Welcome to our community

Be a part of something great, join today!

[SOLVED] find the general solution for y'+2y=2-e^(-4t), y(0)=1

karush

Well-known member
Jan 31, 2012
2,886
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

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Why did you divide the integrating factor by 2 in the second term on the LHS?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

\(\displaystyle e^{2t}y'+2e^{2t}y=2e^{2t}-e^{-2t}\)

And then, this may be written as follows:

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

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

karush

Well-known member
Jan 31, 2012
2,886
I'm not sure what you mean, but you simply want to multiply the given ODE by the integrating factor to get:

\(\displaystyle e^{2t}y'+2e^{2t}y=2e^{2t}-e^{-2t}\)

And then, this may be written as follows:

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

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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,886
$\displaystyle e^{2t}y=\int 2e^{2t}-e^{-2t} \, dt=\frac{e^{-2t}}{2}+e^{2t}+c$
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

Administrator
Staff member
Feb 24, 2012
13,775
Looks correct to me. :)
 

karush

Well-known member
Jan 31, 2012
2,886
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