MHB -b.2.1.14 Solve y'+2y=te^{-2t}, y(1)=0

[karush]

$\tiny{2.1.{14}}$
$\textsf{Find the solution of the given initial value problem}$
$$y'+2y=te^{-2t} \quad y(1)=0$$
$\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 $e^{2t}$ }$
$$y= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}$$

 

[tkhunny]

$\tiny{2.1.{14}}$
$\textsf{Find the solution of the given initial value problem}$
$$y'+2y=te^{-2t} \quad y(1)=0$$
$\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

I'm a bit puzzled by this. You do the hard part and can't do the substitution?

Substitute t = 1 and see where it leads.

 

[karush]

I'm a bit puzzled by this. You do the hard part and can't do the substitution?

Substitute t = 1 and see where it leads.

\begin{align*}\displaystyle
y(1)&= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}=0\\
&=\frac{1}{2e^{2}}+\frac{c_1}{e^{2}}=0\\
\frac{c_1}{e^{2}}&=-\frac{1}{2e^{2}} \\
c_1&=-\frac{1}{2}
\end{align*}

and then

 

[tkhunny]

\begin{align*}\displaystyle
y(1)&= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}=0\\
&=\frac{1}{2e^{2}}+\frac{c_1}{e^{2}}=0\\
\frac{c_1}{e^{2}}&=-\frac{1}{2e^{2}} \\
c_1&=-\frac{1}{2}
\end{align*}

and then

I'm a little unhappy with your notation. After you write y(1), there should be no more appearance of "t".

 

[karush]

I'm a little unhappy with your notation. After you write y(1), there should be no more appearance of "t".

\begin{align*}\displaystyle
y(t)&= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}=0\\
y(1)&=\frac{c_1}{e^{2}}=-\frac{1}{2e^{2}} \\
c_1&=-\frac{1}{2}\\
\text{so then}\\
y(t)&=\frac{t^2}{2e^{2t}} + \frac{1}{2e^{2t}}
\end{align*}

 

[tkhunny]

\begin{align*}\displaystyle
y(t)&= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}=0\\
y(1)&=\frac{c_1}{e^{2}}=-\frac{1}{2e^{2}} \\
c_1&=-\frac{1}{2}\\
\text{so then}\\
y(t)&=\frac{t^2}{2e^{2t}} + \frac{1}{2e^{2t}}
\end{align*}

Remember when you said that one must be SUPER CAREFUL on these?

Put the "=0" on the y(1), not on the y(t). :-)

 

[karush]

Remember when you said that one must be SUPER CAREFUL on these?

Put the "=0" on the y(1), not on the y(t). :-)

\begin{align*}\displaystyle
y(t)&= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}\\
y(1)&=\frac{c_1}{e^{2}}=-\frac{1}{2e^{2}}=0 \\
c_1&=-\frac{1}{2}\\
\text{so then}\\
y(t)&=\frac{t^2}{2e^{2t}} + \frac{1}{2e^{2t}}
\end{align*}

 

[HOI]

\begin{align*}\displaystyle
y(t)&= \frac{t^2}{2e^{2t}} + \frac{c_1}{e^{2t}}\\
y(1)&=\frac{c_1}{e^{2}}=-\frac{1}{2e^{2}}=0 \end{align*}

And that is still wrong!
[math]y(1)= \frac{1}{2e^2}+ \frac{c_1}{e^2}= 0[/math]
so [math]\frac{c_1}{e^2}= -\frac{1}{2e^2}[/math]
and from that [math]c_1= -\frac{1}{2}[/math].

\begin{align*} \displaystyle c_1&=-\frac{1}{2}\\
\text{so then}\\
y(t)&=\frac{t^2}{2e^{2t}} + \frac{1}{2e^{2t}}
\end{align*}

 

[tkhunny]

\begin{align*}\displaystyle
y(1)&=\frac{c_1}{e^{2}}=-\frac{1}{2e^{2}}=0 \\
\end{align*}

SUPER careful. :-)

Your getting this. Keep hitting it.

 

[karush]

SUPER careful. :-)

Your getting this. Keep hitting it.

not in a class yet which starts 8/20

so this forum is my crying shoulder