Toronto Math Forum
MAT2442014F => MAT244 MathTests => Quiz 1 => Topic started by: Victor Ivrii on September 29, 2014, 02:21:19 AM

Please post problem and solution

2.1 p. 40, #18
\begin{gather}
ty' + 2y = \sin t,\\
y(Ï€/2) = 1
\end{gather}
First, change the equation by dividing every term by $t$.
\begin{equation}
y' + (2/t)y = (\sin (t))/t
\end{equation}
Then, solve for the integrating factor.
\begin{equation}
Î¼ = e^{\int (2/t)\,dt} = e^{2\ln(t)\,} = t^2
\end{equation}
Multiply every term in equation by Î¼.
\begin{equation}
t^2y' + 2ty = t\sin(t)
\end{equation}
Integrate both sides of the equation.
\begin{gather}
\int[t^2y]' = \int{t\sin(t)}dt\implies
t^2y = âˆ’t\cos(t) âˆ’ \int{\cos(t)}dt= âˆ’t\cos(t) + \sin(t) + C\implies y = (âˆ’t \cos(t) +\sin (t) + C)/t^2
\end{gather}
Substitute the initial value to solve for C.
\begin{gather}
C = (Ï€/2)^21
\end{gather}
Therefore, the solution is:
\begin{gather}
y = (âˆ’t \cos(t) +\sin (t) + (Ï€/2)^21)/t^2
\end{gather}