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3.1 - 3.4:  Homogeneous  linear differential 
equation
\t
\centerline{$a(t)y'' + b(t)y' + c(t)y = 0$}

\t\t

Thm 3.2.2:  If $y = \phi_1(t)$ and $y = \phi_2(t)$ are two solutions to a \underbar{homogeneous} linear differential 
equation, then 
$$y = c_1\phi_1(t) + c_2\phi_2(t)$$ is also a solution to this {homogeneous} linear differential equation.



Proof of thm 3.2.2:  Plug $y = c_1\phi_1(t) + c_2\phi_2(t)$ to see that it satisfies the {homogeneous} linear differential equation.

\hr

To solve homogeneous  linear differential 
equation with constant coefficients
\t
\centerline{$ay'' + by' + cy = 0$}

Guess $y = e^{rt}$ for HOMOGENEOUS equation and plug in:

$y' = re^{rt}$, $y'' = r^2e^{rt}$

 $ay'' + by' + cy = 0$
 
$  ar^2e^{rt} + bre^{rt}  + ce^{rt} = 0 ~~\Rightarrow~~$
$  e^{rt}(ar^2 + br + c) = 0$

$ e^{rt} \not= 0$, thus can divide both sides by $e^{rt}$: 
~~~{$ ar^2 + br + c = 0$}



Suppose $r = r_1, r_2$ are solutions to $ar^2 + br + c = 0$
\vskip 10pt
\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
{$y = c_1e^{r_1t} + c_2e^{r_2t}$}
\vskip 5pt
\hrule

If $b^2 - 4ac > 0$, general solution is $y = c_1e^{r_1t} + c_2e^{r_2t}$.

\vskip 5pt
\hrule

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 $y = c_1 e^{xt} cos (yt) + c_2 e^{xt} sin (yt)$ where $r = x \pm iy$

\vskip 5pt
\hrule

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

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

\vskip 5pt
\hrule

Example 1:  ~~$4y'' + y' = 0 ~~\Rightarrow~~ 4r^2 + r = 0$  

$4r^2 + r =  r(4r + 1) = 0$.   Thus $r = 0, -{1 \over 4}$  


Hence $y = e^{0t} = 1$ and $y = e^{-{1 \over 4}t}$ are both solutions.

Thus the general solution is $y = c_1(1) + c_2(e^{-{t \over 4}})$ 

\hr

Example 2:  ~~$4y'' + y = 0 ~~\Rightarrow~~ 4r^2 + 1 = 0$  

$4r^2 + 1 = 0  ~~\Rightarrow~~ 4r^2 = -1$.   Thus $r =  \pm {i \over 2} =  0  \pm  {1 \over 2}i$  


Thus the general solution is $y = c_1(cos({t \over 2}) ) + c_2(sin({t \over 2}))$ 


\hr

Example 3:  ~~$4y'' + 2y' + y = 0 ~~\Rightarrow~~ 4r^2 + 2r + 1 = 0$  

Thus $r= {-2 \pm \sqrt{2^2 - 4(4)(1)} \over 2(4)}
= {-2 \pm \sqrt{2^2[1 - (4)(1)]} \over 2(4)}$

$= {-2 \pm 2\sqrt{-3} \over 2(4)}
= {2(-1 \pm i\sqrt{3}) \over 2(4)}
= {-1 \pm i\sqrt{3} \over 4}
= {-1  \over 4} \pm i { \sqrt{3} \over 4}
$


Thus the general solution is $y = c_1(e^{-t  \over 4}cos ( {\sqrt{3} \over 4}t)) + c_2(e^{-t  \over 4}cos ( {\sqrt{3} \over 4}t))$


\hr

Example 4:  ~~$4y'' + 4y' + y = 0 ~~\Rightarrow~~ 4r^2 + 4r + 1 = 0$  

$ 4r^2 + 4r + 1 =  (2r + 1)(2r + 1) = 0$.   Thus $r = -{1 \over 2}, -{1 \over 2}$  


Hence  $y = e^{-{1 \over 2}t}$ is a solution.

NEED 2 linearly independent solutions for a 2nd order linear homogeneous DE.

To get 2nd solution in repeated root case:  multiply by $t$

Thus 2 solutions are $y = e^{-{1 \over 2}t}$  and $y = te^{-{1 \over 2}t}$ 

Thus the general solution is $y = c_1(e^{-{1 \over 2}t}) + c_2(te^{-{1 \over 2}t})$ 



















\end
}%\huge
[dcos(nt)sin(nt) + ncos^2(nt) - dsin(nt)cos(nt)+ nsin^2(nt)]%\huge
)}$
\u\u

$~~= e^{2dt}%\huge
[ ncos^2(nt) + nsin^2(nt)%\huge
]$
\s\s
\rightline{$= ne^{2dt}%\huge
[ cos^2(nt) + sin^2(nt)%\huge
]$
$~=~ ne^{2dt} > 0$ for all $t$.}

\eject
\hsize 7.2truein
\hoffset -0.4truein

\right]
=
\left[\matrix{
0 \cr 0
}\right]$
\vfill
In other words the fundamental set of solutions $\{\po, \pt\}$ to $y'' + p(t) y' + q(t)y =
0$ form a basis for the set of all solutions to this linear homogeneous DE.

\end

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Thm:  Suppose $c_1 \phi_1(t) + c_2 \phi_2(t)$ is a general solution to

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

If $\psi$ is a solution to

\centerline{$ay'' + by' + cy = g(t)$ [*],}
Then $\psi + c_1 \phi_1(t) + c_2 \phi_2(t)$ is also a solution to [*].

Moreover if $\gamma$ is also a solution to [*], then there exist constants $c_1, c_2$ such that 

