# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: NANAC on October 11, 2019, 02:08:56 PM

Title: TUT 0401 Quiz 3
Post by: NANAC on October 11, 2019, 02:08:56 PM
\begin{equation}
y^{\prime \prime}+5 y^{\prime}+3 y=0, y(0)=1, y^{\prime}(0)=0
\end{equation}
\begin{array}{c}{\text { We assume that } y=e^{r t} \text { , and then it follows that } r \text { must }} \\ {\text { be a root of characteristic equation }} \\ {r^{2}+5 r+3=0}\end{array}
\begin{array}{l}{\text { We use the quadratic formula which is }} \\ {\qquad r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}}\end{array}
\begin{equation}
\text { Hence, }\left\{\begin{array}{l}{r_{1}=\frac{-5+\sqrt{13}}{2}} \\ {r_{2}=\frac{-5-\sqrt{13}}{2}}\end{array}\right.
\end{equation}
\begin{equation}
\begin{array}{c}{\text { Since the general solution has the form of }} \\ {y=c_{1} e^{r_{1} t}+c_{2} e^{r_{2} t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{c}{\text { Therefore, the general solution of the given equation is }} \\ {\qquad y=c_{1} e^{\frac{\sqrt{13}-5}{2} t}+c_{2} e^{\frac{-5-\sqrt{13}}{2} t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { since } y(0)=1, y^{\prime}(0)=0} \\ {\qquad\left\{\begin{array}{l}{c_{1}+c_{2}=1} \\ {\frac{-5+\sqrt{13}}{2} c_{1}+\left(\frac{-5-\sqrt{13}}{2}\right) c_{2}=0}\end{array} \Rightarrow\left\{\begin{array}{l}{c_{1}=\frac{13+5 \sqrt{13}}{26}} \\ {c_{2}=\frac{13-5 \sqrt{13}}{26}}\end{array}\right.\right.}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Therefore, the solution of the inital value problem is }} \\ {\qquad y=\frac{13+5 \sqrt{13}}{26} e^{\frac{-5+\sqrt{13}}{2} t}+\frac{13-5 \sqrt{13}}{26} e^{\frac{-5-\sqrt{13}}{2}} t}\end{array}
\end{equation}