ODE4.1

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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)]\,



Ordinary Differential Equations

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