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.

$y^{″}+4y^{\prime }+4y=e^{-x}cos(x)$

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

$y_h=c_1{e}^{\lambda x}+c_{2}x{e}^{\lambda x}$

$y_h=c_1{e}^{-2x}+c_{2}x{e}^{-2x}$

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

${y_p}^{″}+4{y_p}^{\prime }+4y_p=r(x)$

where

${y_p}^{\prime }=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]$

${y_p}^{\prime }=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]$

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

${y_p}^{″}+4{y_p}^{\prime }+4y_p=r(x)$

$e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)$

$e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)$

$2Mcos(x)-2Ksin(x)=cos(x)$

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

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

$y(x)=c_1{e}^{-2x}+c_{2}x{e}^{-2x}+\frac{1}{2}{e}^{-x}sin(x)$

   $y(x)=e^{-2x}(c_1+c_{2}x)+\frac{1}{2}{e}^{-x}sin(x)$

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.

$y^{″}+y^{\prime }+({\pi }^2+\frac{1}{4})y=e^{-\frac{x}{2}}sin(\pi x)$

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 $y_p(x)=x{e}^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]$

differentiating

${y_p}^{\prime }=\pi x{e}^{-\frac{x}{2}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac{x}{2}}-\frac{1}{2}x{e}^{-\frac{x}{2}})[Kcos(\pi x)+Msin(\pi x)]$

${y_p}^{\prime }=e^{-\frac{x}{2}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-\frac{1}{2}x)[Kcos(\pi x)+Msin(\pi x)])$

${y_p}^{\prime }=e^{-\frac{x}{2}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-\frac{1}{2}xKcos(\pi x)-\frac{1}{2}xMsin(\pi x)]$

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

${y_p}^{\prime }=e^{-\frac{x}{2}}[(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK+M-\frac{1}{2}xM)sin(\pi x)]$

next we substitute the above equations into the ODE

${y_p}^{″}+{y_p}^{\prime }+({\pi }^2+\frac{1}{4})y_p=r(x)$

$e^{-\frac{x}{2}}[(2\pi M-\pi xM+(\frac{1}{4}-{\pi }^2)xK-K)cos(\pi x)+((\frac{1}{4}-{\pi }^2)xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-\frac{x}{2}}[(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK+M-\frac{1}{2}xM)sin(\pi x)]+({\pi }^2+\frac{1}{4})x{e}^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]=e^{-\frac{x}{2}}sin(\pi x)$

$e^{-\frac{x}{2}}[(2\pi M-\pi xM+(\frac{1}{4}-{\pi }^2)xK-K)cos(\pi x)+((\frac{1}{4}-{\pi }^2)xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK+M-\frac{1}{2}xM)sin(\pi x)+({\pi }^2+\frac{1}{4})x(Kcos(\pi x)+Msin(\pi x))]=e^{-\frac{x}{2}}sin(\pi x)$

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

$2\pi Mcos(\pi x)-2\pi Ksin(\pi x)=sin(\pi x)$

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

$y_p(x)=-\frac{1}{2\pi }x{e}^{-\frac{x}{2}}cos(\pi x)$

General Solution

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

$y=y_h+y_p$

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

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

Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)