Find the general solution of the equation:

1= [3y*cos(y^2)*dy/dx] + [3e^(-x) * ycos(y^2) * dy/dx]

really really stuck on this, any help would be appreciated!

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- Feb 9th 2010, 01:01 PMbobguyfinidng the general solution
Find the general solution of the equation:

1= [3y*cos(y^2)*dy/dx] + [3e^(-x) * ycos(y^2) * dy/dx]

really really stuck on this, any help would be appreciated! - Feb 9th 2010, 03:35 PMJester
If you factor the RHS you get

$\displaystyle

1 = 3y \cos \left(y^2\right) \left(1 + e^{-x}\right) \frac{dy}{dx}

$

at which point we recognize this is separable, so

$\displaystyle

\frac{dx}{1+e^{-x}} = 3y \cos \left(y^2\right)dy

$

or

$\displaystyle

\frac{e^x\,dx}{e^{x}+1} = 3y \cos \left(y^2\right)dy

$.

Now integrate each side.