# ODE4.1

$y''+2y'+5y=e^{{-x}}\sin(2x)\,$

First, solve the homogeneous equation to get $y_{h}$. Then, guess a particular solution $y_{p}$ and add them together to get the general solution $y=y_{h}+y_{p}$.

$y_{h}''+2y_{h}'+5y_{h}=0\,$

Set up the characteristic equation and solve.

$m^{2}+2m+5=0\,$

So,

$m={\frac {-2\pm {\sqrt {4-20}}}{2}}\,$

$m=-1\pm 2i\,$

Therefore, the solution to the homogeneous equation is

$y_{h}=c_{1}e^{{-x}}\cos(2x)+c_{2}e^{{-x}}\sin(2x)\,$
$y_{h}=e^{{-x}}[c_{1}\cos(2x)+c_{2}\sin(2x)]\,$

Now, check the homogeneous solution to see if any new information can be gained. Start by taking the first and second derivatives of $y_{h}$.

$y_{h}'=e^{{-x}}[(-2c_{1}-c_{2})\sin(2x)+(-c_{1}+2c_{2})\cos(2x)]\,$

$y_{h}''=e^{{-x}}[(-4c_{1}-2c_{2})\cos(2x)-(-2c_{1}+4c_{2})\sin(2x)]-e^{{-x}}[(-2c_{1}-c_{2})\sin(2x)+(-c_{1}+2c_{2})\cos(2x)]\,$
$y_{h}''=e^{{-x}}[(-3c_{1}-4c_{2})\cos(2x)+(4c_{1}-3c_{2})\sin(2x)]\,$

Substitution of these derivatives into the homogeneous version of the original differential equation and collection of the exponential and trig functions gives the following relations.

$e^{{-x}}\cos(2x)[-4c_{1}+c_{1}-4c_{2}-2c_{1}+2c_{2}+5c_{1}]=0\,$
$e^{{-x}}\sin(2x)[4c_{1}-3c_{2}-4c_{1}-2c_{2}+5c_{2}]=0\,$

So from the first equation, $c_{1}$ is arbitrary and $-2c_{2}=0$, so $c_{2}=0$. From the second equation, no new information is obtained. Therefore, a better homogeneous equation is

$y_{h}=c_{1}e^{{-x}}\cos(2x)\,$

Now a particular solution should be guessed. Based on the nonhomogeneous part of the original DE, $e^{{-x}}\sin(2x)$ , a first guess is

$y_{p}=Ae^{{-x}}[c_{3}\cos(2x)+c_{4}\sin(2x)]\,$, with $A$, $c_{3}$, and $c_{4}$ undetermined constants.

But, the homogeneous solution appears as one of the terms of the guessed particular solution, so that term should be multiplied by $x$. Now, a guess for the particular solution is

$y_{p}=Ae^{{-x}}[c_{3}x\cos(2x)+c_{4}\sin(2x)]\,$

Find the first and second derivatives of $y_{p}$.

$y_{p}'=Ae^{{-x}}[-2c_{3}x\sin(2x)+c_{3}\cos(2x)+2c_{4}\cos(2x)]-Ae^{{-x}}[c_{3}x\cos(2x)+c_{4}\sin(2x)]\,$
$y_{p}'=Ae^{{-x}}[-c_{3}x\cos(2x)-2c_{3}x\sin(2x)+(c_{3}+2c_{4})\cos(2x)-c_{4}\sin(2x)]\,$

$y_{p}''=Ae^{{-x}}[2c_{3}x\sin(2x)-c_{3}\cos(2x)-4c_{3}x\cos(2x)-2c_{3}\sin(2x)-2(c_{3}+2c_{4})\sin(2x)-2c_{4}\cos(2x)]\,$
$-Ae^{{-x}}[-c_{3}x\cos(2x)-2c_{3}x\sin(2x)+(c_{3}+2c_{4})\cos(2x)-c_{4}\sin(2x)]\,$
$y_{p}''=Ae^{{-x}}[-3c_{3}x\cos(2x)+4c_{3}x\sin(2x)+(-2c_{3}-4c_{4})\cos(2x)+(-4c_{3}-3c_{4})\sin(2x)]=e^{{-x}}\sin(2x)\,$

Substitute these derivatives into the original DE to get these relations:

$Ae^{{-x}}[(-3c_{3}-2c_{3}+5c_{3})x\cos(2x)+(2c_{3}+2c_{3}-4c_{3})x\sin(2x)+\,$
$(-c_{3}-2c_{4}-c_{3}-2c_{4}+2c_{3}+4c_{4})\cos(2x)+(-4c_{3}-4c_{4}+c_{4}-2c_{4}+5c_{4})\sin(2x)]\,$

So comparing the coefficients of $x\sin(2x)$ and $x\cos(2x)$ gives no information about $c_{3}$ except that it is arbitrary so far. The other terms give the relations

$-4c_{3}=1\,$ so that $c_{3}={\frac {-1}{4}}\,$, and
$c_{4}$ is arbitrary, so that the particular solution is

$y_{p}=e^{{-x}}[-{\frac {1}{4}}x\cos(2x)+c_{4}\sin(2x)]\,$

Finally, the general solution is

$y(x)=y_{h}+y_{p}=e^{{-x}}[-{\frac {1}{4}}x\cos(2x)+c_{1}\cos(2x)+c_{4}\sin(2x)]\,$