1. We can write it 4y 3 dy = (x3 + x) dx. Integrating it, we get y 4 = x4 /4 +
x2 /2 + C.
2. The characteristic polynomial is λ2 − 2λ − 3, hence the eigenvalues are
λ = −1, 3.
(
)
−1 1
Subtracting −1 from the diagonal, we get
, hence an eigenvector
−5 5
is (1, 1).
(
)
−5 1
Subtracting 3 from the diagonal, we get
, hence an eigenvector
−5 1
is (1, 5).
( )
(
)
1
1
It follows that the general solution is c1 e−t
+ c2 e3t
. Substitut1
5
ing the initial value, we get
{
c1 + c2 = 1
c1 + 5c2 = 1
Subtracting the first equation from the second, we get 4c2 = 0, hence c2 = 0
and c1 = 1.
15
3. The characteristic polynomial is λ2 + 12 λ − 16
+ 2 = λ2 + 12 λ + 17
16 , hence the
√
1
eigenvalues are (− 21 ± 14 − 17
4 )/2 = − 4 ± i.
(
)
1−i
−2
1
Subtracting the eigenvalue − 4 +i from the diagonal we get
,
1
−1 − i
(
)
1+i
hence an eigenvector is
. It follows that a complex solution is
1
(
(− 14 +i)t
e
1+i
1
)
− 14 t
=e
((
(cos t + i sin t)
1
1
)
(
+i
1
0
))
.
Its real and imaginary parts are
(
(
)
(
))
1
1
− 14 t
e
cos t
− sin t
,
1
0
and
(
(
)
(
))
1
1
1
e− 4 t cos t
+ sin t
.
0
1
The general solution is c1 times the real part plus c2 times the imaginary
part. Substituting t = 0 and the initial condition we get
(
)
(
) ( )
1
1
1
c1
+ c2
=
,
1
0
0
or
{
c1 + c2
c1
1
= 1
= 0
Then c1 = 0 and c2 = 1. Consequently, the solution of the initial value problem
is
)
(
))
(
(
1
1
1
+ sin t
.
e− 4 t cos t
0
1
4. The roots of the characteristic polynomial are λ = ±3i. It follows that the
general solution of the homogeneous equation is C1 cos 3t + C2 sin 3t. It also
follows that we have to look for a particular solution of the non-homogeneous
equation in the form At cos 3t+Bt sin 3t. Its first order derivative is −3At sin 3t+
3Bt cos 3t + A cos 3t + B sin 3t = (3Bt + A) cos 3t + (−3At + B) sin 3t. Its second
order derivative is −(9Bt+3A) sin 3t+3B cos 3t+(−9At+3B) cos 3t−3A sin 3t =
(−9At + 6B) cos 3t − (9Bt + 6A) sin 3t. Substituting it into the equation we get
(−9At + 6B) cos 3t − (9Bt + 6A) sin 3t + 9(At cos 3t + Bt sin 3t) = sin 3t
or
6B cos 3t − 6A sin 3t = sin 3t,
hence B = 0 and A = −1/6. Hence, the general solution of the non-homogeneous
equation is
1
C1 cos 3t + C2 sin 3t − t cos 3t.
6
Its derivative is −3C1 sin 3t + 3C2 cos 3t −
initial conditions, we get
C1 = 1,
1
6
cos 3t + 31 t sin 3t. Substituting the
3C2 −
1
= 0,
6
1
hence C1 = 1 and C2 = 18
.
5. Let us find A, B, and C such that
1
A
Bs + C
=
+
.
(s + 1)(s2 + 2s + 2)
s + 1 s2 + 2s + 2
It is equivalent to
1 = As2 + 2As + 2A + Bs2 + Cs + Bs + C,
hence
A + B = 0, 2A + B + C = 0, 2A + C = 1.
Substituting B = −A and C = 1 − 2A into the second equation we get 2A −
A + 1 − 2A = 0, i.e, A = 1, hence B = −1, and C = −1. It follows that
1
s+1
1
=
−
,
(s + 1)(s2 + 2s + 2)
s + 1 (s + 1)2 + 1
hence its inverse Laplace transform is
e−t − e−t cos t.
2
6. Applying the Laplace transform we get
Y 2 + 4Y + 4 =
hence
Y =
1
1
+
,
s + 2 s2 + 1
1
1
+ 2
.
(s + 2)3
(s + 1)(s + 2)2
1
Inverse Laplace transform of (s+2)
3 is
Laplace transform of the second fraction:
1 2 −2t
.
2t e
Let us find the inverse
1
As + B
C
D
(As + B)(s + 2)2 + C(s2 + 1)(s + 2) + D(s2 + 1)
=
+
+
=
.
(s2 + 1)(s + 2)2
s2 + 1 s + 2 (s + 2)2
(s2 + 1)(s + 2)2
Substituting s = −2 into the denominator, we get 1 = 5D. Substituting s =
i, we get 1 = (Ai+B)(i+2)2 = (Ai+B)(3+4i) = (3B −4A)+(3A+4B)i, hence
3A + 4B = 0, while 3B − 4A = 1. We get then B = − 34 A, and − 94 A − 4A = 1,
4
3
hence A = − 25
, and B = 25
. Taking the coefficient at s3 , we get A + C = 0. It
4
follows that C = 25 .
4
1
Consequently, the inverse Laplace transform of (s2 +1)(s+2)
2 is − 25 cos t +
3
4 −2t
1 −2t
+ 5 te . It follows that the solution of the equation is
25 sin t + 25 e
y=−
4
3
4
1
1
cos t +
sin t + e−2t + te−2t + t2 e−2t .
25
25
25
5
2
7.–8. Are very similar to the previous training homework.
3