Toronto Math Forum
MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: EroSkulled on October 11, 2019, 02:00:00 PM
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Solve $y''+5y'+3y=0$ with initial condition $y(0)=1, y'(0)=0$
Characteristic equation: $r^2+5r+3=0$
Then we get two solutions:$r_1=\frac{-5+\sqrt{13}}{2}$, $r_2=\frac{-5-\sqrt{13}}{2}$
Hence general solution:
\begin{equation}
y(t)=Ae^{t\frac{-5+\sqrt{13}}{2}}+Be^{t\frac{-5-\sqrt{13}}{2}}
\end{equation}
\begin{equation}
y’(t)= \frac{-5+\sqrt{13}}{2} Ae^{t\frac{-5+\sqrt{13}}{2}}-\frac{5+\sqrt{13}}{2} Be^{t\frac{-5-\sqrt{13}}{2}}
\end{equation}
Since $y(0)=1, y'(0)=0$
Solve equation sets above for A and B, we get $A=\frac{5+\sqrt{13}}{2\sqrt{13}}$, $B=\frac{-5+\sqrt{13}}{2\sqrt{13}}$
Hence the solution of equation is:
\begin{equation}
y(t)= \frac{5+\sqrt{13}}{2\sqrt{13}} e^{t\frac{-5+\sqrt{13}}{2}}+ \frac{-5+\sqrt{13}}{2\sqrt{13}} e^{t\frac{-5-\sqrt{13}}{2}}
\end{equation}