Author Topic: TUT 0301 Quiz 1  (Read 641 times)

LLY

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TUT 0301 Quiz 1
« on: September 27, 2019, 04:02:03 PM »
\begin{equation}
y^{\prime}-y=2 t e^{2 t}, \quad y(0)=1
\end{equation}
\begin{equation}
\begin{array}{l}{p(t)=-1, \quad g(t)=2 t e^{2 t}} \\ {u(t)=e^{\int-1 d t}=e^{-t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{e^{-t} y^{\prime}-e^{-t} y=2 t e^{t}} \\ {\left(e^{-t} y\right)^{\prime}=2 t e^{t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{d\left(e^{-t} y\right)=2 t e^{t} d t} \\ {e^{-t} y=\int 2 t e^{t} d t} \\ {e^{-t} y=2 e^{t}(t-1)+C} \\ {y=2 e^{2 t}(t-1)+C e^{t}}\end{array}
\end{equation}
\begin{equation}
y(0)=1
\end{equation}
\begin{equation}
1=2 \times e^{0}(0-1)+C e^{0}
\end{equation}
\begin{equation}
C=3
\end{equation}
« Last Edit: October 07, 2019, 06:24:52 PM by LLY »

Wang Jingyao

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Re: TUT 0301 Quiz 1
« Reply #1 on: September 27, 2019, 05:03:03 PM »
Question:Find general solution for y'-y=2te^2t,y(0)=1