2.1.13 ok this isn't uv'+u'v

[karush]

$\tiny{2.1.{13}}$
$\textsf{Find the solution of the given initial value problem}$
$$y'-y=2te^{2t}, \quad y(0)=1$$
$\textit{Find u(x)}$
$$\displaystyle\exp\int -1 dt =e^{ t^{-1}}$$
$\textit{multiply thru with $e^{ t^{-1}}$} $
$$ e^{ t^{-1}}y'- e^{ t^{-1}}y=2te^{t}$$
ok this isn't uv'+u'v
$\textit{W|A}$
$$\color{red}{c_1e^t+2e^{2t}t-2e^{2t}}$$

 

[MarkFL]

The integrating factor is:

\(\displaystyle \mu(t)=\exp\left(-\int\,dt\right)=e^{-t}\)

Try that...:)

 

[karush]

$\tiny{2.1.{13}}$
$\textsf{Find the solution of the given initial value problem}$
$$y'-y=2te^{2t}, \quad y(0)=1$$
$\textit{Find u(x)}$
$$\displaystyle\exp\int -1 dt =e^{ -t}$$
$\textit{multiply thru with $e^{- t}$} $
$$ e^{-t}y'- e^{-t}y=2te^{t}$$
but still this isn't uv'+u'v
$\textit{W|A}$
$$\color{red}{c_1e^t+2e^{2t}t-2e^{2t}}$$

 

[Sudharaka]

$\tiny{2.1.{13}}$
$\textsf{Find the solution of the given initial value problem}$
$$y'-y=2te^{2t}, \quad y(0)=1$$
$\textit{Find u(x)}$
$$\displaystyle\exp\int -1 dt =e^{ -t}$$
$\textit{multiply thru with $e^{- t}$} $
$$ e^{-t}y'- e^{-t}y=2te^{t}$$
but still this isn't uv'+u'v
$\textit{W|A}$
$$\color{red}{c_1e^t+2e^{2t}t-2e^{2t}}$$

Hi,

Notice that, $\frac{d}{dt}\left(e^{-t}y\right)=e^{-t}y'-ye^{-t}$. I hope you can continue with this hint. :)

 

[karush]

Hi,

Notice that, $\frac{d}{dt}\left(e^{-t}y\right)=e^{-t}y'-ye^{-t}$. I hope you can continue with this hint. :)

$\tiny{2.1.{13}}$
$\textsf{Find the solution of the given initial value problem}$
$$y'-y=2te^{2t}, \quad y(0)=1$$
$\textit{Find u(x)}$
$$\displaystyle\exp\int -1 dt =e^{ t^{-1}}$$
$\textit{multiply thru with $e^{ t^{-1}}$} $
$$ e^{ t^{-1}}y'- e^{ t^{-1}}y=2te^{t}$$
$$(e^{-t}y)'=2te^{t}$$
$\textit{integrate}$
$$e^{-t}y=\int 2te^{t} \, dt =2e^t (t-1)+c_1$$
$\textit{Divide thru by $e^{-t}y$ and distribute}$
$$c _1e^t+2e^{2t}t-2e^{2t}$$
$$y(0)=c _1e^{(0)}+2e^{2(0)}t-2e^{2(0)}=1+2-2=1$$