MAT244--2019F > Quiz-3

TUT0401 QUIZ3

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EroSkulled:
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}

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