Ordinary Differential Equations, Spring 2015
Solutions to Homework 6
Maximal grade for HW6: 100 points
Section 3.5: 3. Solve the equation y 00 − y 0 − 2y = −2t + 4t2 .
Solution: The characteristic equation has the form:
r2 − r − 2 = (r + 1)(r − 2) = 0,
its roots are r1 = −1 and r2 = 2. Therefore 0 is not a root, and we can
find a specific solution of the nonhomogeneous equation in the form y(t) =
A + Bt + Ct2 . We have:
y 0 = B + 2Ct, y 00 = 2C,
so
y 00 − y 0 − 2y = −2Ct2 − (2C + 2B)t + (2C − B − 2A) = −2t + 4t2 ,
and
−2C = 4, 2B + 2C = 2, 2C − B − 2A = 0 ⇒ C = −2, B = 3, A = −7/2.
Finally,
7
y(t) = − + 3t − 2t2 + C1 e−t + C2 e2t .
2
18. Solve the initial value problem:
y 00 − 2y 0 − 3y = 3te2t , y(0) = 1, y 0 (0) = 0.
Solution: The characteristic equation has the form:
r2 − 2r − 3 = (r + 1)(r − 3) = 0,
its roots are r1 = −1 and r2 = 3. Therefore 2 is not a root, and we can
find a specific solution of the nonhomogeneous equation in the form y(t) =
(A + Bt)e2t . We have:
y 0 (t) = Be2t +2(A+Bt)e2t , y 00 (t) = 2Be2t +2Be2t +4(A+Bt)e2t = 4(A+B+Bt)e2t ,
1
so
y 00 −2y 0 −3y = (4A+4B+4Bt−4A−4Bt−2B−3A−3Bt)e2t = (2B−3A−3Bt)e2t ,
and
−3B = 3, 2B − 3A = 0 ⇒ A = −2/3, B = −1.
We conclude that the general solution has the form
2
y(t) = −( + t)e2t + C1 e−t + C2 e3t .
3
Finally,
2
4
y(0) = − + C1 + C2 = 1, y 0 (0) = − − 1 − C1 + 3C2 = 0,
3
3
so
5
7
2
C1 + C2 = , −C1 + 3C2 =
⇒ C1 = , C2 = 1.
3
3
3
and
2
2
y(t) = −( + t)e2t + e−t + e3t .
3
3
19. Solve the initial value problem:
y 00 + 4y = 3 sin 2t, y(0) = 2, y 0 (0) = −1.
Solution: The characteristic equation has the form r2 + 4 = 0, so r1 = 2i
and r2 = −2i. The general solution of the homogeneous equation y 00 + 4y = 0
can be written as y(t) = C1 cos 2t + C2 sin 2t. Since the argument in the right
hand side coincides with the argument of the general solution (that is, 2i
coincides with the root of the characteristic equation), we need to look for a
solution of the nonhomogeneous equation in the form
y(t) = At sin(2t) + Bt cos(2t).
We have:
y 0 (t) = A sin(2t) + 2At cos(2t) + B cos(2t) − 2Bt sin(2t),
y 00 (t) = 2A cos(2t)+2A cos(2t)−4At sin(2t)−2B sin(2t)−2B sin(2t)−4Bt cos(2t),
2
so
y 00 (t) + 2y(t) = 4A cos(2t) − 4B sin(2t).
Therefore A = 0 and B = − 34 , hence the general solution of the nonhomogeneous equation has the form
3
y(t) = − t cos(2t) + C1 cos 2t + C2 sin 2t.
4
Now
3
3
y 0 (t) = − cos(2t) + t sin(2t) − 2C1 sin(2t) + 2C2 cos(2t),
4
2
3
1
y(0) = C1 = 2, y 0 (0) = − + 2C2 = −1 ⇒ C2 = − .
4
8
Finally,
1
3
y(t) = − t cos(2t) + 2 cos 2t − sin 2t.
4
8
Section 3.8: 15. Solve the initial value problem
y 00 + y = F (t), y(0) = y 0 (0) = 0,
where


0≤t≤π
F0 t,
F (t) = F0 (2π − t), π < t ≤ 2π


0,
t > 2π.
Solution: The characteristic equation is r2 + 1 = 0, with roots at ±i,
so the general solution of the homogeneous equation has the form y(t) =
C1 cos t + C2 sin t. Note that for a linear function F (t) we can take F (t) itself
as a particular solution of the nonhomogeneous equation y 00 + y = F (t). Let
us solve the equation separately on each interval:
1) On [0, π] the general solution equals y(t) = F0 t+C1 cos t+C2 sin t. Since
0
y (t) = F0 − C1 sin t + C2 cos t, we have y(0) = C1 = 0, y 0 (0) = F0 + C2 = 0,
so C2 = −F0 and y(t) = F0 t − F0 sin t. Note that y(π) = F0 π, y 0 (π) = 2F0 .
2) On [π, 2π] the general solution equals y(t) = F0 (2π − t) + D1 cos t +
D2 sin t. Since y 0 (t) = −F0 − D1 sin t + D2 cos t, we have y(π) = F0 π − D1 =
F0 π, y 0 (π) = −F0 −D2 = 2F0 , so D2 = −3F0 and y(t) = F0 (2π −t)−3F0 sin t.
Note that y(2π) = 0 and y 0 (2π) = −4F0 .
3
3) On [2π, +∞) the general solution equals y(t) = E1 cos t+E2 sin t. Since
y (t) = −E1 sin t + E2 cos t, we have y(2π) = E1 = 0, y 0 (2π) = E2 = −4F0 , so
y(t) = −4F0 sin t.
Combining these three intervals, we get


0≤t≤π
F0 t − F0 sin t,
y(t) = F0 (2π − t) − 3 sin t, π < t ≤ 2π


−4 sin t,
t > 2π.
0
18(a). Solve the initial value problem for ω 6= 1:
y 00 + y = 3 cos ωt, y(0) = 0, y 0 (0) = 0.
Solution: The characteristic equation is r2 + 1 = 0, with roots at ±i,
so the general solution of the homogeneous equation has the form y(t) =
C1 cos t + C2 sin t. Since ω 6= 1, we can take y = A cos ωt + B sin ωt, then
y 0 (t) = −Aω sin ωt + Bω cos ωt, y 00 (t) = −Aω 2 cos ωt + Bω 2 sin ωt,
so
y 00 + y = A(1 − ω 2 ) cos ωt + B(1 − ω 2 ) sin ωt.
Since ω > 0, ω 6= 1, we get 1 − ω 2 6= 0, so
A = 0, B =
3
3
⇒ y(t) =
cos ωt + C1 cos t + C2 sin t.
2
1−ω
1 − ω2
Now
y(0) =
3
+ C1 = 0, y 0 (0) = C2 = 0,
1 − ω2
so
y(t) =
3
(cos ωt − cos t).
1 − ω2
19(a). Solve the initial value problem for ω 6= 1:
y 00 + y = 3 cos ωt, y(0) = 1, y 0 (0) = 1.
Solution: In previous problem, we checked that the general solution has
the form
3
y(t) =
cos ωt + C1 cos t + C2 sin t.
1 − ω2
4
Since
y(0) =
3
+ C1 = 1, y 0 (0) = C2 = 1,
1 − ω2
we get
y(t) =
3
3
cos ωt + (1 −
) cos t + sin t = .
2
1−ω
1 − ω2
3
(cos ωt − cos t) + (cos t + sin t).
1 − ω2
5