Author Topic: Q1: TUT 0801  (Read 1709 times)

Victor Ivrii

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Q1: TUT 0801
« 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*}

Wenhan Sheng

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Re: Q1: TUT 0801
« Reply #1 on: September 28, 2018, 04:52:41 PM »
Solution in the following PDF file

Victor Ivrii

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Re: Q1: TUT 0801
« Reply #2 on: September 29, 2018, 07:53:50 PM »
Waiting for a typed solution.

Wei Cui

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Re: Q1: TUT 0801
« Reply #3 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: $y = 2e^{2t}(t-1)+3e^{t}$
« Last Edit: September 29, 2018, 09:57:57 PM by Wei Cui »