University of Florida/Egm4313/s12.team8.dupre/R2.3

Problem Statement
Find a general solution. Check your answer by substitution. a) $$\displaystyle y+6y'+8.96y=0$$    (3-1)  b)$$\displaystyle y+4y'+(\pi^2+4)y=0$$    (3-2)

Solution
The quadratic formula is necessary for these solutions: $$\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Part a
Plugging into the quadratic formula: $$\displaystyle \frac{-6\pm \sqrt{6^2-(4)(1)(8.96)}}{2(1)}=\frac{-6\pm .4}{2}$$ This shows us that the roots of the equation are: $$\displaystyle \lambda_{1} =-2.8, \lambda _{2}=-3.2$$ Therefore, the general equation is: $$\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}$$    (3-3)

Substitution
We need to first find the first and second derivatives of equation (3-3): $$\displaystyle y'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}$$ $$\displaystyle y''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}$$ Plugging into equation (3-1), we find: $$\displaystyle (10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x})+6(-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x})+8.96(c_{1}e^{-3.2x}+c_{2}e^{-2.8x})=0$$    (3-4) Continuing to solve: $$\displaystyle 19.2c_{1}e^{-3.2x}+16.8c_{2}e^{-2.8x} -19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}=0$$    (3-5) This shows that the general equation is correct, since everything cancels out to 0.

Part b
Plugging into the quadratic formula: $$\displaystyle \frac{4\pm\sqrt{4^{2}-4(1)(\pi^2 +4)}}{2(1)}=\frac{-4\pm (-4)(1))(pi^2+4)}{2(1)} $$ The roots are, therefore:  $$\displaystyle \lambda _{1}=-2-\pi i, \lambda _{2}=-2+\pi i$$  Therefore, the general solution to (3-2) is:  $$\displaystyle y=c_{1}\cos(\pi x)e^{-2x} + c_{2}\sin(\pi x)e^{-2x}$$     (3-6)

Substitution
We must first find the first and second derivatives of equation (3-6): $$\displaystyle y'= -2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} + (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}$$ $$\displaystyle y''= - 2(-c_{1} \pi \sin (\pi x) + c_{2}\pi \cos (\pi x))e^{-2x} + (-c_{1} \pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x}$$ Plugging into equation (3-2): $$\displaystyle - 2(-c_{1} \pi \sin (\pi x) + c_{2}\pi \cos (\pi x))e^{-2x} + (-c_{1} \pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x}...$$ $$\displaystyle+4[-2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} + (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}]+(\pi^2+4)(c_{1}\cos(\pi x)e^{-2x} + c_{2}\sin(\pi x)e^{-2x})=0$$ Finally, plugging (3-6) and it's first and second derivatives into equation (3-2), we find: $$\displaystyle 4(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} - 2(-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x} - 2(-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}...$$ $$\displaystyle+(-c_{1}\pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x} +4 -2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x}) +...$$ $$\displaystyle (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}+(\pi^{2}+4)((c_{1}\cos(\pi x) + c_{2}\sin(\pi x))e^{-2x})=0$$  Since this equals 0, we know that the general equation (3-6) is correct.