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

$y^{″}+6y^{\prime }+8.96y=0$

Solution

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

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

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.

$\frac{dy}{dx}=y^{\prime }=-2.8x{c}_1{e}^{-2.8x}+-3.2x{c}_2{e}^{-3.2x}$

$\frac{d^{2}y}{d{x}^2}=y^{″}=7.84x{c}_1{e}^{-2.8x}+10.24c_2{e}^{-3.2x}$

Substituting the values of $y,y^{\prime }$ 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

$y{\text{'}}^{\prime }+4y^{\prime }+({\pi }^2+4)y=0$

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 $y\text{'}$ and $y{\text{'}}^{\prime }$ respectively:

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

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

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

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

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.