User:Egm4313.s12.team9.ehlers.km

= Problem 2.4: 5 =

Initial Information
Problem Statement: Find a general solution and check by substitution.

$$ \; y'' + 2\pi y' + \pi^2 y = 0 $$

Methods
$$ \; y'' + 2 \pi y' + \pi^2 y = 0 \to \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi^2 y = 0 $$

$$ \; \lambda = \frac{d}{dx}\ $$

$$ \; y'' + 2 \pi y' + \pi^2 y = 0 \to \lambda^2 + 2 \pi \lambda + \pi^2 = 0 $$

Solve for $$ \; \lambda $$, through factoring.

$$ \; (\lambda + \pi) (\lambda + \pi) = 0 $$

Roots of $$ \; \lambda = -\pi, -\pi $$

Solution
General Solution:

$$ \; y = c_1 e^{ \lambda _1 x} + c_2e^{ \lambda _2 x} $$

So the general solution becomes:

$$ \; y = c_1 e^{-\pi x} + c_2 x e^{-\pi x} $$

$$ \; y = e^{-\pi x} (c_1 + c_2 x) $$

Check by Substitution:

To check by substitution, you need to find $$ \; y, y', y'' $$ and plug them into the original ODE.

$$ \; y= e ^ {-\pi x} (c_1 + c_2 x) $$

$$ \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi x e ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2 x) + c_2 e^ {- \pi x} $$

$$ \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2 e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi x e^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) -2 \pi c_2 e^ {- \pi x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$ \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2 x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1 + c_2 x)]=0 $$

After substitution, it proves the general solution because the substitution equals zero.

= Problem 2.4: 6 =

Initial Information
Problem Statement: Find a general solution and check by substitution.
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle 10 y'' - 32 y' + 25.6 y = 0$$
 * $$\displaystyle $$
 * }

Methods
$$ \; 10 y'' - 32 y' + 25.6 y = 0 \to 10 \frac{d^2y}{dx^2} - 32 \frac{dy}{dx} + 25.6 y = 0 $$

$$ \; \lambda = \frac{d}{dx}\ $$

$$ \; 10 y'' - 32 y' + 25.6 y = 0 \to 10 \lambda^2 - 32 \lambda + 25.6 = 0 $$

Solve for $$ \; \lambda $$, by using the quadratic equation.

$$ \; \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

$$ \; \lambda = \frac{8}{5} $$

Solution
General solution:

$$ \; y = c_1 e^{ \lambda _1 x} + c_2e^{ \lambda _2 x} $$

The general solution then becomes:

$$ \; y = e^{\frac{8}{5}x}({c_1 + c_2 x}) $$

Check by Substitution:

To check by substitution, you need to find $$ \; y, y', y'' $$ and plug them into the original ODE.

$$ \; y = e^{\frac{8}{5}x}({c_1 + c_2 x}) $$

$$ \; y' = \frac{8}{5} e^{\frac{8}{5}x}(c_1 + c_2 x) + c_2 e^{\frac{8}{5}x} $$

$$ \; y'' = \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + \frac{16}{5}c_2 e^{\frac{8}{5}x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; 10 y - 32 y' + 25.6 y = 0 $$ :

$$ \; \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) - \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + 32 c_2 e^{\frac{8}{5}x} - 32 c_2 e^{\frac{8}{5}x} = 0 $$

After substitution, it proves the general solution because the substitution equals zero.