Quote:

2.

$\displaystyle y" - 16y = 2e^4^x$

auxiliary equation

$\displaystyle m^2 - 16 = 0$

$\displaystyle Yc = ce^-^4^x + ce^4^x$

$\displaystyle Y1 = e^-4^x$

$\displaystyle Y2 = e^4^x$

Wronskian = 8

$\displaystyle f(x) = 2e^4^x$

$\displaystyle u1' = -(1/4)e^8^x$

$\displaystyle u1 = -(1/32)e^8^x$

$\displaystyle u2' = 1/4$

$\displaystyle u2 = (1/4)x$

$\displaystyle Yp = (1/4)xe^4^x - (1/32)e^4^x$

so $\displaystyle Y = ce^-^4^x + ce^4^x + (1/4)xe^4^x - (1/32)e^4^x$

Not sure if I did something wrong because when I used undetermined coefficients to solve the problem, I only got $\displaystyle Yp = (1/4)x*e^4^x$...?

Thanks in advance.

It looks good. Note that we can say that $\displaystyle ce^{4x}-\frac{1}{32}e^{4x} \sim ke^{4x};\,\,k=c-\frac{1}{32}$, so it doesn't really add "new" information.