University of Florida/Egm4313/s12.team11.perez.gp/R2.3

Problem Statement (K 2011 p.59 pb. 3)

Find a general solution. Check your answer by substitution.

Given

$y''+6y'+8.96y=0\!$ Solution

We can write the above differential equation in the following form:

${\frac {d^{2}y}{dx^{2}}}+6{\frac {dy}{dx}}+8.96y=0$ Let ${\frac {d}{dx}}=\lambda .$ The characteristic equation of the given DE is

$\lambda ^{2}+6\lambda +8.96=0\!$ Now, in order to solve for $\lambda$ , we can use the quadratic formula:

$\lambda ={\frac {-(6)\pm {\sqrt {(6)^{2}-4(1)(8.96)}}}{2(1)}}$ $\lambda ={\frac {-6\pm {\sqrt {36-35.84}}}{2}}$ $\lambda ={\frac {-6\pm {\sqrt {0.16}}}{2}}$ $\lambda ={\frac {-6\pm 0.4}{2}}$ Therefore, we have:

$\lambda _{1}={\frac {-6+0.4}{2}}={\frac {-5.8}{2}}=-2.8$ and

$\lambda _{1}={\frac {-6-0.4}{2}}={\frac {-6.4}{2}}=-3.2$ Thus, we have found that the general solution of the DE is actually:

$y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}\!$ Check:

To check if $y$ is indeed the solution of the given DE, we can differentiate the

what we found to be the general solution.

${\frac {dy}{dx}}=y'=-2.8xc_{1}e^{-2.8x}+-3.2xc_{2}e^{-3.2x}\!$ ${\frac {d^{2}y}{dx^{2}}}=y''=7.84xc_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}\!$ Substituting the values of $y,y'$ and $y''$ in the given equation, we get:

$7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}+6(-2.8c_{1}e^{-2.8x}-3.2c_{2}e^{-3.2x})+8.96(c_{1}e^{-2.8x}+c_{2}e^{-3.2x})=0\!$ $7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}-16.8c_{1}e^{-2.8x}-19.2c_{2}e^{-3.2x}+8.96c_{1}e^{-2.8x}+8.96c_{2}e^{-3.2x}=0\!$ and thus:

$0\equiv 0.\!$ Therefore, the solution of the given DE is in fact:

$y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}\!$ Problem Statement (K 2011 p.59 pb. 4)

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given

${y}''+4{y}'+(\pi ^{2}+4)y=0\!$ Solution

The characteristic equation of this ODE is therefore:

$\lambda ^{2}+a\lambda +b=\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!$ Evaluating the discriminant:

$a^{2}-4b=4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0\!$ Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))\!$ Where: $\omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {\pi ^{2}+4-{\frac {1}{4}}(4)^{2}}}={\sqrt {\pi ^{2}}}=\pi \!$ And finally we find the general homogenous solution:

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$ Check:

We found that:

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$ Differentiating $y\!$ to obtain ${y}'\!$ and ${y}''\!$ respectively:

${y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!$ ${y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))\!$ ${y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$ ${y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$ Substituting these equations into the original ODE yields:

$4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$ $+4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x)))+(\pi ^{2}+4)(e^{-2x}(Acos(\pi x)+Bsin(\pi x)))=0\!$ $4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$ $-8e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!$ $(4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x))+(-4+4)\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+(\pi ^{2}-\pi ^{2})e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!$ $0\equiv 0\!$ Therefore, the solution is correct.