integrating factor would be $\displaystyle e^{8x}$Solve the inital value problem for y(x):

$\displaystyle xy' + 8y = 3x^4$

with the initial condition: $\displaystyle y(1) = 4$

$\displaystyle \frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)$

$\displaystyle e^{8x}y = \int e(3x^4)^{8x}$

not sure if what im doing so far is right, having a hard time finishing it.