Solutions of HW7
1. The characteristic polynomial of the homogeneous equation is 2λ2 + 3λ + 1.
Its roots are (−3±1)/4, i.e., −1/2 and −1. It follows that the general solution of
the homogeneous equation is C1 e−t + C2 e−t/2 . Let us find particular solutions
of the non-homogeneous equations with the right-hand side t2 and 3 sin t using
the method of undetermined coefficients.
Let y = At2 + Bt + C, substituting it into the equation we have
2(2A) + 3(2At + B) + At2 + Bt + C = t2
so that
A = 1,
6A + B = 0,
4A + 3B + C = 0.
Then B = −6, and 4 − 18 + C = 0, hence C = 14. We get y = t2 − 6t + 14.
Consider now y = A cos t + B sin t. We need
2(−A cos t − B sin t) + 3(−A sin t + B cos t) + (A cos t + B sin t) = 3 sin t,
so that
−2B − 3A + B = 3,
−2A + 3B + A = 0,
or
−3A − B = 3,
−A + 3B = 0,
Substituting A = 3B into the first equation we get −9B − B = 3, so that
B = −0.3, and A = −0.9. We get the function y = −0.9 cos t − 0.3 sin t.
Consequently, the general solution of the equation is
y = C1 e−t + C2 e−t/2 + t2 − 6t + 14 − 0.9 cos t − 0.3 sin t.
2. The characteristic polynomial has roots λ = ±i. We are looking therefore
for a particular solution in the form y = (At + B) cos 2t + (Ct + D) sin 2t.
We have then y 0 = A cos 2t − 2(At + B) sin 2t + C sin 2t + 2(Ct + D) cos 2t =
(2Ct + 2D + A) cos 2t + (−2At − 2B + C) sin 2t. Then y 00 = 2C cos 2t − 2(2Ct +
2D + A) sin 2t − 2A sin 2t + 2(−2At − 2B + C) cos 2t = (−4At − 4B + 4C) cos 2t +
(−4Ct − 4D − 4A) sin 2t. We need then
(−3At − 3B + 4C) cos 2t + (−3Ct − 3D − 4A) sin 2t = t cos 2t + 3 sin 2t,
or
−3A = 1,
−3B + 4C = 0,
−3C = 0,
−3D − 4A = 3.
Then A = −1/3, C = 0, B = 0, −3D + 4/3 = 3, D = −5/9.
The general solution of the equation is
5
1
y = C1 cos t + C2 sin t − t cos 2t − sin 2t.
3
9
3. The general solution of the homogeneous equation is C1 cos 2t + C2 sin 2t,
therefore we have to look for a particular solution of the equation in the form y =
1
At cos 2t + Bt sin 2t. Then y 00 = −4At cos 2t − 4A sin 2t − 4Bt sin 2t + 4B cos 2t,
so that after substituting into the equation we have
−4A sin 2t + 4B cos 2t = 3 sin 2t
so that A = −3/4 and B = 0. Hence the general solution of the equation is
y = C1 cos 2t+C2 sin 2t− 43 t cos 2t. Its derivative is y 0 = −2C1 sin 2t+2C2 cos 2t−
3
3
4 cos 2t + 2 t sin 2t. Substituting the initial conditions, we get
2C2 −
C1 = 2,
3
= −1
4
hence C2 = −1/8. Hence the solution of the initial value problem is
y = 2 cos 2t −
1
3
sin 2t − cos 2t.
8
4
4. The characteristic polynomial λ2 +2λ+5 has roots −1±2i. So, the general solution of the corresponding homogeneous equation is C1 e−t cos 2t + C2 e−t sin 2t,
and we have to look for a particular solution of the non-homogeneous equation in the form y = Ate−t cos 2t + Bte−t sin 2t. Then y 0 = Ae−t cos 2t −
Ate−t cos 2t − 2Ate−t sin 2t + Be−t sin 2t − Bte−t sin 2t + 2Bte−t cos 2t = ((−A +
2B)t+A)e−t cos 2t+((−2A−B)t+B)e−t sin 2t, and y 00 = (−A+2B)e−t cos 2t−
((−A+2B)t+A)e−t cos 2t−2((−A+2B)t+A)e−t sin 2t+(−2A−B)e−t sin 2t−
((−2A − B)t + B)e−t sin 2t + 2((−2A − B)t + B)e−t cos 2t = ((−3A − 4B)t −
2A + 4B)e−t cos 2t + ((4A − 3B)t − 4A − 2B)e−t sin 2t. Substituting this into
the equation we get
((−3A − 4B)t − 2A + 4B)e−t cos 2t + ((4A − 3B)t − 4A − 2B)e−t sin 2t+
2(((−A + 2B)t + A)e−t cos 2t + ((−2A − B)t + B)e−t sin 2t)+
5(Ate−t cos 2t + Bte−t sin 2t) = 4e−t cos 2t,
or, after all cancellations:
4Be−t cos 2t − 4Ae−t sin 2t = 4e−t cos 2t
so that A = 0 and B = 1.
