) + c2 sin(2x) −
cos(2x) + C + x sin(2x) −
cos(4x)
12
4
12
1
1
1
= c1 cos(2x) + c2 sin(2x) −
cos(2x) + x sin(2x) −
cos(4x)
12
4
12
4