Solve.

$\displaystyle \int 7x cos(x) = \int (2y + e^{3y}) y'$, $\displaystyle y(0) = 0$

I'm pretty sure my work is right, but as usual Webassign does not care for my answer.

$\displaystyle

-7 cos(x) - 7x sin(x) + C = y^2 + \frac{e^{3y}}{3}$

*Now, webassign is having me give the answer based on what and how they provide HALF of the answer. So, here is what WA has on the left:

$\displaystyle e^{3y} + 3y^2$ = __my answer________

So, before I solved for C, I went ahead and multiplied through by the 3 so that I could just work it through in the form it has to be in anyway:

$\displaystyle 3y^2 + e^{3y} = 21 (-xsin(x) - cos(x)) + 3C$

yadda yadda $\displaystyle C = \frac{22}{3}$

So, the right side answer that I am getting is:

$\displaystyle -21x sin(x) - 21 cos(x) + \frac{22}{3}$

Anyone? Anyone? Thanks!