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

solved by Luca Imponenti

Complete the solution to the following problem

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

where

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

and

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

Find the overall solution ${\displaystyle y(x)\!}$ corresponds to the initial condition:

${\displaystyle y(0)=1\ ,\ y'(0)=0\!}$

Plot the solution over 3 periods.

Taking the derivatives of the particular solution ${\displaystyle y_{p}(x)\!}$

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

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

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

Plugging these into the ODE yields

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

Equating like terms allows us to solve for M and N

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

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

${\displaystyle -6M=0\!}$

${\displaystyle 6N=2\!}$

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

So the particular solution is

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

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

${\displaystyle y(x)=y_{h}(x)+y_{p}(x)\!}$

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

To find A and B we apply the initial conditions

${\displaystyle y(0)=1\ ,\ y'(0)=0\!}$

${\displaystyle y(0)=1=A\!}$

Taking the derivative

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

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

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

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

Giving us the overall solution

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

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

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

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