TU Wien:Analysis 2 UE (diverse)/Übungen SS12/Beispiel 42

Man bestimme die Lösung des folgenden linearen Anfangswertproblems mittels Laplace-Transformation:

$y^{‴}+y^{″}-5y^{\prime }+3y=6\sinh (2x)$, $y(0)=y^{\prime }(0)=0$, $y^{″}(0)=4$.

Hilfreiches

$L\{\sinh (\omega t)\}=\frac{\omega }{s^2-{\omega }^2}$

Lösungsvorschlag

$ℒ\left\{y^{‴}\right\}=s^{3}Y(s)-s^2\underset{=0}{\underbrace{y(0)}}-s\underset{=0}{\underbrace{y^{\prime }(0)}}-\underset{=4}{\underbrace{y^{″}(0)}}=s^{3}Y(s)-4$

$ℒ\left\{y^{″}\right\}=s^{2}Y(s)-s\underset{=0}{\underbrace{y(0)}}-\underset{=0}{\underbrace{y^{\prime }(0)}}=s^{2}Y(s)$

$ℒ\left\{y^{\prime }\right\}=sY(s)-\underset{=0}{\underbrace{y(0)}}=sY(s)$

$ℒ\left\{y\right\}=Y(s)$

$ℒ\{6\sinh (2x)\}=6\frac{2}{s^2-2^2}=\frac{12}{s^2-4}$

$\begin{array}{cc}\frac{4s^2-4}{s^5+s^4-9s^3-s^2+20s-12}& =\frac{A}{s+3}+\frac{B}{s+2}+\frac{C}{s-1}+\frac{D}{s-2}\\ 4s^2-4& =A(s+2)(s-1)(s-2)+B(s+3)(s-1)(s-2)+C(s+3)(s+2)(s-2)+D(s+3)(s+2)(s-1)\\ 4s^2-4& =A(s^3-s^2-4s+4)+B(s^3-7s+6)+C(s^3+3s^2-4s-12)+D(s^3+4s^2+s-6)\\ 4s^2-4& =(A+B+C+D)s^3+(-A+3C+4D)s^2+(-4A-7B-4C-D)s+(4A+6B-12C-6D)\\ \end{array}$

Koeffizentenvergleich ergibt $A+B+C+D=0$, $-A+3C+4D=4$, $-4A-7B-4C-D=0$ und $4A+6B-12C-6D=-4$. Aufgelöst ergibt das $A=\frac{2}{5}$, $B=-\frac{1}{3}$, $C=-\frac{2}{3}$ und $D=\frac{3}{5}$

$\begin{array}{cc}{s}^{3}Y(s)-4+s^{2}Y(s)-5sY(s)+3Y(s)& =\frac{12}{s^2-4}\\ s^{3}Y(s)+s^{2}Y(s)-5sY(s)+3Y(s)& =\frac{12}{s^2-4}+4\\ Y(s)(s^3+s^2-5s+3)& =\frac{12}{s^2-4}+4\\ Y(s)& =(\frac{12}{s^2-4}+4)\cdot \frac{1}{s^3+s^2-5s+3}\\ Y(s)& =\frac{12+4s^2-16}{s^5+s^4-5s^3+3s^2-4s^3-4s^2+20s-12}\\ Y(s)& =\frac{4s^2-4}{s^5+s^4-9s^3-s^2+20s-12}\\ Y(s)& =\frac{2}{5}\frac{1}{s+3}+\frac{1}{3}\frac{1}{s+2}+\frac{2}{3}\frac{1}{s-1}+\frac{3}{5}\frac{1}{s-2}\\ {ℒ}^{-1}\{Y(s)\}& =\frac{2}{5}{e}^{-3x}-\frac{1}{3}{e}^{-2x}-\frac{2}{3}{e}^x+\frac{3}{5}{e}^{2x}\end{array}$

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