11.1. The characteristic polynomial of the matrix is λ2 + 4λ + 3. Its roots
are −1 and −3. Let us find the eigenvectors.
For λ = −1:
−a + b = 0
a−b = 0
1
We can take
.
1
For λ = −3:
a+b = 0
a+b = 0
1
and we can take
.
−1
It follows that the general solution is
1
1
x = c1 e−t
+ c2 e−3t
.
1
−1
11.2. The characteristic polynomial is (1−λ) λ2 − 2λ + 5 , hence the eigenvalues are λ = 1, and λ = 1 ± 2i.
The eigenvector for λ = 1 is found from the system

 0·a+0·b+0·c = 0
2a + 0 · b − 2c = 0

3a + 2b + 0 · c = 0
3
The second equation gives us a = c. The thirdequation
 gives us b = − 2 a.
2
The first equation is void. We get an eigenvector  −3 . The corresponding
2
solution is


2
x = et  −3  .
2
The eigenvector of λ = 1 + 2i is found from the system

−2ia = 0

2a − 2ib − 2c = 0

3a + 2b − 2ic = 0
From the first equation we get a = 0. The other
 two
 equation are both equivalent
0
to b = ic. It follows, that an eigenvector is  i . The corresponding solution
1
1
is




0
0
x = e(1+2i)t  i  = et (cos 2t + i sin 2t)  i  =
1
1

 



0
0
0
 −et sin 2t + iet cos 2t  =  −et sin 2t  + i  et cos 2t  .
et cos 2t + iet sin 2t
et cos 2t
et sin 2t
Consequently, the general solution of the system is






