Undetermined Coefficients Practice Problems
1. (3.5 #5) Find the general solution: y 00 + 9y = t2 e3t + 6
First we solve the homogeneous DE:
y 00 + 9y
r2 + 9
r
yc
=
=
=
=
0
0
±3i
C1 cos(3t) + C2 sin(3t)
Now solve y 00 +9y = t2 e3t . The guess for the particular solution is yp1 = (At2 +Bt+C)e3t .
No term in the guess solves the homogeneous DE, so we can proceed.
yp1 = (At2 + Bt + C)e3t
yp0 1 = (2At + B)e3t + 3(At2 + Bt + C)e3t
yp001 = (2A)e3t + 6(2At + B)e3t + 9(At2 + Bt + C)e3t
Plug into the DE:
(2A)e3t + 6(2At + B)e3t + 9(At2 + Bt + C)e3t + 9(At2 + Bt + C)e3t
18At2 e3t + (12A + 18B)te3t + (2A + 6B + 18C)e3t
18A
12A + 18B
2A + 6B + 18C
A
B
C
=
=
=
=
=
=
=
=
t2 e3t
t2 e3t
1
0
0
1/18
−6/162
1/162
Then yp1 = ((1/18)t2 − (6/162)t + (1/162))e3t .
Now solve y 00 + 9y = 6. The guess for the particular solution is yp2 = A. This does not
solve the homogeneous DE, so we can continue.
yp2
yp0 2
0 + 9A
A
=
=
=
=
A
0 = yp002
6
2/3
Undetermined Coefficients Practice Problems
Then yp2 = 2/3.
Finally, the general solution to y 00 + 9y = t2 e3t + 6 is
y = C1 cos(3t) + C2 sin(3t) + ((1/18)t2 − (6/162)t + (1/162))e3t + (2/3)
2. (3.5 #9) Find the general solution: u00 + a2 u = cos(wt) where a2 6= w2 .
First solve u00 + a2 u = 0.
r 2 + a2 = 0
r = ±ai
yc = C1 cos(at) + C2 sin(at)
Now find a particular solution to u00 +a2 u = cos(wt). The guess for yp is yp = A cos(wt)+
B sin(wt). Since a 6= ±w, no term in this guess solves the homogeneous DE, and we can
continue.
yp = A cos(wt) + B sin(wt)
yp0 = −wA sin(wt) + wB cos(wt)
yp00 = −w2 A cos(wt) − w2 B sin(wt)
Plugging into the DE yields:
−Aw2 cos(wt) − Bw2 sin(wt) + a2 A cos(wt) + a2 B sin(wt) = cos(wt)
Thus
−w2 A + a2 A
−w2 B + a2 B
A
B
=
=
=
=
1
0
1/(a2 − w2 )
0
Then the particular solution is:
yp =
a2
1
cos(wt)
− w2
and the general solution is:
y = C1 cos(at) + C2 sin(at) +
a2
1
cos(wt).
− w2
Undetermined Coefficients Practice Problems
3. (3.5 #16) Solve the initial value problem:
y 00 − 2y 0 − 3y = 3te2t
y 0 (0) = 0.
y(0) = 1,
y 00 − 2y 0 − 3y
r2 − 2r − 3
(r − 3)(r + 1)
r
yc
=
=
=
=
=
0
0
0
−1, 3
C1 e−t + C2 e3t
The guess foryp is yp = (At + B)e2t . Then we have
yp0 = (A)e2t + 2(At + B)e2t
yp00 = 4Ae2t + 4(At + B)e2t
Plug in:
4Ae2t + 4(At + B)e2t − 2(A)e2t − 4(At + B)e2t − 3(At + B)e2t = 3te2t
Then we have:
−3Ate2t + (4A − 3B)e2t
−3A
2A − 3B
A
B
=
=
=
=
=
3te2t
3
0
−1
−2/3
Thus yp = (−t − 2/3)e2t and the general solution is y = C1 e−t + C2 e3t + (−t − 2/3)e2t .
Finally, we need to apply the initial conditions: We have
y = C1 e−t + C2 e3t + (−t − 2/3)e2t
and
y 0 = −C1 e−t + 3C2 e3t − e2t + 2(−t − 2/3)e2t ,
so the initial conditions give us:
1
0
C1
C2
=
=
=
=
C1 + C2 − 2/3
−C1 + 3C2 − 1 − 4/3
2/3
1
Undetermined Coefficients Practice Problems
Then the solution is y = (2/3)e−t + e3t + (−t − 2/3)e2t .
For each of the following, determine the form of the general solution, but do not solve
for the coefficients.
Example:
y 00 + y = t + sin(t)
The general solution has the form y = C1 cos(t)+C2 sin(t)+A1 t+A2 +B1 sin(t)+B2 cos(t).
4. y 00 + 3y 0 = 4t3 + tet
r2 + 3r = 0
r = 0, −3
yc = C1 + C2 e−3t
The particular solution for y 00 + 3y 0 = 4t3 has the form A1 t3 + A2 t2 + A3 t + A4 , but the
constant term A4 is a solution of the homogeneous DE, so we must multiply by t. Then
we get A1 t4 + A2 t3 + A3 t2 + A4 t.
The particular solution for y 00 + 3y 0 = tet has the form (B1 t + B2 )et .
All together, the general solution has the form
y = C1 + C2 e−3t + A1 t4 + A2 t3 + A3 t2 + A4 t + (B1 t + B2 )et .
5. (3.5 #21) y 00 − 5y 0 + 6y = et cos(2t) + e2t (3t + 4) sin(t)
r2 − 5r + 6
(r − 2)(r − 3)
r
yc
=
=
=
=
0
0
2, 3
C1 e2t + C2 e3t
The particular solution to y 00 −5y 0 +6y = et cos(2t) has the form A1 et cos(2t)+A2 et sin(2t).
The particular solution to y 00 − 5y 0 + 6y = e2t (3t + 4) sin(t) has the form (B1 t +
B2 )e2t cos(t) + (B3 t + B4 )e2t sin(t).
The general solution has the form:
y = C1 e2t + C2 e3t + A1 et cos(2t) + A2 et sin(2t) + (B1 t + B2 )e2t cos(t) + (B3 t + B4 )e2t sin(t)