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Name: Solutions Mathematics University of Washington
November 13, 2019
MIDTERM 2 REVIEW SOLUTIONS
Problem 1. Determine a suitable form for Yp(t), the particular solution, if the method of undeter- mined coefficients is to be used. You don’t need to find the coefficients, just write down the form.
(a) y ′′
- 3y ′ = 2t 3
- t 2 e − 3 t
- 4 sin(3t)
Solutions:
(1) Find the solution to the homogeneous equation y′′^ + 3y′^ = 0: The characteristic polynomial is given by
r 2
Hence, the roots of the polynomial are given by r = 0 and r = −3. These are distinct real roots so the solution is y = c 1 e 0 t
- c 2 e − 3 t = c 1 + c 2 e − 3 t .
(2) Break the problem into separate differential equations:
(a) y ′′
(b) y ′′
(c) y ′′
(3) Guess the form for each of the differential equations:
For (a) Yp = t s
A 3 t 3
For (b) Yp = t s
B 2 t 2
e − 3 t
For (c) Yp = t s
C sin(3t) + D cos(3t)
(4) Determine s:
For (a):
- If s = 0, then Yp = (A 3 t 3 + A 2 t 2 + A 1 t + A 0 ) which contains a multiple of the homogeneous equation
- If s = 1, then Yp = t(A 3 t 3 + A 2 t 2 + A 1 t + A 0 ) which doesn’t contain a multiple
Therefore, Yp = t(A 3 t 3
For (b):
- If s = 0, then Yp = (B 2 t^2 + B 1 t + B 0 )e−^3 t^ which contains a multiple of the homogeneous equation
- If s = 1, then Yp = t(B 2 t^2 + B 1 t + B 0 )e−^3 t^ which doesn’t contain a multiple
Therefore, Yp = t(B 2 t 2
For (c):
- If s = 0, then Yp = C sin(3t) + D cos(3t) which is not a mul- tiple.
Therefore, Yp = C sin(3t) + D cos(3t).
The form of the particular solution is then
For (a) Yp = t 1
A 3 t 3
For (b) Yp = t 1
B 2 t 2
e − 3 t
For (c) Yp = C sin(3t) + D cos(3t)
The form of the particular solution is then
For (a) Yp = Ae −t
For (b) Yp = Be −t cos(2t) + Ce −t sin(2t)
For (c) Yp = t
D 2 t 2
e −t sin(t) +
F 2 t 2
e −t cos(t)
(c) y ′′
- 2y ′
- y = t cos(2t) + e −t sin(t) + e −t
Solutions:
(1) Find the solution to the homogeneous equation y ′′
- 2y + y = 0: The characteristic polynomial is given by
r 2
- 2r + 1 = 0 ⇒ (r + 1) 2 = 0.
Hence the roots are r = −1. This is a repeated root so the solution is
y = c 1 e
−t
−t .
(2) Break the problem into separate differential equations:
(a) y ′′
(b) y ′′
(c) y ′′
(3) Guess the form for each of the differential equations:
For (a) Yp = t s
(A 1 t + A 0 ) cos(2t) + (B 1 t + B 0 ) sin(2t)
For (b) Yp = t s
Ce −t cos(t) + De −t sin(t)
For (c) Yp = t s F e −t
(4) Determine s:
For (a):
(A 1 t + A 0 ) cos(2t) + (B 1 t + B 0 ) sin(2t)
does not contains a multiple of the homogeneous equation.
Therefore, Yp = (A 1 t + A 0 ) cos(2t) + (B 1 t + B 0 ) sin(2t)
For (b):
- If s = 0, then Yp = Ce −t cos(t)+De −t sin(t) which not contain a multiple.
Therefore, Yp = Ce −t cos(t) + De −t sin(t)
For (c):
- If s = 0, then Yp = F e −t which is a multiple of the homoge- neous so s 6 = 0
- If s = 1, Yp = tF e −t which is a multiple of the homogeneous so s 6 = 1
- If s = 0, Yp = t 2 F e −t which is not a multiple of the homoge- neous so s = 2.
Therefore, Yp = t^2 F e−t
The form of the particular solution is then
For (a) Yp = Yp = (A 1 t + A 0 ) cos(2t) + (B 1 t + B 0 ) sin(2t)
For (b) Yp = Ce −t cos(t) + De −t sin(t)
For (c) Yp = t 2 F e −t
