# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: Shang Wu on October 18, 2019, 02:21:01 PM

Title: TUT5103 Quiz4
Post by: Shang Wu on October 18, 2019, 02:21:01 PM
Find the solution of the given initial value problem.
\begin{align*}
y''+y&=0, \\
y(\pi /3)&=2, \\
y'(\pi /3)&=-4
\end{align*}
Solution:
First we solve the characteristic equation:
\begin{align*}
r^2+1&=0\\
r&=\pm i
\end{align*}
so here $\lambda = 0, \mu = 1$, the general solution for the homogeneous equation is
\begin{align*}
y_1=c_1\cos(t)+c_2\sin(t)
\end{align*}
We can check the Wronskian is $1\neq 0$.
Differentiate y, we get
\begin{align*}
y'=-c_1\sin(t)+c_2\cos(t)
\end{align*}
Now substitute the initial values:
\begin{align*}
y(\pi /3)&=2 \\
c_1\cos(\pi /3) +c_2\sin(\pi/3) &=2\\
\frac{1}{2}c_1+\frac{\sqrt{3}}{2}c_2 &=2\\
y'(\pi /3) &=-4\\
-c_1\sin(\pi/3)+c_2\cos(\pi/3)&=-4\\
-\frac{\sqrt{3}}{2}c_1+\frac{1}{2}c_2 &=-4
\end{align*}
Solve for $c_1, c_2$, we get
\begin{align*}
c_1 = 1+2\sqrt{3}, c_2 = \sqrt{3}-2
\end{align*}
Then the solution of the initial value problem is
\begin{align*}
y = (1+2\sqrt{3})\cos(t) + (\sqrt{3}-2)\sin(t)
\end{align*}