University of Florida/Egm4313/s12.team11.imponenti/R5.5

R5.6
solved by Luca Imponenti

Problem Statement
Complete the solution to the following problem


 * $$y''+4y'+13y=2e^{-2x}cos(3x)\!$$

where

$$y_h=e^{-2x}[Acos(3x)+Bsin(3x)]\!$$

and

$$y_p=xe^{-2x}[Mcos(3x)+Nsin(3x)]\!$$

Find the overall solution $$y(x)\!$$ corresponds to the initial condition:
 * $$y(0)=1 \, \ y'(0)=0\!$$
 * Plot the solution over 3 periods.

Taking the derivatives of the particular solution $$y_p(x)\!$$

Particular Solution
$$y_p=xe^{-2x}[Mcos(3x)+Nsin(3x)]\!$$

$$y'_p=e^{-2x}[sin(3x)(N-2Nx-3Mx)+cos(3x)(3Nx+M-2Mx)]\!$$

$$y''_p=e^{-2x}[sin(3x)(12Mx-6M-5Nx-4N)+cos(3x)(6N-5Mx-4M-12Nx)]\!$$

Plugging these into the ODE yields

$$sin(3x)(6M-12Mx+5Nx+4N)+cos(3x)(5Mx+4M+12Nx-6N)+\!$$ $$4[sin(3x)(N-2Nx-3Mx)+cos(3x)(3Nx+M-2Mx)]+$$ $$13x[Mcos(3x)+Nsin(3x)]=2cos(3x)\!$$

Equating like terms allows us to solve for M and N

$$sin(3x)[(12Mx-6M-5Nx-4N)+4(N-2Nx-3Mx)+13Nx]=0\!$$

$$cos(3x)[(6N-5Mx-4M-12Nx)+4(3Nx+M-2Mx)+13Mx]=2cos(3x)\!$$

$$-6M=0\!$$

$$6N=2\!$$

$$M=0 \, \ N=\frac{1}{3}\!$$

So the particular solution is

$$y_p=\frac{1}{3}xe^{-2x}sin(3x)\!$$

Overall Solution
The overall solution in the sum of the homogeneous and particular solutions

$$y(x)=y_h(x)+y_p(x)\!$$

$$y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]+\frac{1}{3}xe^{-2x}sin(3x)\!$$

To find A and B we apply the initial conditions

$$y(0)=1 \, \ y'(0)=0\!$$

$$y(0)=1=A\!$$

Taking the derivative

$$y'(x)=\frac{d}{dx}[e^{-2x}[cos(3x)+Bsin(3x)]+\frac{1}{3}xe^{-2x}sin(3x)]\!$$

$$y'(x)=e^{-2x}[(3B+x-2)cos(3x)-(2B+\frac{2}{3}x+\frac{8}{3})sin(3x)]\!$$

$$y'(0)=0=3B-2\!$$

$$B=\frac{2}{3}\!$$

Giving us the overall solution

$$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)+\frac{1}{3}xsin(3x)]\!$$

Plot
The period for $$cos(3x) \, \ sin(3x)\!$$ is $$\frac{2\pi}{3}$$

Plotting the solution $$y(x)\!$$ over 3 periods yields



Egm4313.s12.team11.imponenti (talk) 01:56, 30 March 2012 (UTC)