# University of Florida/Egm4313/s12.team11.imponenti/R3.8

## Problem 3.8

solved by Luca Imponenti

### Kreyszig 2011 pg.84 problem 5

#### Problem Statement

Find a (real) general solution. State which rule you are using. Show each step of your work.

$y''+4y'+4y=e^{-x}cos(x)\!$ #### Homogeneous Solution

To find the homogeneous solution, $y_{h}\!$ , we must find the roots of the equation

$\lambda ^{2}+4\lambda +4=0\!$ $(\lambda +2)(\lambda +2)=0\!$ $\lambda =-2\!$ We know the homogeneous solution for the case of a double root to be

$y_{h}=c_{1}e^{\lambda x}+c_{2}xe^{\lambda x}\!$ $y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}\!$ #### Particular Solution

We have the following excitation

$r(x)=e^{-x}cos(x)\!$ From table 2.1, K 2011, pg. 82, we have

$y_{p}(x)=e^{-x}[Kcos(x)+Msin(x)]\!$ Since this does not correspond to our homogeneous solution we can use the Basic Rule (a), K 2011, pg. 81 to solve for the particular solution

$y_{p}''+4y_{p}'+4y_{p}=r(x)\!$ where

$y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!$ $y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$ and

$y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$ $y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!$ $y_{p}''=e^{-x}[2Ksin(x)-2Mcos(x)]\!$ Plugging these equations back into the differential equation

$y_{p}''+4y_{p}'+4y_{p}=r(x)\!$ $e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!$ $e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!$ $2Mcos(x)-2Ksin(x)=cos(x)\!$ from the above equation it is obvious that $K=0\!$ and $M={\frac {1}{2}}\!$ therefore the particular solution to the differential equation is

$y_{p}(x)={\frac {1}{2}}e^{-x}sin(x)\!$ #### General Solution

The general solution will be the summation of the homogeneous and particular solutions

$y(x)=y_{h}(x)+y_{p}(x)\!$ $y(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}+{\frac {1}{2}}e^{-x}sin(x)\!$ $y(x)=e^{-2x}(c_{1}+c_{2}x)+{\frac {1}{2}}e^{-x}sin(x)\!$ The coefficients $c_{1}\!$ and $c_{2}\!$ can be readily solved for given either initial conditions or boundary value conditions.

### Kreyszig 2011 pg.84 problem 6

#### Problem Statement

Find a (real) general solution. State which rule you are using. Show each step of your work.

$y''+y'+(\pi ^{2}+{\frac {1}{4}})y=e^{-{\frac {x}{2}}}sin(\pi x)\!$ #### Homogeneous Solution

To find the homogeneous solution, $y_{h}\!$ , we must find the roots of the equation

$\lambda ^{2}+\lambda +(\pi ^{2}+{\frac {1}{4}})=0\!$ $\lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\!$ with $a=1,b=1,c=(\pi ^{2}+{\frac {1}{4}})\!$ $\lambda ={\frac {-1\pm {\sqrt {1^{2}-4*1*(\pi ^{2}+{\frac {1}{4}})}}}{2*1}}\!$ $\lambda =-\alpha \pm i\omega =-{\frac {1}{2}}\pm \pi i\!$ We know the homogeneous solution for the case of a double root to be

$y_{h}=e^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!$ $y_{h}=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]\!$ #### Particular Solution

We have the following excitation

$r(x)=e^{-{\frac {x}{2}}}sin(\pi x)\!$ From table 2.1, K 2011, pg. 82, we have

$y_{p}(x)=e^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!$ Since this corresponds to our homogeneous solution we must use the Modification Rule (b), K 2011, pg. 81 to solve for the particular solution

so $y_{p}(x)=xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!$ differentiating

$y_{p}'=\pi xe^{-{\frac {x}{2}}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac {x}{2}}-{\frac {1}{2}}xe^{-{\frac {x}{2}}})[Kcos(\pi x)+Msin(\pi x)]\!$ $y_{p}'=e^{-{\frac {x}{2}}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-{\frac {1}{2}}x)[Kcos(\pi x)+Msin(\pi x)])\!$ $y_{p}'=e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!$ and

$y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)]-{\frac {1}{2}}e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!$ $y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{4}}xKcos(\pi x)+{\frac {1}{4}}xMsin(\pi x)]\!$ grouping cosine and sine terms we get

$y_{p}''=e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]\!$ and

$y_{p}'=e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]\!$ next we substitute the above equations into the ODE

$y_{p}''+y_{p}'+(\pi ^{2}+{\frac {1}{4}})y_{p}=r(x)\!$ $e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]+(\pi ^{2}+{\frac {1}{4}})xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!$ $e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)+(\pi ^{2}+{\frac {1}{4}})x(Kcos(\pi x)+Msin(\pi x))]=e^{-{\frac {x}{2}}}sin(\pi x)\!$ $e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K+\pi xM+K-{\frac {1}{2}}xK+(\pi ^{2}+{\frac {1}{4}})xK)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K-\pi xK+M-{\frac {1}{2}}xM+(\pi ^{2}+{\frac {1}{4}})xM)sin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!$ $2\pi Mcos(\pi x)-2\pi Ksin(\pi x)=sin(\pi x)\!$ after cancelling terms; we can equate cosine and sine coefficients to get two equations

$2\pi M=0\!$ $-2\pi K=1\!$ so $M=0\!$ and $K=-{\frac {1}{2\pi }}\!$ and the particular solution to the ODE is

$y_{p}(x)=-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!$ #### General Solution

The general solution will be the summation of the homogeneous and particular solutions

$y=y_{h}+y_{p}\!$ $y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!$ $y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)-{\frac {1}{2\pi }}xcos(\pi x)]\!$ Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)