Disguised Bernoulli.

**blackhug** · October 27th 2010, 12:28 PM

I'm sure this is a very basic differential equation but I cant figure out how to separate it to begin with.

$\frac{dy}{dx} = e^{x + y} + e^x\sin{x}$

**Ackbeet** · October 27th 2010, 12:42 PM

Note that $e^{x+y}=e^{x}e^{y}.$ Then substitute $z=e^{y}.$ I think you'll find that the resulting DE is Bernoulli.

**blackhug** · October 27th 2010, 12:46 PM

Thanks, thought it was that.

**Ackbeet** · October 27th 2010, 12:49 PM

You're welcome. Incidentally, I don't think you'll be able to find a closed-form solution of this DE. That is, you'll still have integrals in your final solution, because there are no known antiderivatives for the integrals you get. That's ok. You still get an explicit solution (all the x's on one side), which is nice.

**blackhug** · October 27th 2010, 01:09 PM

Oh sorry I mistyped the problem in the OP, it's actually:
$\frac{dy}{dx} = e^{x + y} + e^y\sin{x}$
from which I get...

$\int \frac{dy}{e^{2y}} = \int e^x + sinx dx$

$\frac{e^{-2y}}{2} = e^x - cosx$
and finally...

$2y = \ln{\cos{x}} - x$

Is that correct?

**Ackbeet** · October 27th 2010, 01:14 PM

The two sign errors that you made evidently corrected themselves in your final answer. Don't forget the constant, too.

[EDIT]: Also, I don't think you're handling the factor of 2 correctly.

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