# Lineare Differentialgleichung/Zweite Ordnung/Rechte Seite/Ansatz/4/Aufgabe/Lösung

Zur Navigation springen Zur Suche springen

Wegen

${\displaystyle {}5^{2}-3\cdot 5+9=19\,}$

ist ${\displaystyle {}5}$ keine Nullstelle des charakteristischen Polynoms der Differentialgleichung. Somit kann man nach Fakt den Ansatz

${\displaystyle {}y(t)=(at^{2}+bt+c)e^{5t}\,}$

machen. Die linke Seite wird dann zu

{\displaystyle {}{\begin{aligned}&\,\,\,\,\,\,\,y^{\prime \prime }(t)-3y'(t)+9y(t)\\&={\left((at^{2}+bt+c)e^{5t}\right)}^{\prime \prime }-3{\left((at^{2}+bt+c)e^{5t}\right)}'+9(at^{2}+bt+c)e^{5t}\\&={\left((5at^{2}+5bt+5c+2at+b)e^{5t}\right)}^{\prime }-3(5at^{2}+5bt+5c+2at+b)e^{5t}+9(at^{2}+bt+c)e^{5t}\\&={\left((5at^{2}+(5b+2a)t+5c+b)e^{5t}\right)}'-3(5at^{2}+(5b+2a)t+5c+b)e^{5t}+9(at^{2}+bt+c)e^{5t}\\&=(5(5at^{2}+(5b+2a)t+5c+b)+10at+5b+2a)e^{5t}-3(5at^{2}+(5b+2a)t+5c+b)e^{5t}+9(at^{2}+bt+c)e^{5t}\\&={\left({\left(25a-15a+9a\right)}t^{2}+{\left(25b+10a+10a-15b-6a+9b\right)}t+25c+5b+5b+2a-15c-3b+9c\right)}e^{5t}\\&={\left(19at^{2}+{\left(14a+19b\right)}t+2a+7b+19c\right)}e^{5t}.\,\end{aligned}}}

Der Vergleich

${\displaystyle {}19at^{2}+{\left(14a+19b\right)}t+2a+7b+19c=t^{2}-8\,}$

liefert

${\displaystyle {}a={\frac {1}{19}}\,,}$
${\displaystyle {}14a+19b=0\,,}$

also

${\displaystyle {}b=-{\frac {14}{19}}a=-{\frac {14}{361}}\,}$

und

${\displaystyle {}2a+7b+19c=-8\,,}$

also

{\displaystyle {}{\begin{aligned}c&={\frac {-8-2a-7b}{19}}\\&={\frac {-8-2{\frac {1}{19}}+7\cdot {\frac {14}{361}}}{19}}\\&={\frac {-8\cdot 361-2\cdot 19+7\cdot 14}{6859}}\\&={\frac {-2888-38+98}{6859}}\\&=-{\frac {2828}{6859}}.\end{aligned}}}

Somit ist

${\displaystyle {\left({\frac {1}{19}}t^{2}-{\frac {14}{361}}t-{\frac {2828}{6859}}\right)}e^{5t}}$
eine Lösung der Differentialgleichung.