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{\bf Second order differential equation:}
\v
Linear equation with constant coefficients:
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\hskip 10pt
If the second order differential equation is 

\centerline{$ay'' + by' + cy = 0$,} 

\centerline{then $y = e^{rt}$ is a solution}

Need to have two independent solutions.


$ay'' + by' + cy = 0$, ~~$y = e^{rt}$, then 
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$ar^2e^{rt} + bre^{rt} + ce^{rt} = 0$ implies $ar^2 + br + c = 0$,

Suppose $r = r_1, r_2$ are solutions to $ar^2 + br + c = 0$
\v

\centerline{$r_1, r_2 = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$$


If $r_1 \not= r_2$, then $b^2 - 4ac \not= 0$.  Hence a general solution is
{$y = c_1e^{r_1t} + c_2e^{r_2t}$}
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If $b^2 - 4ac > 0$, general solution is $y = c_1e^{r_1t} + c_2e^{r_2t}$.

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If $b^2 - 4ac < 0$, change format to linear combination of real-valued functions instead of 
complex valued functions by using Euler's formula.

general solution is $y = c_1 e^{dt} cos (nt) + c_2 e^{dt} sin (nt)$ where $r = d \pm in$

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If $b^2 - 4ac = 0$, $r_1 = r_2$, so need 2nd (independent) solution:  $te^{r_1t}$

Hence general solution is $y = c_1e^{r_1t} + c_2te^{r_1t}$.

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Initial value problem:  use $y(0) = y_0$, $y'(0) = y_0'$ to solve for $c_1, c_2$ to find unique 
solution. 

Derivation of general solutions:

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If $b^2 - 4ac > 0$ we guessed $e^{rt}$ is a solution and noted that any linear combination of 
solutions is a solution to a homogeneous linear differential equation.

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If $b^2 - 4ac < 0$, :

Changed format of $y = c_1e^{r_1t} + c_2e^{r_2t}$ to linear combination of real-valued 
functions 
instead of
complex valued functions by using Euler's formula:

\centerline{$e^{it} = cos(t) + i sin(t)$}

Hence $e^{(d + in)t} =e^{dt}e^{int} = e^{dt}[cos (nt) + i sin (nt)]$

Let $r_1 = d + in$, $r_2 = d - in$

$y = c_1e^{r_1t} + c_2e^{r_2t} $
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$= 
c_1e^{dt}[cos (nt) + i sin (nt)] +c_2e^{dt}[cos (-nt) + i sin (-nt)] $
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$=c_1e^{dt}cos (nt) + ic_1 e^{dt} sin (nt) + c_2e^{dt}cos (nt) - ic_2 e^{dt} sin (nt) $
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=$(c_1 + c_2)e^{dt}cos (nt) + i(c_1 - c_2) e^{dt}sin (nt)$
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\hskip 28pt$=k_1 e^{dt} cos (nt) + k_2e^{dt} sin (nt)$

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If $b^2 - 4ac = 0$, then $r_1 = r_2$.  \hfil \break
Hence one solution is $y = e^{r_1t}$  Need second 
solution.
\u
If $y = e^{rt}$ is a solution, $y = ce^{rt}$ is a solution.
\u
How about $y = v(t)e^{rt}$?

$y' = v'(t)e^{rt} + v(t)re^{rt}$

$y'' = v''(t)e^{rt} +  v'(t)re^{rt} + v'(t)re^{rt} +  v(t)r^2e^{rt}$
\u
$\hskip 14pt = 
v''(t)e^{rt} +  2v'(t)re^{rt} +  v(t)r^2e^{rt} 
$


$ay'' + by' + cy = 0$

$a(v''(t)e^{rt} + 2v'(t)re^{rt} + v(t)r^2e^{rt}) + b(v'(t)e^{rt} + v(t)re^{rt}) + c(v(t)e^{rt})
= 0$

$a(v''(t) + 2v'(t)r + v(t)r^2) + b(v'(t) + v(t)r) + cv(t)
= 0$

$av''(t) + 2av'(t)r + av(t)r^2 + bv'(t) + bv(t)r + cv(t)
= 0$


$av''(t) 
+ (2ar+b)v'(t) 
+ (ar^2
+ br + c)v(t)
= 0$

$av''(t) 
+ (2a({-b \over 2a})+b)v'(t) 
+ 0
= 0$

Since
$ar^2 + br + c = 0$
and $r = {-b \over 2a}$


$av''(t) 
+ (-b + b)v'(t) 
= 0$


$av''(t)  = 0$

Hence 
$v''(t)  = 0$

$v'(t)  = k_1$

$v(t)  = k_1t + k_2$


Hence $v(t)e^{r_1t}   = (k_1t + k_2)e^{r_1t}$ is a soln

Hence $te^{r_1t}$ is a nice second solution.

Hence general solution is $y = c_1 e^{r_1t}+ c_2 te^{r_1t}$

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3.6  Nonhomogeneous Equations:  $ay'' + by' + cy = g(t)$

Method of Undetermined Coefficients.


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