I am having trouble this this problem.

$\displaystyle y' + 6y = e^4^t, y(0) = 2$

Heres the work I have up until where I get stuck:

$\displaystyle SY(s) - y(0) + 6Y(s) = \frac{1}{s-4} $

$\displaystyle SY(s) - 2 + 6Y(s) = \frac{1}{s-4} $

$\displaystyle SY(s) + 6Y(s) = \frac{1}{s-4} + 2 $

$\displaystyle Y(s)(s+6) = \frac{1}{s-4} + 2 $

$\displaystyle Y(s) = \frac{1}{(s-4)(s+6)} + \frac{2}{s+6} $

partial fractions:

$\displaystyle \frac{1}{(s-4)(s+6)} + \frac{2}{s+6} = \frac{A}{s-4} + \frac{B}{s+6} $

How do I solve this partial fraction equation now? I have tried it and I am ending up with a different answer than the solutions manual for B.

My other question is:

Find the fundeamental set of solutions using only real coefficients: $\displaystyle y^(^4^) + 4y = 0$

I get the steps:

$\displaystyle m^4 + 4 = 0$

$\displaystyle m = \sqrt[4]{-4} $

$\displaystyle y1 = e^xcosx$

$\displaystyle y2 = e^xsinx$

$\displaystyle y3 = e^-^xcosx$

$\displaystyle y4 = e^-^xsinx$

I am lost as to where you get $\displaystyle m = \sqrt[4]{-4}$ is equal to $\displaystyle \pm 1 \pm i $ and how this works on a graph....

Thanks for the help guys, I really appreciate it!