In this exercise you will solve the initial value problem v$^(\text{'}\text{'})-12$y$^(\text{'})+36$y$=($e$^(-6$x$))/(1+$x$^(2)),$v$(0)=1,$y$(0)=-5.[1]$ Let C$\_(1)$ and C$\_(2)$ be arbitrary constants. The general solution to the related homogeneous differentlal equation y$^(\text{'}\text{'})-12$y$^(\text{'})+36$y$=0$ is the function y$\_($h$)($x$)=$C$\_(1)$y$\_(1)($x$)+$C$\_(2)$y$\_(2)($x$)=$C$\_(1)$e$^(6$x$)+$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$^(\text{'}\text{'})+12$y$^(\text{'})+36$y$=($e$^(-10))/(1-$x$^(2))$ is of the form v$\_($p$)($x$)=$y$\_(1)($x$)$u$\_(1)($x$)+$y$\_(2)($x$)$u$\_(2)($x$)$ where u$\_(1)^(\text{'})($x$)=|$xe$^(-12$x$)|$ and u$\_(2)^(\text{'})($x$)=\backslash$Delta $-12$x $(3)$ The most general solution to the non homogeneous differential equation y$^(\text{'}\text{'})=12$y$^(\text{'})+36$y$=($e$^(-$t$))/(1-$x$^(2))$ is y$=$e$^(6$x$)\backslash$int$\_0^0+$xe$^(6$x$)(0)$dt$+($e$^(-17$t$))/(1+$z$^(2))\backslash$times $($e$^(-12$t$))/(1+$z$^(2))$dt Note: You can eam partial credit on this problem.