User:Egm4313.s12.team11.sheider/R1.5: K 2011 p.59 pb. 4

Problem: 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:

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

Checking:

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 this solution is correct.