A differential equation: y' - 3y = e^(-2x) + 2

[As time decides]

Hi people

I would like some help with this diff. equation:

\(\displaystyle y' - 3y = e^{-2x} + 2\)

I really don't know anything about solving this kind of differential equation when we got \(\displaystyle y' + ny = e^{nx}\)

All i can say is that we'll get \(\displaystyle y = Ce^{3x} + something\)

Thanks for all replies.

 

[tkhunny]

It's already in Standard Form. You just need an integrating factor.

\(\displaystyle f(x) = e^{\int(-3)dx} = e^{-3x}\)

Multiply
Simplify
Find Antiderivatives (Don't forget the arbitrary constant.)
Simplify