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

R5.6

solved by Luca Imponenti

Problem Statement

Complete the solution to the following problem

$y^{″}+4y^{\prime }+13y=2e^{-2x}cos(3x)$

where

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

and

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

Find the overall solution $y(x)$ corresponds to the initial condition:

$y(0)=1,y^{\prime }(0)=0$

Plot the solution over 3 periods.

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

Particular Solution

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

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

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

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

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

Overall Solution

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

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

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

To find A and B we apply the initial conditions

$y(0)=1,y^{\prime }(0)=0$

$y(0)=1=A$

Taking the derivative

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

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

$y^{\prime }(0)=0=3B-2$

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

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

File:Plot56.jpg

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