Showing $\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=e^{ax}\cos(bx).$ I've got:
$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}.$$
Could someone show me how it simplifies to:
$e^{ax} \cos(bx)$?
It looks like the denominator is canceled by the terms that are being added, but then how do I get rid of one of the cosines?

You use the distributive law, which says that $(X+Y)\cdot Z=(X\cdot Z)+(Y\cdot Z)$ for any $X$, $Y$, and $Z$. In your case, we have $X=b^2$, $Y=a^2$, and $Z=\cos(bx)$, and so
$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=\frac{e^{ax}((b^2+a^2)\cos(bx))}{(a^2+b^2)}=\frac{e^{ax}((a^2+b^2)\cos(bx))}{(a^2+b^2)}=e^{ax}\cos(bx).$$