Thus the general solution of the equation is y = C1 e−t cos 2t + C2 e−t sin 2t +
−t
te sin 2t. Then y 0 = −C1 e−t cos 2t−2C1 e−t sin 2t−C2 e−t sin 2t+2C2 e−t cos 2t+
e−t sin 2t − te−t sin 2t + 2te−t cos 2t. Substituting the initial conditions, we get
1 = C1 ,
0 = −C1 + 2C2 ,
so that C1 = 1, C2 = 1/2, and the solution of the initial value problem is
1
y = e−t cos 2t + e−t sin 2t + te−t sin 2t.
2
5. Equation describing motion of the spring mass system without damping and
external force is my 00 + ky = 0, where w = mg = kL for displacement L of
2
the spring by the mass. We have then k = w/L = 2 lb/0.5 ft = 4 lb/ft and
m = w/g = 2 lb/32 ft/s2 . We get hence the equation
1 00
y + 4y = 0
16
with the initial conditions y(0) = 1/4 ft, y 0 (0) = 0.
1 2
The characteristic polynomial of the equation is 16
λ +4, hence the roots are
λ = ±8i, and the general solution is C1 cos 8t+C2 sin 8t. Then y 0 = −8C1 sin 8t+
8C2 cos 8t. Substituting the initial condition we get C1 = 1/4 and 8C2 = 0.
Consequently, at moment of t seconds the spring is by 41 cos 8t feet down.
6. A fundamental system of solutions of the
corresponding homogeneous equa cos t sin t = cos2 t+sin2 t = 1.
tion is {cos t, sin t}. Its Wronskian is W = − sin t cos t Then a particular solution of the non-homogeneous equation is
Z
Z
1
1
− cos t
sin t tan t dt + y2
cos t tan t dt =
W
W
Z
Z
sin2 t
dt + sin t sin t dt =
− cos t
cos t
Z
sin2 t
− cos t
d sin t − sin t cos t =
cos2 t
Z 1
1
1
− cos t
−1 +
+
d sin t − sin t cos t =
2 1 − sin t 1 + sin t
1 1 − sin t
cos t 1 − sin t
cos t sin t + ln
− sin t cos t =
ln
.
2 1 + sin t
2
1 + sin t
The general solution is then equal to the sum of this function with C1 cos t +
C2 sin t.
Solutions of HW8
1
1. The eigenvalues of the matrix are −1 and 1. An eigenvector of −1 is
.
0
−1
An eigenvector of 1 is
. Hence a fundamental matrix of the homoge1
neous system is
−t
e
−et
X(t) =
.
0
et
Its determinant (the Wronskian) is W = 1. Its inverse is
t
et
e
X −1 =
.
0 e−t
−t e
−et
,
Then a particular solution of the system is equal to u1 (t)
+u2 (t)
0
et
where
Z
Z
1 e−t −et u1 (t) =
dt = (1 + tet ) dt = t + tet − et ,
et W t
3
Z
u2 (t) =
1 e−t
W 0
Z
e−t dt
=
te−t dt = −te−t − e−t .
t It follows that the general solution of the system is
−t −t
e
−et
te + 2t
c1
+ c2
+
.
0
et
−t − 1
2. The roots of the denominator are −2 and 3. We are looking for a decomposition
A
B
3s
=
+
,
s2 − s − 6
s+2 s−3
or
3s = A(s − 3) + B(s + 2).
Substituting s = 3 we get 9 = 5B, or B = 9/5. Substituting s = −2 we get
−6 = −5A, or A = 6/5.
