MHB -find general solution y'+2y=2-e^(-4t), y(0)=1

[karush]

2000
find the general solution for
$y' + 2y = 2 - {{\bf{e}}^{ - 4t}} \hspace{0.25in}y\left( 0 \right) = 1$
find
$\displaystyle u(x)=\exp \left(\int 2 \, dt\right) =e^{2t}$
$e^{2t}y' + 2\frac{e^{2t}}{2}y =2 e^{2t} - {{\bf{e}}^{ - 4t}}e^{2t}$
$(e^{2t}y)'=2e^{2t}-e^{-2t}$
i continued but didn't get the answer which is

$y\left( t \right) = 1 + \frac{1}{2}{{\bf{e}}^{ - 4t}} - \frac{1}{2}{{\bf{e}}^{ - 2t}}$

 

[MarkFL]

Why did you divide the integrating factor by 2 in the second term on the LHS?

 

[MarkFL]

isn't that the dt

I'm not sure what you mean, but you simply want to multiply the given ODE by the integrating factor to get:

$$e^{2t}y'+2e^{2t}y=2e^{2t}-e^{-2t}$$

And then, this may be written as follows:

$$\frac{d}{dt}\left(e^{2t}y\right)=2e^{2t}-e^{-2t}$$

And now you're ready to integrate both sides. ;)

 

[karush]

I'm not sure what you mean, but you simply want to multiply the given ODE by the integrating factor to get:

$$e^{2t}y'+2e^{2t}y=2e^{2t}-e^{-2t}$$

And then, this may be written as follows:

$$\frac{d}{dt}\left(e^{2t}y\right)=2e^{2t}-e^{-2t}$$

And now you're ready to integrate both sides. ;)

$$e^{2t}y=\int 2e^{2t}-e^{-2} \, dt=\frac{e^{-2t}}{2}+e^{2t}+c$$
hopefully

 

[MarkFL]

Looks good so far, with the exception of a minor typo where the \(t\) in the exponent of one of the terms in the integrand was omitted.

 

[karush]

$\displaystyle e^{2t}y=\int 2e^{2t}-e^{-2t} \, dt=\frac{e^{-2t}}{2}+e^{2t}+c$
dividing thru by $e^{2t}$
$y=\frac{1}{2}e^{-4t}+1+ce^{-2t}$
applying $y(0)=1$
$\frac{1}{2}+1+c=1$
$c=-\frac{1}{2}$
really ?

 

[MarkFL]

Looks correct to me. :)

 

[karush]

the formal view...
$y(t)=1+\frac{1}{2}{{\bf{e}}^{- 4t}}-\frac{1}{2}{{\bf{e}}^{- 2t}}$

want to continue this applying eulers method
but feel i should start a new post.

lots of advice on the internet but too many sacred cows
better just do step by step here