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:

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

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.

$$ \frac{dy}{dx}=y'=-2.8x c_1 e^{-2.8x} + -3.2x c_2 e^{-3.2x} \! $$

$$ \frac{d^2y}{dx^2}=y''=7.84x c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} \! $$

Substituting the values of $$ y, y' $$ and $$ y'' $$ in the given equation, we get:

$$ 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}+6(-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x})+ 8.96(c_1e^{-2.8x}+c_2e^{-3.2x}) = 0 \! $$

$$ 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}-16.8c_1e^{-2.8x}-19.2c_2e^{-3.2x}+ 8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x} = 0 \! $$

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.