Variation der Konstanten
B Man löse die Differentialgleichung
√
y 00 − 8y 0 + 16y = e4x x
| {z }
y 0 (1) = −2
y(1) = 1,
f (x)
1. Homogene Lösung:
λ2 − 8λ + 16 = 0
λ1,2 = 4 (Resonanz!)
⇒
yh (x) = c1 e4x + c2 x e4x
2. Partikulärlösung, Variation der Konstanten:
4x
yp (x) = c1 (x) |{z}
e4x +c2 (x) xe
|{z}
y1 (x)
wobei
c1 (x) =
Z
y2 (x)
−y2 (x)f (x) dx
W (x)
y1 (x)
mit W (x) = y10 (x)
c2 (x) =
y2 (x) e4x
=
0
y2 (x) 4e4x
xe4x
(1 − 4x)e4x
Z
y1 (x)f (x) dx
W (x)
= (1 − 4x)e8x − 4xe8x = e8x
√
Z
−xe4x e4x x dx
2x5/2
3/2
c1 (x) =
=
−
x
dx
=
−
e8x
5
Z
Z 4x 4x √
√
x dx
e e
2x3/2
=
c2 (x) =
x dx =
8x
e
3
Z
yp (x) = −
4 5/2 4x
2x5/2 4x 2x3/2 4x
e +
xe =
x e
5
3
15
3. Allgemeine Lösung:
y(x) = c1 e
4x
+ c2 xe
4x
4
+ x5/2 e4x =
15
4 5/2 4x
e
c1 + c2 x + x
15
4. Lösung des Anfangswertproblems:
4
y(1) = c1 + c2 +
e4 = 1
15
4 5 3/2 4x
4
2
16
y 0 (x) = c2 +
e + c1 + c2 x + x5/2 ·4e4x = 4c1 + c2 + 4c2 x + x3/2 + x5/2 e4x
· x
15 2
15
3
15
2 16 4
y 0 (1) = 4c1 + 5c2 + +
e = −2
3 15
4
.
= −0.248
15
2 16 .
4c1 + 5c2 = −2e−4 − −
= −1.770
3 15
.
.
⇒ c1 = 0.530
c2 = −0.778
c1 + c2 =
y(x) =
e−4 −
4 5/2 4x
e
0.530 − 0.778x + x
15