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)