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Math 331: Differential Equations Solutions of Final Practice Exam (Corrected version) Department of Mathematics and Statistics University of Massachusetts
Question 1: (a) Separable.
(3y − 1)dy = (4 + sinh(2x))dx =⇒
y^2 − y = 4x +
cosh 2x 2
+ C
=⇒ (y −
)^2 =
x +
cosh(2x) 3
+ C
=⇒ y =
x + cosh(2x) 3
+ C.
(b) Linear.
t^3 y′^ + 2t^2 y = te−^2 t^ =⇒ y′^ + 2t−^1 y = t−^2 e−^2 t
=⇒
d dt
t^2 y
= e−^2 t^ (Integrating factor: μ(t) = t^2 )
=⇒ t^2 y = −
e−^2 t 2
+ C
=⇒ y = −
t−^2 2 e−^2 t^ + Ct−^2.
(c) Exact since My = cos y cosh x = Nx. There exists ψ(x, y) such that ψx = M and ψy = N.
ψx = M =⇒ ψ(x, y) =
M dx + h(y) = sin y sinh x dx + h(y)
ψy = N =⇒ cos y sinh x + h′(y) = sinh x cos y =⇒ h′(y) = 0 =⇒ h(y) = c
Thus, ψ(x, y) = sin y sin hx and the implicit solution is sin y sinh x = C, i.e. y = arcsin(Ccosechx).
(d) Homogeneous. Let v = y/x. Then
y′^ =
2 xy + 3y^2 2 xy + x^2
⇐⇒ xv′^ + v =
2 v + 3v^2 2 v + 1
=⇒ xv′^ =
v^2 + v 2 v + 1 =⇒
2 v + 1 v^2 + v
dv =
dx x =⇒ ln(v^2 + v) = ln(x) + ln C =⇒ v^2 + v = Cx
=⇒ v = −
1 + Cx 2
=⇒ y = − x 2
x
1 + Cx 2
Question 2: (a) Homogeneous: The characteristic eq. is λ^2 − λ − 2 = 0. The roots are λ = 2 and λ = −1. Thus, the general sol of the homogenous eq is yh(t) = C 1 e^2 t^ + C 2 e−t.
Particular solution: We use method of variation of parameters to get
Yp(t) =
∫ (^) t
0
e^2 se−t^ − e−se^2 t W [e^2 s, e−s] sinh 2s ds
∫ (^) t
0
e^2 se−t^ − e−se^2 t − 3 es
e^2 s^ − e−^2 s 2
ds
e^2 t 6
∫ (^) t
0
e−^2 s(e^2 s^ − e−^2 s) ds −
e−t 6
∫ (^) t
0
es(e^2 s^ − e−^2 s) ds
e^2 t 6
t +
e−^4 t 4
e−t 6
e^3 t 3
t 6
e^2 t^ −
e−^2 t^ − e^2 t 72
e−t
Thus, the general solution is y(t) = yh(t) + Yp(t) = C 1 e^2 t^ + C 2 e−t^ +
te^2 t^ −
e−^2 t^.
(b) Homogeneous: The characteristic eq. is λ^2 + 2λ + 5 = 0. The roots are λ = −1 + 2i and λ = − 1 − 2 i. Thus, the general sol of the homogenous eq is yh(t) = C 1 e−t^ cos 2t + C 2 e−t^ sin 2t.
Particular solution: Let YP (t) = at^2 + bt + c + A cos t + B sin t. Plugging it into the equation yields
t^2 + 3 sin t = Y
′′ p + 2Y^
′ p + 5Yp = 2 a + 4 A cos (t) + 4 B sin (t) + 4 at + 2 b − 2 A sin (t) + 2 B cos (t) + 5 at^2 + 5 bt + 5 c = (4 B − 2 A) sin (t) + (4 A + 2 B) cos (t) + 5 at^2 + (4 a + 5 b) t + 2 a + 2 b + 5 c
We obtain the system:
4 B − 2 A = 1
4 A + 2B = 0
5 a = 1 4 a + 5b = 0 2 a + 2b + 5c = 0
=⇒ A = −
, B =
, a =
, b = −
, c = −
Thus, Yp(t) = t
2 5 −^
4 t 25 −^
2 125 −^
cos t 10 +^
sin t
- The general solution is
y(t) = yh(t) + YP (t) = C 1 e−t^ cos 2t + C 2 e−t^ sin 2t +
t^2 5
4 t 25
cos t 10
sin t 5
Question 3: Homogeneous: The characteristic eq is λ^2 + 6λ + 13 = 0 whose solutions are λ = − 3 ± 2 i. Thus,
xh(t) = c 1 e−^3 t^ cos 2t + c 2 e−^3 t^ sin 2t
Particular: We use the MUC. Let Xp(t) = A cos 2t + B sin 2t. We get
sin 2t − cos 2t = X
′′ p + 6X
′ p + 13Xp = 9 A cos (2 t) + 9 B sin (2 t) − 12 A sin (2 t) + 12 B cos (2 t) = (9 B − 12 A) sin (2 t) + (9 A + 12 B) cos (2 t)
We obtain the system: { 9 A + 12B = − 1 − 12 A + 9B = 1
=⇒ A = −
, B = −
x
2t
Question 4: (a) First note that
g(t) = (sin t) uπ (t) + (t + 1)(1 − uπ (t) = − sin(t − π) uπ (t) + t + 1 − (t − π)uπ (t) − (1 + π) uπ (t)
We take the Laplace transform on both sides of the ODE to get:
2 s^2 Y (s) + 2sY (s) + Y (s) = L{g(t)}(s)
(2s^2 + 2s + 1)Y (s) = −
e−πs s^2 + 1
s^2
s
e−πs s^2
− πe−πs
Y (s) = −
e−πs (2s^2 + 2s + 1)(s^2 + 1)
s^2 (2s^2 + 2s + 1)
s(2s^2 + 2s + 1)
− e−πs s^2 (2s^2 + 2s + 1)
(1 + π)e−πs s(2s^2 + 2s + 1) We use the partial fraction decomposition method to simplify the right hand side: 1 (2s^2 + 2s + 1)(s^2 + 1)
As + B 2 s^2 + 2s + 1
Cs + D s^2 + 1
[
4 s + 6 2 s^2 + 2s + 1
2 s + 1 s^2 + 1
]
s^2 (2s^2 + 2s + 1)
A
s
B
s^2
Cs + D 2 s^2 + 2s + 1
s
s^2
4 s + 6 2 s^2 + 2s + 1 1 s(2s^2 + 2s + 1)
A
s
Bs + C 2 s^2 + 2s + 1
s
2 s + 2 2 s^2 + 2s + 1 Next, we use the fact that 2s^2 + 2s + 1 = 2((s + 1/2)^2 + 1/4) to compute the inverse Laplace transform to recover y(t).