It follows that the inverse Laplace transform of the function is 6/5e−2t +
9/5e3t .
3. We are looking for a decomposition
4s2 − 12s + 44
A(s − 2) + 2B
C
=
+
,
− 4s + 8)(s + 3)
(s − 2)2 + 22
s+3
(s2
or
4s2 − 12s + 44 = (A(s − 2) + 2B)(s + 3) + C((s − 2)2 + 4)
Substituting s = −3 we get 36+36+44 = 29C, hence C = 116
29 = 4. Substituting
s = 2 we get 16−24+44 = 10B +4C, hence 36 = 10B +16, i.e., B = 2. Looking
at the coefficient at s2 we get 4 = A + C, therefore A = 0. Thus, our function is
1
. Its inverse Laplace transform is 2e2t sin 2t + 4e−3t .
equal to 2 · (s−2)22 +22 + 4 s+3
4. Applying Laplace transform to the equation we get (for Y equal to the
Laplace transform of the solution):
s2 Y − s + 3sY − 3 + 2Y =
hence
Y =
1
s2
s+3
1
+
.
s2 + 3s + 2 s2 (s2 + 3s + 2)
We have s2 + 3s + 2 = (s + 1)(s + 2). Let us decompose the fractions into sums
of partial fractions:
s+3
A
B
=
+
s2 + 3s + 2
s+1 s+2
or
s + 3 = A(s + 2) + B(s + 1)
We can take A = 2 and B = −1.
For the second fraction take
1
A
B
C
D
= 2+ +
+
,
s2 (s2 + 3s + 2)
s
s
s+2 s+1
4
or
1 = A(s + 2)(s + 1) + Bs(s + 2)(s + 1) + Cs2 (s + 1) + Ds2 (s + 2).
Substituting s = 0 we get 1 = 2A, hence A = 1/2. Substituting s = −1 we get
1 = D. Substituting s = −2 we get 1 = −4C, hence C = −1/4. Looking at the
coefficient at s3 we get 0 = B + C + D = B − 1/4 + 1, hence B = −3/4.
Putting everything together we get
Y =2
1
1 1 31 1 1
1
1
5 1
1 1 3 1
1
−
+ · − −
+
= 3·
−
+ · − · .
s + 1 s + 2 2 s2 4 s 4 s + 2 s + 1
s + 1 4 s + 2 2 s2 4 s
Then
5
1
3
y = 3e−t − e−2t + t − .
4
2
4
5. Applying the Laplace transform to the equation we get
s4 Y − s3 − s − Y = 0,
or
Y =
s3 + s
s(s2 + 1)
s
A
B
=
=
=
+
.
4
2
s −1
(s − 1)(s + 1)(s + 1)
(s − 1)(s + 1)
s−1 s+1
The coefficients are found then from
s = A(s + 1) + B(s − 1)
Substituting s = −1 we get −1 = −2B, hence B = 1/2. Substituting s = 1, we
get 1 = 2A, hence A = 1/2. Applying the inverse Laplace transform we get
y=
1 t 1 −t
e + e .
2
2
6. Applying the Laplace transform to the system we get
sY1 − 1 = 5Y1 − 2Y2
sY2 = 6Y1 − 2Y2
or
(5 − s)Y1 − 2Y2 = −1,
Then
6Y1 + (−2 − s)Y2 = 0.
−1
−2 0 −2 − s 2+s
=
Y1 = 2
s − 3s + 2
−2 5−s
6
−2 − s 5 − s −1 6
0 6
=
Y2 = 2 − 3s + 2
s
5
−
s
−2
6
−2 − s 5
We have s2 − 3s + 2 = (s − 1)(s − 2) and we express the solutions as
s2
2+s
A
B
=
+
− 3s + 2
s−1 s−2
s2
6
C
D
=
+
.
− 3s + 2
s−1 s−2
and
For the first expression we have 2 + s = A(s − 2) + B(s − 1). Substituting s = 1
and s = 2 we get 3 = −A and 4 = B, hence A = −3 and B = 4. For the second
expression we have 6 = C(s − 2) + D(s − 1). Substituting s = 1 and s = 2 we
get 6 = −C and 6 = D. Applying the inverse Laplace transform we get
y1 = −3et + 4e2t ,
y2 = −6et + 6e2t .
6