Hallo,
\( x^{\prime \prime}+x^{\prime}-6 x=e^{2 t}+e^{3 t} \)
------>homogene Gleichung:
x'' +x'-6x=0
k^2 +k -6=0
k1,2= -1/2 ±√(1/4 +6)
k1,2= -1/2 ± 5/2
k1= 2 → x1= C1 e^(2t)
k2= -3 -----> x2= C2 e^(-3t)
xh= x1+x2
Ansatz part.Lösung:
xp1= t * A *e^(2t)
xp2= B e^(3t)
xp=xp1+xp2
usw.
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\( x^{\prime \prime}-2 x^{\prime}+5 x=(5 t-1) \cdot \sin t, \quad x(0)=1, \quad x^{\prime}(0)=1 \)
k1,2= 1± 2i
xh= C1 e^t cos(2t) +C2 e^t sin(2t)
xp=A cos(t) +B *t cos(t) +C sin(t) +D *t sin(t)