# [SOLVED] Solve for a DE

• Feb 21st 2010, 04:11 PM
arbolis
[SOLVED] Solve for a DE
I must solve a lot of DE's. I'm finishing my vacations and the upper level undergraduate courses of physics are coming very soon. I just begin with DE's. I hope to improve quickly.

Find the general solution of this DE:
$\displaystyle y'+3y=x+e^{-2x}$.
I don't expect a full solution. First, I'd be glad to know what kind of DE it is and the way to solve it. Maybe a reference about this thread: http://www.mathhelpforum.com/math-he...-tutorial.html can be nice.
Thanks for the help.
• Feb 21st 2010, 04:53 PM
pickslides
Quote:

Originally Posted by arbolis

Find the general solution of this DE:
$\displaystyle y'+3y=x+e^{-2x}$.
I don't expect a full solution. First, I'd be glad to know what kind of DE it is and the way to solve it.

It is a first order linear non homogenous D.E. It is not separable.

You need to employ the integrating factor method.

Quote:

Originally Posted by arbolis

Thanks for the help.

There is a section near the beginning of this thread, let us know how you go.
• Feb 21st 2010, 05:10 PM
arbolis
Thank you very much!
I'm not sure I'm doing well. I reach $\displaystyle y=e^{-\int 3 dx} \left [ \int (x+e^{-2x})e^{-\int 3 dx} dx \right ] =e^{-3x+C_0} \left [ \int (x+e^{-2x})e^{-3x+C_1} dx \right ]$. Is it right until here?
• Feb 21st 2010, 06:30 PM
pickslides
I get an itegrating factor of $\displaystyle e^{\int 3~dx} = e^{ 3x}$

Now you multiply this through both sides of your equation

$\displaystyle y'+3y=x+e^{-2x}$

should give you

$\displaystyle e^{ 3x}y'+3e^{ 3x}y=e^{ 3x}(x+e^{-2x})$

then integrating both sides you get

$\displaystyle e^{ 3x}y=\int e^{ 3x}(x+e^{-2x})~dx$

$\displaystyle y=\frac{\int e^{ 3x}(x+e^{-2x})~dx}{e^{ 3x}}$

$\displaystyle y=\frac{\int xe^{ 3x}+e^{x}~dx}{e^{ 3x}}$

Use integration by parts for the numerator and you will be finished.
• Feb 22nd 2010, 01:08 PM
arbolis
I reached $\displaystyle y=\frac{1}{3} \left ( x-1-\frac{e^{2x}}{3}+C \right )$ where $\displaystyle C$ is a constant which depends on initial conditions not given here. I don't know if I made an arithmetic error, but at least I've understood this method. (Bow)
• Feb 22nd 2010, 01:19 PM
pickslides
• Feb 22nd 2010, 01:26 PM
arbolis
Quote:

Originally Posted by pickslides

Ok I made a mistake. (Wink) I will correct it.