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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:  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 1in

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

\vskip 2in

\eject

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 1in
\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
$\gamma = \psi + c_1 \phi_1(t) + c_2 \phi_2(t)$.

\eject

Thm:  Suppose that  
$f_1$ is a a solution to
$ay'' + by' + cy = g_1(t)$
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:

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 $ay'' + by' + cy = g_1(t) + g_2(t)$.
\vskip 0.65in

\vfill
\vfill

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

Examples:

Find a suitable form for $\psi$ for the following differential equations:

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

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


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

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


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

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

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

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


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

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

12.) $y'' - 4y' - 5y = 4e^{-t}$

\eject
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 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 $c_1, c_2$).}



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