# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: Coollight on October 11, 2019, 02:05:13 PM

Title: Q3: TUT0801
Post by: Coollight on October 11, 2019, 02:05:13 PM
Find the solution of the initial value problem
\begin{align*}
\ 2y”- 3y’+y=0, \ y(0)=2, \ y’(0)=\frac{1}{2} \\
\end{align*}
Solution:
\begin{align*}
\ Assume \ y=e^{rt} \ a \ solution \ of \ the \ equation. \\
\ Then \ we \ will \ have \\
\ y&=e^{rt} \\
\ y’&=re^{rt} \\
\ y”&=r^{2}e^{rt} \\
\ Then \ 2y”-3y’+y=0 \ will \ be \\
\ 2(r^{2}e^{rt})-3(re^{rt})+e^{rt}&=0 \\
\ e^{re}(2r^{2}-3r+1)&=0 \ where \ e^{rt} \neq 0 \\
\ 2r^2-3r+1 &=0 \\
\ (2r-1)(r-1)&=0\\
\ r_{1}=\frac{1}{2}&,\ r_{2}=1 \\
\ Then \ we \ will \ have \ the \ general \ solution \\
\ y=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}&=c_{1}e^{\frac{1}{2}t}+c_{2}e^{t} \\
\ Since \ y(0)=2 \\
\ c_{1}e^0 + c_{2}e^0 &= 2 \\
\ c_{1} + c_{2} &=2 \\
\ Since \ y’(0)=\frac{1}{2} \\
\ y’&=\frac{1}{2}c_{1}e^{\frac{1}{2}t}+c_{2}e^t \\
\ \frac{1}{2}c_{1}e^0 + c_{2}e^0 &= \frac{1}{2}\\
\ \frac{1}{2}c_{1}+c_{2}&= \frac{1}{2}\\
\ So \ solve \ for \ c_{1}, \ c_{2}\ to \ get: \\
\ c_{1}= 3, \ c_{2}=-1\\
\ So \ the \ solution \ will \ be\ y&=3e^{\frac{1}{2}t}-e^{t} \\
\end{align*}