# Linear DE

nateshoe
1. Homework Statement [/b]
y'=4x–y+4; y(0)=8

2. The attempt at a solution[/b]
This is simply a first-order DE
So:

y'+y=4x+4
P(x)=1
Q(x)=4x+4
p(x)=e^x
So that ends up being:

ye^x=(4xe^x+4e^x)dx

Integration:
ye^x=4xe^x-4e^x+4e^x

ye^x=4xe^x
divide by e^x
y=4x+C
y=4x+8

However it tells me I'm wrong. either I'm very stupid or I'm going about this completely wrong.

Thanks,
Nate

$$y e^x = 4 x e^x + C$$