User:Egm4313.s12.team8.kyle/R2.4

Problem Statement
Find a general solution. Check your answer by substitution. (Advanced Engineering Mathematics, Tenth Edition, Kresyzig, Problem set 2.2 pg.59)

5. $$y''+2{\pi}y'+{\pi}^2y=0$$
Find the characteristic equation:

$$r^2+2{\pi}r+{\pi}^2=0$$

A single real root exists at $$r={-\pi}$$ leading to a general solution of:

$$y=C_1e^{rx}+C_2xe^{rx}$$

Substituting $$r={-\pi}$$ gives the general solution:



To check by substitution first express $$y$$, $$y'$$, and $$y''$$:

$$y=C_1e^{{-\pi}x}+C_2xe^{{-\pi}x}$$

$$y'=-C_1{\pi}e^{{-\pi}x}-C_2x{\pi}e^{{-\pi}x}+C_2e^{{-\pi}x}$$

$$y''=C_1{\pi}^2e^{{-\pi}x}+C_2x{\pi}^2e^{{-\pi}x}-2C_2{\pi}e^{{-\pi}x}$$

Substitute into the original equation:

$$C_1{\pi}^2e^{{-\pi}x}+C_2x{\pi}^2e^{{-\pi}x}-2C_2{\pi}e^{{-\pi}x}+2{\pi}(-C_1{\pi}e^{{-\pi}x}-C_2x{\pi}e^{{-\pi}x}+C_2e^{{-\pi}x})+{\pi}^2C_1e^{{-\pi}x}+C_2xe^{{-\pi}x}=0$$

Canceling like terms leaves 0=0, showing the general solution is correct.

6. $$10y''-32y'+25.6y=0$$
First divide each term by the $$y''$$ coefficient of 10 to get the equation in standard form:

$$y''-3.2y'+2.56y=0$$

Find the characteristic equation:

$$r^2-3.2r+2.56=0$$

A single real root exists at $$r=1.6$$ leading to a general solution of:

$$y=C_1e^{rx}+C_2xe^{rx}$$

Substituting $$r=1.6$$ gives the general solution:



To check by substitution first express $$y$$, $$y'$$, and $$y''$$:

$$y=C_1e^{1.6x}+C_2xe^{1.6x}$$

$$y'=C_1{1.6}e^{1.6x}+C_2x{1.6}e^{1.6x}+C_2e^{1.6x}$$

$$y''=C_1{1.6}^2e^{1.6x}+C_2x1.6^2e^{1.6x}+2C_21.6e^{1.6x}$$

Substitute into the original equation:

$$C_1{1.6}^2e^{1.6x}+C_2x1.6^2e^{1.6x}+2C_21.6e^{1.6x}-3.2(C_1{1.6}e^{1.6x}+C_2x{1.6}e^{1.6x}+C_2e^{1.6x})+2.56(C_1e^{1.6x}+C_2xe^{1.6x})=0$$

Canceling like terms leaves 0=0, showing the general solution is correct.