User:Egm4313.s12.team13.Neisha.R1

Problem Statement
5. Find a general solution. Check by substitution.

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

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi ^{2}y=0   $$ (5.0b)

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

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$ (5.1b)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$ (5.2b)

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

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x}  $$ (5.3b)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$ (5.4b)

Check by Substitution:

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

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$(5.5b)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$ (5.6b)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Section 2 Notes