First order exact diff. equation.

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In this solution for an exact differential equation I can't reconcile my answer with the book's answer. Please help.

Re: First order exact diff. equation.

I've never seen the solution method that you have presented so I can't find if there's an error in it. On the other hand if we look at the solution method I know I do not get your book answer either. The fix is trivial and I suspect a typo. I'll run through how I know how to do it and maybe you can compare the two methods to find where your problem is.

I won't bother to prove that your original equation is exact. I'm assuming you already checked that. So I will start with the assumption that there is a function F(x, y) such that
$\frac{\partial F}{\partial x} = P(x, y) = e^x sin(y) + e^{-y}$

which means that
$F(x, y) = \int P(x, y) \partial x + \phi (y)$
(The $\partial x$ in the integrand is merely to specify we are keeping the value of y fixed during the integration.)

This leads to
$F(x, y) = e^x sin(y) + xe^{-y} + \phi (y)$

We also know that
$\frac{\partial F}{\partial y} = Q(x, y)$

Integrating this out we get that $\phi (y) = C$

So the final solution will be
$F(x, y) = e^x sin(y) + xe^{-y} + C$

whereas the book claims $F(x, y) = C = e^x sin(y) + xe^{-y}$ . Hence the suspicion of a typo.

-Dan

Re: First order exact diff. equation.

thank you, it makes sense.