Download MTH 219 HOMEWORK 14 SOLUTIONS

MTH 219
HOMEWORK 14
SOLUTIONS
This homework covers solutions of nonhomogeneous linear equations with constant coefficients.
d3 y
d2 y
dy
+ 2 −6
= 10e2x + 24x − 10.
3
dx
dx
dx
Solution. The complementary equation can be written as (D3 + D2 − 6D)y = 0. Now D3 + D2 − 6D =
D(D2 + D − 6) = D(D + 3)(D − 2), so the complementary solution is yc = c1 + c2 e−3x + c3 e2x . The annihilator
of e2x , x, and a constant is D2 (D − 2). These are also factors of the differential operator of the equation,
so there are common solutions. The particular solution, yp , must be a solution of D3 (D − 2)2 (D + 3)y = 0.
When we eliminate the common functions, we get yp = Ax + Bx2 + Cxe2x .
1. Find the general solution of
yp = Ax + Bx2 + Cxe2x
yp0 = A + 2Bx + Ce2x + 2Cxe2x
yp00 = 2B + 2Ce2x + 2Ce2x + 4Cxe2x
= 2B + 4Ce2x + 4Cxe2x
yp000 = 8Ce2x + 4Ce2x + 8Cxe2x
= 12Ce2x + 8Cxe2x
Now replace y by yp in the differential equation.
12Ce2x + 8Cxe2x + 2B + 4Ce2x + 4Cxe2x − 6A − 12Bx − 6Ce2x − 12Cxe2x
= 10Ce2x − 12Bx + 2B − 6A = 10e2x + 24x − 10
By comparing coefficients we get the following equations.
10C = 10
−12B = 24
2B − 6A = −10
Therefore C = 1 and B = −2. The last equation becomes −4 − 6A = −10, so A = 1. The general solution is
y = c1 + c2 e−3x + c3 e2x + x − 2x2 + xe2x
2. Solve the following initial value problem.
y 00 − 4y 0 + 3y = 20 cos x
y(0) = 1
y 0 (0) = −9
Solution. The complementary equation can be written as (D2 −4D+3)y = 0. Now D2 −4D+3 = (D−1)(D−3),
so the complementary solution is yc = c1 ex + c2 e3x . The annihilator of cos x is D2 + 1, so there are no common
solutions. The particular solution is yp = A sin x + B cos x.
yp = A sin x + B cos x
yp0 = A cos x − B sin x
yp00 = −A sin x − B cos x
Replace y by yp in the differential equation.
−A sin x − B cos x − 4A cos x + 4B sin x + 3A sin x + 3B cos x = 20 cos x
(2A + 4B) sin x + (2B − 4A) cos x = 20 cos x
By comparing coefficients we get the following equations.
2A + 4B = 0
2B − 4A = 20
From the first equation we see that A = −2B. The second equation then becomes 2B + 8B = 20, so B = 2.
Then A = −4. The general solution is
y = c1 ex + c2 e3x − 4 sin x + 2 cos x. To solve the initial value problem we also need y 0 = c1 ex + 3c2 e3x −
4 cos x − 2 sin x. Then
y(0) = c1 + c2 + 2 = 1
y 0 (0) = c1 + 3c2 − 4 = −9
so that
c1 + c2 = −1
c1 + 3c2 = −5
Subtract the first equation from the second to get 2c2 = −4 or c2 = −2. Then
c1 = 2 − 1 = 1 and the solution of the initial value problem is
y = ex − 2e3x − 4 sin x + 2 cos x
3. Solve the following boundary value problem.
y 00 + 4y = 8 sin 2x
y(0) = 0
y(π/4) = 3
Solution. The complementary equation can be written as (D2 + 4)y = 0 so the complementary solution
is yc = c1 sin 2x + c2 cos 2x. The annihilator of sin 2x is D2 + 4, so there are common solutions. The
particular solution must be a solution of (D2 + 4)2 . When we eliminate the common functions we get
yp = Ax sin 2x + Bx cos 2x.
yp = Ax sin 2x + Bx cos 2x
yp0 = A sin 2x + 2Ax cos 2x + B cos 2x − 2Bx sin 2x
yp00 = 2A cos 2x + 2A cos 2x − 4Ax sin 2x − 2B sin 2x − 2B sin 2x − 4Bx cos 2x
= 4A cos 2x − 4B sin 2x − 4Ax sin 2x − 4Bx cos 2x
Replace y by yp in the differential equation.
4A cos 2x − 4B sin 2x − 4Ax sin 2x − 4Bx cos 2x + 4Ax sin 2x + 4Bx cos 2x = 8 sin 2x
4A cos 2x − 4B sin 2x = 8 sin 2x
By comparing coefficients we see that −4B = 8 or B = −2, and A = 0. Therefore the general solution is
y = c1 sin 2x + c2 cos 2x − 2x cos 2x.
y(0) = c2 = 0
y(π/4) = c1 = 3
The solution of the boundary value problem is y = 3 sin 2x − 2x cos 2x.