In this exercise you will solve the initial value problem

$y^{\text{'}\text{'}}-18y^{\text{'}}+81y=\frac{e^{-9x}}{1+x^2},y(0)=-2,y^{\text{'}}(0)=-10$

$(1)$ Let $C_1$ and $C_2$ be arbitrary constants. The general solution to the related homogeneous differential equation $y^{\text{'}\text{'}}-18y^{\text{'}}+81y=0$ 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$,C_{1}f(x)+C_{2}g(x)C_{1}g(x)+C_{2}f(x).$

$(2)$ The particular solution $y_p(x)$ to the differential equation $y^{\text{'}\text{'}}+18y^{\text{'}}+81y=\frac{e^{-9x}}{1+x^3}$ is of the form $y_p(x)=y_1(x)u_1(x)+y_2(x)u_2(x)$ where $u_1^{\text{'}}(x)=$ and $u_2^{\text{'}}(x)=$

$(3)$ The most general solution to the non$-$homogeneous differential equation $y^{\text{'}\text{'}}-18y^{\text{'}}+81y=\frac{e^{-n}}{1+x^2}$ is

$y=$

$$

$$

$$

$*dt+$

$dt$