need help solving this diff plz
3y´ - y = 5x
You can use an integrating factor:
First
$\displaystyle y' - \frac{1}{3}y = \frac{5}{3}x$
Then
$\displaystyle M(x) = e^{\int(-dx/3)} = e^{-x/3}$
So
$\displaystyle e^{-x/3}y' - \frac{1}{3}e^{-x/3}y = \frac{5}{3}xe^{-x/3}$
Notice that the left hand side of the equation is a "perfect derivative"
$\displaystyle \frac{d}{dx} \left ( e^{-x/3}y \right ) = \frac{5}{3}xe^{-x/3}$
$\displaystyle e^{-x/3}y = \int \frac{5}{3}xe^{-x/3}~dx$
I'll let you finish out from there.
-Dan