# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-1 => Topic started by: Victor Ivrii on September 28, 2018, 03:37:29 PM

Title: Q1: TUT 0801
Post by: Victor Ivrii on September 28, 2018, 03:37:29 PM
Find the solution of the given initial value problem
\begin{equation*}
y' - y = 2te^{2t},\qquad y(0)=1.
\end{equation*}
Title: Re: Q1: TUT 0801
Post by: Wenhan Sheng on September 28, 2018, 04:52:41 PM
Solution in the following PDF file
Title: Re: Q1: TUT 0801
Post by: Victor Ivrii on September 29, 2018, 07:53:50 PM
Waiting for a typed solution.
Title: Re: Q1: TUT 0801
Post by: Wei Cui on September 29, 2018, 09:35:40 PM
Question: $y^{'} - y = 2te^{2t}$,   $y(0) = 1$

$p(t) = -1$,   $g(t) = 2te^{2t}$

$u(t) = e^{\int -1dt} = e^{-t}$

multiply both sides with $u$, then we get:

$e^{-t}y^{'}-e^{-t}y=2te^{t}$

$(e^{-t}y)^{'} = 2te^{t}$

$d(e^{-t}y)= 2te^{t}dt$

$e^{-t}y=\int 2te^{t}dt$

$e^{-t}y = 2e^{t}(t-1)+C$

$y = 2e^{2t}(t-1)+Ce^{t}$

Since $y(0) = 1 \implies 1= 2\times e^{0}(0-1)+Ce^{0}$, then we get $C =3$

Therefore, general solution is: $y = 2e^{2t}(t-1)+3e^{t}$