Willkommen hier:
Aus dem Differentationssatz folgt:
y=F(s)
y'= -y(0) +s F(s) = -1 +s F(s)
y'' = - s y(0) -y'(0) +s^2 F(s) = -s +1/4 +s^2 F(s)
y''' = - s^2 y(0) -s y'(0) -y''(0) +s^3 F(s) = -s^2 +s/4 +3/2 +s^3 F(s)
--->in die DGL einsetzen ,
\( \begin{aligned} \Rightarrow-s^{2}+\frac{s}{4}+\frac{3}{2}+s^{3} F(s)-3\left(-s+\frac{1}{4}+s^{2} F(s)\right) & +2(-1+s F(s) \\ & =LT\{x+e^ x\}\end{aligned} \)
\( \begin{array}{l}-s^{2}+\frac{s}{4}+3/2 +s^{3} F(s)+3 s-\frac{3}{4}- 3 s^2 F(s)-2 +2 s F(s) =\frac{1}{s^{2}}+\frac{1}{s-1} \\ F(s)\left[s^{3}-3 s^{2}+2 s\right]-s^{2}+\frac{13s}{4 }-\frac{5}{4}=\frac{1}{s^ 2}+\frac{1}{s-1} \\ F(s)\left[s^{3}-3 s^{2}+2 s\right]=\frac{1}{s^2}+\frac{1}{s-1}+s^{2}-\frac{13}{4} s +\frac{5}{4}\end{array} \)
\( F(s)[s(s-1)(s-2)]=\frac{4(s-1)+4 s^{2}+4 s^{4}(s-1)-13 s^{3}(s-1)+5 s^{2}(s-1)}{4 s^{2}(s-1)} \)
\( F(s)[s(s-1)(s-2)]=\frac{4 s-4+4 s^{2}+4 s^{5}-4 s^{4}-13 s^{4}+13 s^{3}+5 s^{3}-5 s^{2}}{4 s^{2}(s-1)} \)
\( F(s)[s(s-1)(s-2)]=\frac{4 s^{5}-17 s^{4}+18 s^{3}-s^{2}+4 s-4}{4 s^{2}(s-1)} \)
\( F(s)[s(s-1)(s-2)]=\frac{(s-2)\left(4 s 4-9 s^{3}-s+2\right)}{4 s^{2}(s-1)} \)
\( F(s)=\frac{\left.(s-2) (4 s^{4}-9 s^{3}-s+2\right)}{4 s^{2}(s-1) \cdot s(s-1)(s-2)} \)
\( F(s)=\frac{4 s^{4}-9 s^{3}-s+2}{4s^3(s-1)^{2}} \)
Partialbruchzerlegung:
Ansatz:
\( \frac{4 s^{4}-9 s^{3}-s+2}{4s^3(s-1)^{2}} \) =\( \frac{A}{4(s-1)} \) + \( \frac{B}{4(s-1)^2} \)+\( \frac{C}{4s} \) +\( \frac{D}{4s^2} \) +\( \frac{E}{4 s^3} \)
A=0
B= -4
C=4
D=3
E=2
-->Rücktransformation ->Tabelle
\( y(t)=\frac{1}{4}\left(t^{2}+3 t-4 e^{t} t+4\right) \)
