# 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)