2
0
0
x = c1 et  −3  + c2  −et sin 2t  + c3  et cos 2t  .
2
et cos 2t
et sin 2t
11.3. The characteristic polynomial of the matrix is λ2 − 6λ + 8. Its roots
are λ = 2 and 4. An eigenvector for λ = 2 is found from the system
3a − b = 0
3a − b = 0
1
Hence, we can take
.
3
An eigenvector for λ = 4 is given by
a−b
3a − 3b
=
=
0
0
1
.
1
A fundamental system of solutions is then
1
1
2t
4t
e
,
e
,
3
1
so that we can take
hence a fundamental matrix is
Ψ(t) =
e2t
3e2t
e4t
e4t
.
−1/2 1/2
−1
We have Ψ(0) =
, and Ψ(0) =
.
3/2 −1/2
2t
4t
−1/2
e
e
It follows that Φ(t) = Ψ(t)Ψ(0)−1 =
3e2t e4t
3/2
!
2t
4t
2t
4t
−e +3e
2
−3e2t +3e4t
2
1
3
e −e
2
3e2t −e4t
2
1
1
.
2
1/2
−1/2
=
Then the solution of the initial value problem x(0) =
Φ(t)
1
1
=
e4t
e4t
1
1
is
.
11.4. The characteristic polynomial is λ2 + 6λ + 9, hence we have one
repeated root λ = −3. The eigenvector is found from the system
4a − 4b = 0
4a − 4b = 0
−3t 1
e
We can take v =
. This gives us a solution x(t) = e−3t v =
.
1
e−3t
a
The next vector w =
is found from
b
4a − 4b = 1
4a − 4b = 1
1/4
We can take, for instance, w =
. Then a solution is x(t) = e−3t w +
0
1/4 + t
te−3t v = e−3t
.
t
Consequently, the general solution is
1
1/4 + t
−3t
−3t
x = c1 e
+ c2 e
.
1
t
11.5. The characteristic polynomial
The roots are λ = ±1.
For λ = 1:
a−b
3a − 3b
1
hence an eigenvector is
.
1
For λ = −1:
3a − b
3a − b
1
and we can take
.
3
of the homogeneous system is λ2 − 1.
=
=
0
0
=
=
0
0
Then a fundamental matrix of the system is Ψ(t) =
Its determinant is 2. Consequently,
−t
3e /2
−1
Ψ(t) =
−et /2
3
−e−t /2
et /2
.
et
et
e−t
3e−t
.
The a solution of the non-homogeneous system is given by
Z
Ψ(t)
Ψ(t)
−1
et
t
dt =
t Z −t
3e /2 −e−t /2
e
Ψ(t)
dt =
−et /2
et /2
t
R
3/2 − te−t /2 dt
=
Ψ(t) R
−1/2 + tet /2 dt
t
e
e−t
3/2t + te−t /2 + e−t /2 + c1
=
et 3e−t
−t/2 + tet /2 − et /2 + c2
1 −t
3 t
+t
c1 et + c2 e−t
2 te − 2 te
+
.
3 t
3 −t
c1 et + 3c2 e−t
+ 2t − 1
2 te − 2 te
We used integration by parts:
Z
Z
Z
te−t dt = − t de−t = −te−t + e−t dt = −te−t − e−t + c
and
Z
tet dt =
Z
t det = tet −
11.6. We have
y 00 =
Z
6
4x 0
y −
y.
1 + x2
1 + x2
We will use the Taylor’s formula: an = y
Then a0 = y(0) = 1, a1 = y 0 (0) = 0,
(n)
(x0 )
.
n!
y 00 (0) = −6y(0) = −6,
y 000 (x) =
et dt = tet − et .
a2 = −3
4 − 4x2 0
4x 00
12x
6
y +
y +
y−
y0 =
2
2
2
2
2
(1 + x )
1+x
(1 + x )
1 + x2
2 + 10x2 0
12x
4x 00
y
−
y +
y.
2
2
2
1+x
(1 + x )
(1 + x2 )2
hence y 000 (0) = 0, and a3 = 0.
Hence the first four terms a0 + a1 x + a2 x2 + a3 x3 of the series solution are
1 − 3x2 .
11.7. We are looking for solutions in the form y(x) = xr . Substituting, we
get
r(r − 1)xr − 5rxr + 9xr = 0,
r(r − 1) − 5r + 9 = 0,
or r2 − 6r + 9 = 0. The only (double) root is r = 3. It follows that the general
solution is
y(x) = c1 x3 + c2 x3 ln x,
x > 0.
4
11.8. Dividing the equation by 2x, we get
y 00 +
1 0 1
y + y = 0.
2x
2
The coefficients go to infinity as x → 0, hence 0 is a singular point. If we multiply
the first coefficient by x and the second coefficient by x2 , we get continuous
rational functions at 0, hence the singular point is regular.
Substituting y(x) = a0 xr + a1 xr+1 + a2 xr+2 + · · · into the original equation,
we get
2r(r − 1)a0 xr−1 + 2(r + 1)ra1 xr + · · · + 2(r + n + 1)(r + n)an+1 xr+n + · · ·
ra0 xr−1 + (r + 1)a1 xr + · · · + (r + n + 1)an+1 xr+n + · · ·
a0 xr+1 + a1 xr+2 + · · · + an−1 xr+n + · · · = 0.
The coefficients have to be equal to zero, i.e.,
2r(r − 1)a0 + ra0 = 0
2(r + 1)ra1 + (r + 1)a1 = 0
2(r + 2)(r + 1)a2 + (r + 2)a2 + a0 = 0
..
.
2(r + n + 1)(r + n)an+1 + (r + n + 1)an+1 + an−1
=
0
From the first equation we get the indicial equation:
2r(r − 1) + r = 0,
r(2r − 2 + 1) = 0 r(2r − 1) = 0,
hence we have either r = 0, or r = 1/2. Then from the second equation, we get
a1 = 0. (Note that a0 6= 0, since otherwise we can replace r by r + 1.)
Then the last equation is
(r + n + 1)(2r + 2n + 1)an+1 = −an−1
or
an+1 =
or
an =
−an−1
,
(r + n + 1)(2r + 2n + 1)
−an−2
,
(r + n)(2n + 2r − 1)
which is the recurrent relation (starting from n = 2). In the case r = 0 it is
an =
−an−2
.
n(2n − 1)
In the case r = 1/2 it is
an =
−an−2
−an−2
=
.
(n + 1/2)2n
n(2n + 1)
5
11.9.
Euler’s method for this equation is
yn+1 = yn + h · (2yn − 3tn ),
tn+1 = tn + h,
We have in our case y0 = 1 and h = 0.1. Then
y1 = 1 + 0.1 · 2 = 1.2,
and
y2 = 1.2 + 0.1 · (2 · 1.2 − 0.3) = 1.2 + 0.1 · 2.1 = 1.41.
6