Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz1 => Topic started 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*}

Solution in the following PDF file

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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:
$e^{t}y^{'}e^{t}y=2te^{t}$
$(e^{t}y)^{'} = 2te^{t}$
$d(e^{t}y)= 2te^{t}dt$
$e^{t}y=\int 2te^{t}dt$
$e^{t}y = 2e^{t}(t1)+C$
$y = 2e^{2t}(t1)+Ce^{t}$
Since $y(0) = 1 \implies 1= 2\times e^{0}(01)+Ce^{0}$, then we get $C =3$
Therefore, general solution is: $y = 2e^{2t}(t1)+3e^{t}$