\centerline{$\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)$}

Or in other words, $\psi + c_1 \phi_1(t) + c_2 \phi_2(t)$ is a general solution to [*].


Proof:  

Define   $L(f) = af'' + bf' + cf$.

Recall $L$ is a linear function.

 Let $h = c_1 \phi_1(t) + c_2 \phi_2(t)$.  Since $h$ is a solution to the differential 
equation, $ay'' + by' + cy = 0$,

\vskip 0.5in

Since $\psi$ is a solution to 
 $ay'' + by' + cy = g(t)$,

\vskip 0.7in

We will now show that $\psi + c_1 \phi_1(t) + c_2 \phi_2(t) = \psi + h$ is also a solution to 
[*].

\vskip 0.7in



Since $\gamma$ a solution to  $ay'' + by' + cy = g(t)$,

\vskip 0.5in

We will first show that $\gamma - \psi$ is a solution to the differential equation
 $ay'' + by' + cy = 0$.

\vskip 0.7in
\vfill

Since $\gamma - \psi$ is a solution to $ay'' + by' + cy = 0$ and 

$c_1 \phi_1(t) + c_2 \phi_2(t)$ is a general solution to

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

there exist constants $c_1, c_2$ such that

\vskip 10pt
\centerline{$\gamma - \psi =\underline{\hskip 2in}$}


Thus
$\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)$.

%\vskip 5pt
%\hrule
%\vskip -5pt

Thm: \hfil \break
 Suppose  
$f_1$ is a a solution to
$ay'' + by' + cy = g_1(t)$  \hfil \break
and $f_2$ is a a solution to
$ay'' + by' + cy = g_2(t)$, then $f_1 + f_2$ is a solution to
$ay'' + by' + cy = g_1(t) + g_2(t)$


Proof: Let  $L(f) = af'' + bf' + cf$.

Since $f_1$ is a solution to 
 $ay'' + by' + cy = g_1(t)$,

\vskip 0.5in

Since $f_2$ is a solution to 
 $ay'' + by' + cy = g_2(t)$,

\vskip 0.5in

We will now show that $f_1 + f_2$ is a solution to \break
 $ay'' + by' + cy = g_1(t) + g_2(t)$.
\vskip 0.5in

\vfill


Sidenote:  The proofs above work even if $a, b, c$ are functions of $t$ instead of constants.

\eject

{\bf Examples:} Find a suitable form for $\psi$ for the following differential equations:

1.)  
$y'' - 4y' - 5y = 4e^{2t}$
\vskip 0.3in
\vfill

2.)  $y'' - 4y' - 5y = t^2 - 2t + 1$
\vskip 0.3in
\vfill


3.)  $y'' - 4y' - 5y = 4sin(3t)$
\vskip 0.3in
\vfill


4.)  $y'' - 5y  = 4sin(3t)$
\vskip 0.3in
\vfill

5.)  $y'' - 4y'  = t^2 - 2t + 1$
\vskip 0.3in
\vfill

6.)  
$y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}$
\vskip 0.3in
\vfill
\eject

7.)  
$y'' - 4y' - 5y = 4 sin(3t) e^{2t}$
\vskip 0.3in
\vfill
8.)  
$y'' - 4y' - 5y = 4 (t^2 - 2t - 1)sin(3t) e^{2t}$
\vskip 0.3in
\vfill

9.) $y'' - 4y' - 5y =4sin(3t) +  4 sin(3t) e^{2t}$
\vskip 0.3in
\vfill
10.)  $y'' - 4y' - 5y$

\rightline{ $= 4 sin(3t) e^{2t} + 4(t^2 - 2t - 1)e^{2t} + t^2 - 2t - 1$}
\vskip 0.85in
\vfill
\eject
11.) $y'' - 4y' - 5y =4sin(3t) +  5cos(3t)$
\vskip 0.3in
\vfill
12.) $y'' - 4y' - 5y = 4e^{-t}$
\vskip 0.3in
\vfill
%\eject
\hrule
To solve $ay'' + by' + cy = g_1(t) + g_2(t) + ... g_n(t)$ [**]

1.)  Find the general solution to $ay'' + by' + cy = 0$:  

\centerline{$c_1\phi_1 + c_2\phi_2$}

2.)  For each $g_i$, find a solution to $ay'' + by' + cy = g_i$:  

\centerline{$\psi_i$}

This includes plugging guessed solution {$\psi_i$} into \break
$ay'' + by' + cy = g_i$.
% to find constant(s).
%%\vskip 10pt

The general solution to [**] is

\vskip 10pt
\centerline{$c_1\phi_1 + c_2\phi_2 + \psi_1 + \psi_2 + ... \psi_n$}

%%\vskip 10pt

3.)  If initial value problem:

{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find $c_1, c_2$).}



\eject

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Solve $y'' - 4y' - 5y = 4sin(3t)$, ~$y(0) = 6$, $y'(0) = 7$.

1.) {\bf First solve homogeneous equation:}

  Find the general solution to $y'' - 4y' - 5y  = 0$:  

Guess $y = e^{rt}$ for HOMOGENEOUS equation:

$y' = re^{rt}$, $y' = r^2e^{rt}$

 $y'' - 4y' - 5y = 0$
 
$  r^2e^{rt} -  4re^{rt} - 5 e^{rt} = 0$

$  e^{rt}(r^2 -  4r - 5) = 0$

$ e^{rt} \not= 0$, thus can divide both sides by $e^{rt}$: 
\vskip 5pt
\centerline{$ r^2 -  4r - 5 = 0$}

$(r +1)(r  - 5) = 0$.  Thus $r = -1, 5$. 

  Thus $y = e^{-t}$ and $y = e^{5t}$ are both solutions to \hb
  LINEAR HOMOGENEOUS equation.


