In this exercise you will solve the initial value problem $$y''-8 y' +16 y =\frac{e^{-4x}}{1+x^{2}},\ \ y(0) = 2,\ \ y'(0) = -7.$$

(1) Let $C_1$ and $C_2$ be arbitrary constants. The general solution to the related homogeneous differential equation $y''-8 y' +16 y =0$ is the function $y_h(x)=C_1\ y_1(x)+C_2\ y_2(x)=C_1$ $+C_2$.

NOTE: The order in which you enter the answers is important; that is, $C_1 f(x) + C_2 g(x) \neq C_1 g(x) + C_2 f(x)$.

(2) The particular solution $y_p(x)$ to the differential equation $y''+8 y' +16 y =\frac{e^{-4x}}{1+x^{2}}$ is of the form $y_p(x)=y_1(x)\ u_1(x)+y_2(x)\ u_2(x)$ where $u_1'(x)=$ and $u_2'(x)=$ .

(3) The most general solution to the non-homogeneous differential equation $y''-8 y' +16 y =\frac{e^{-4x}}{1+x^{2}}$ is

$y=$

$+$

$\displaystyle\int$

$dt+$

$\displaystyle\int$

$dt$