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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

$\tiny{2.1.{14}}$

$\textsf{Find the solution of the given initial value problem}$

$$y'+2y=te^{-2t} \quad y(1)=0$$

$\textit{obtain $u(x)$}$

$$\exp\int 2 dt=e^{2t}$$

$\textit{multiply thru by $e^{2t}$}$

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

$\textit{Integrate}$

$$e^{2t}y= \frac{t^2}{2} + c_1$$

$\textit{Divide thru by $e^{2t}$ }$

$$y= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}$$

$\textit{W|A}$

$ \color{red}{y(t) =c_1e^{-2t}+\frac{1}{2}e^{-2t}t^2 }$

not sure about intial value

$\textsf{Find the solution of the given initial value problem}$

$$y'+2y=te^{-2t} \quad y(1)=0$$

$\textit{obtain $u(x)$}$

$$\exp\int 2 dt=e^{2t}$$

$\textit{multiply thru by $e^{2t}$}$

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

$\textit{Integrate}$

$$e^{2t}y= \frac{t^2}{2} + c_1$$

$\textit{Divide thru by $e^{2t}$ }$

$$y= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}$$

$\textit{W|A}$

$ \color{red}{y(t) =c_1e^{-2t}+\frac{1}{2}e^{-2t}t^2 }$

not sure about intial value

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