L−^1
e−πs (2s^2 + 2s + 1)(s^2 + 1)
L−^1
(2s + 3)e−πs (s + 1/2)^2 + 1/ 4
L−^1
(2s + 1)e−πs s^2 + 1
L−^1
(2(s + 1/2) + 2)e−πs (s + 1/2)^2 + 1/ 4
L−^1
(2s + 1)e−πs s^2 + 1
uπ (t)e−(t−π)/^2 cos(t/ 2 − π/2) +
uπ (t)e−(t−π)/^2 sin(t/ 2 − π/2)
−
uπ (t) cos(t − π) −
uπ (t) sin(t − π)
= uπ (t)
e−(t−π)/^2
sin t 2
cos t 2
cos t +
sin t
L−^1
s^2 (2s^2 + 2s + 1)
= L−^1
s
+ L−^1
s^2
− L−^1
4 s + 6 2 s^2 + 2s + 1
= L−^1
s
+ L−^1
s^2
− 2 L−^1
2(s + 1/2) + 2 (s + 1/2)^2 + 1/ 4
= 2 + t − 2 e−t/^2 cos(t/2) − 8 e−t/^2 sin(t/2)
L−^1
s(2s^2 + 2s + 1)
= L−^1
s
− L−^1
2 s + 2 2 s^2 + 2s + 1
= L−^1
s
− L−^1
(s + 1/2) + 1/ 2 (s + 1/2)^2 + 1/ 4
= 1 − e−t/^2 cos(t/2) − e−t/^2 sin(t/2)
L−^1
e−πs s^2 (2s^2 + 2s + 1)
= L−^1
2 e−πs s
+ L−^1
e−πs s^2
− L−^1
(4s + 6)e−πs 2 s^2 + 2s + 1
= L−^1
2 e−πs s
+ L−^1
e−πs s^2
− 2 L−^1
(2(s + 1/2) + 2)e−πs (s + 1/2)^2 + 1/ 4
= uπ (t)
2 + (t − π) − 2 e−(t−π)/^2 cos((t − π)/2) − 8 e−(t−π)/^2 sin((t − π)/2)
= uπ (t)
2 + t − π − 2 e−(t−π)/^2 (sin(t/2) − 4 cos(t/2))
L−^1
e−πs s(2s^2 + 2s + 1)
= L−^1
e−πs s
− L−^1
(2s + 2)e−πs 2 s^2 + 2s + 1
= L−^1
e−πs s
− L−^1
(s + 1/2)e−πs^ + 1/ 2 e−πs (s + 1/2)^2 + 1/ 4
= uπ (t) − uπ (t)
e−(t−π)/^2 cos(t/ 2 − π/2) − e−(t−π)/^2 sin(t/ 2 − π/2)
= uπ (t) − uπ (t)e−(t−π)/^2 (sin(t/2) + cos(t/2))
Combining the last five results, we finally deduce that
y(t) = 3 + t − 3 e−t/^2 (cos(t/2) + 3 sin(t/2))
−uπ (t)
3 + t − π +
cos t +
sin t − e−(t−π)/^2
sin(t/2) −
− π
cos(t/2)
(b) Here, we must use convolution integral. Taking the Laplace transform on both sides, we get
s^2 Y (s) + 4Y (s) = L{sinh t cosh t}(s) (s^2 + 4)Y (s) = L{sinh t cosh t}(s)
Y (s) =
s^2 + 4 · L{sinh t cosh t}(s)
= L{sin(2t)} · L{sinh t cosh t} = L{sin(2t) ∗ sinh t cosh t}
Thus, y(t) = sin(2t) ∗ sinh t cosh t =
∫ (^) t 0 sin(2t^ −^2 τ^ ) sinh^ τ^ cosh^ τ dτ^.
Question 5: (a) In matrix form, the system is
−→ X ′(t) = A
X (t)
where
X (t) =
x 1 (t) x 2 (t)
and A =
The eigenvalues and eigenvectors of A are:
λ 1 = − 1 /2 + i and
ξ 1 =
−i 1
and
λ 2 = λ 1 = − 1 / 2 − i and
ξ 2 =
ξ 1 =
i 1
~c = A−^1 ~b − A−^1
Thus, we obtain the particular solution
−→ Xp(t) = t^2
The general solution is then
−→ X (t) =
Xh(t) +
Xp(t) = c 1 e^2 t
From the initial condition
X (0) =
0 , we get:
c 1
2 c 1 + c 2 + 1/2 = 0 c 1 + 2c 2 + 1/4 = 0
=⇒ c 1 = − 1 / 4 , c 2 = 0.
The final solution is then
−→ X (t) = −
e^2 t
