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

${\displaystyle y''+6y'+8.96y=0\!}$

Solution

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

${\displaystyle {\frac {d^{2}y}{dx^{2}}}+6{\frac {dy}{dx}}+8.96y=0}$

Let ${\displaystyle {\frac {d}{dx}}=\lambda .}$

The characteristic equation of the given DE is

${\displaystyle \lambda ^{2}+6\lambda +8.96=0\!}$

Now, in order to solve for ${\displaystyle \lambda }$, we can use the quadratic formula:

${\displaystyle \lambda ={\frac {-(6)\pm {\sqrt {(6)^{2}-4(1)(8.96)}}}{2(1)}}}$

${\displaystyle \lambda ={\frac {-6\pm {\sqrt {36-35.84}}}{2}}}$

${\displaystyle \lambda ={\frac {-6\pm {\sqrt {0.16}}}{2}}}$

${\displaystyle \lambda ={\frac {-6\pm 0.4}{2}}}$

Therefore, we have:

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

and

${\displaystyle \lambda _{1}={\frac {-6-0.4}{2}}={\frac {-6.4}{2}}=-3.2}$

Thus, we have found that the general solution of the DE is actually:

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

Check:

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

what we found to be the general solution.

${\displaystyle {\frac {dy}{dx}}=y'=-2.8xc_{1}e^{-2.8x}+-3.2xc_{2}e^{-3.2x}\!}$

${\displaystyle {\frac {d^{2}y}{dx^{2}}}=y''=7.84xc_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}\!}$

Substituting the values of ${\displaystyle y,y'}$ and ${\displaystyle y''}$ in the given equation, we get:

${\displaystyle 7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}+6(-2.8c_{1}e^{-2.8x}-3.2c_{2}e^{-3.2x})+8.96(c_{1}e^{-2.8x}+c_{2}e^{-3.2x})=0\!}$

${\displaystyle 7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}-16.8c_{1}e^{-2.8x}-19.2c_{2}e^{-3.2x}+8.96c_{1}e^{-2.8x}+8.96c_{2}e^{-3.2x}=0\!}$

and thus:

${\displaystyle 0\equiv 0.\!}$

Therefore, the solution of the given DE is in fact:

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

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

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

Solution

The characteristic equation of this ODE is therefore:

${\displaystyle \lambda ^{2}+a\lambda +b=\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!}$

Evaluating the discriminant:

${\displaystyle a^{2}-4b=4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0\!}$

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

${\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))\!}$

Where: ${\displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {\pi ^{2}+4-{\frac {1}{4}}(4)^{2}}}={\sqrt {\pi ^{2}}}=\pi \!}$

And finally we find the general homogenous solution:

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

Check:

We found that:

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

Differentiating ${\displaystyle y\!}$ to obtain ${\displaystyle {y}'\!}$ and ${\displaystyle {y}''\!}$ respectively:

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

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

${\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

                 ${\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

Substituting these equations into the original ODE yields:

${\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

${\displaystyle +4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x)))+(\pi ^{2}+4)(e^{-2x}(Acos(\pi x)+Bsin(\pi x)))=0\!}$

${\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

${\displaystyle -8e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}$

${\displaystyle (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x))+(-4+4)\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+(\pi ^{2}-\pi ^{2})e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}$

${\displaystyle 0\equiv 0\!}$

Therefore, the solution is correct.