# Toronto Math Forum

## MAT244-2014F => MAT244 Math--Tests => Quiz 1 => Topic started by: Victor Ivrii on September 29, 2014, 02:21:19 AM

Title: Q1 problem 1 (day section)
Post by: Victor Ivrii on September 29, 2014, 02:21:19 AM
Please post problem and solution
Title: Re: Q1 problem 1 (day section)
Post by: Rhoda Lam on October 04, 2014, 09:42:22 PM
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 $t$.

y' + (2/t)y = (\sin (t))/t

Then, solve for the integrating factor.

Î¼ = e^{\int (2/t)\,dt} = e^{2\ln(t)\,} = t^2

Multiply every term in equation by Î¼.

t^2y' + 2ty = t\sin(t)

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)^2-1
\end{gather}
Therefore, the solution is:
\begin{gather}
y = (âˆ’t \cos(t) +\sin (t) + (Ï€/2)^2-1)/t^2
\end{gather}