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\nopagenumbers

 


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To solve linear DE  $ay'' + by' + cy = g_1 + g_2 + g_3$

{\bf Step 1:}  Solve homogeneous version:  $ay'' + by' + cy = 0$ implies 

\rightline{$ar^2 +br +c =  0$ implies ~~~~~~~~ ... ~~~~~~~~ $y = c_1\phi_1 + c_2\phi_2$.}

{\bf Step 2a}:  Find one non-homogeneous solution, $y = f_1$, to $ay'' + by' + cy = g_1$\hfil \break
{\bf Step 2b}:   Find one non-homogeneous solution, $y = f_2$, to $ay'' + by' + cy = g_2$\hfil \break
{\bf Step 2c}:  Find one non-homogeneous solution, $y = f_3$, to $ay'' + by' + cy = g_3$

{\bf Step 3:}  Combine all solutions to create the general solution to the non-homogeneous DE:
$$y =  c_1\phi_1 + c_2\phi_2 +  f_1 + f_2 + f_3$$

{\bf Last step:}  If IVP, plug in initial values to find the constants $c_1$ and $c_2$.

\w
\hrule

Guess a possible non-homog soln for the  following DEs:
\u
Note homogeneous solution to $y'' + 2y' + y = 0$ is \hb$y = c_1e^{-t} + c_2te^{-t}$
since $r^2 +2r +1 = (r+1)(r+1) = 0$
\w
\hrule

1.)  
$y'' +2y' + y = 4e^{2t}$
%%\vskip 0.3in
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\v


2.)  
$y'' +2y' + y = 4e^{t}$
%%\vskip 0.3in
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\v


3.)  
$y'' +2y' + y = 4e^{-t}$
%%\vskip 0.3in
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill

\eject
4.)  $y'' +2y' + y = t $
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill




5.)  $y'' +2y' + y  = t + 1$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill


6.)  $y'' +2y' + y   = 4sin(2t)$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill
7.) $y'' +2y' + y   =4sin(2t) +  5cos(2t)$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}

\vfill
8.) $y'' +2y' + y   =4sin(2t) +  5cos(3t)$
\v
\centerline{Guess for step 2a: $\underline{\hskip 5truein}$}
\v
\centerline{Guess for step 2b: $\underline{\hskip 5truein}$}
\vfill

9.) $y'' +2y' + y   =4sin(2t) +   t + 1$
\v
\centerline{Guess for step 2a: $\underline{\hskip 5truein}$}
\v
\centerline{Guess for step 2b: $\underline{\hskip 5truein}$}
\vfill
10.) $y'' +2y' + y   =4tsin(2t) $
\v\v\v\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill
11.)  $y'' + y   = 4sin(2t)$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill

12.)  $y'' + y   = 4sin(t)$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill
\end



8.)  
$y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill

\hr\hr

Note homogeneous solution to $y'' - 6y' +9y = 0$ is \hb$y = c_1e^{3t} + c_2te^{3t}$
since $r^2 - 6r +9 = (r-3)(r-3) = 0$

9.)  $y'' - 6y' +9y = 7e^{3t}$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill

10.)  $y'' - 6y' +9y = 7e^{-3t}$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\vfill

\hr\hr
Some special cases:

11.)  $y'' - 5y  = 4sin(3t)$
\v
\centerline{Best Guess: $\underline{\hskip 6truein}$}
\vfill

12.)  $y'' - 4y'  = t^2 - 2t + 1$
\v
\centerline{Guess: $\underline{\hskip 6truein}$}
\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
\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.

Thus the general solution to the 2nd order LINEAR \hb HOMOGENEOUS equation is 

\centerline{$y = c_1e^{-t} + c_2e^{5t}$}


\eject

2.) {\bf Find one solution to non-homogeneous eq'n:}
\vskip -7pt
  Find a solution to $ay'' + by' + cy = 4sin(3t)$:  



Guess $y = A sin(3t) + B cos(3t)$
\vskip -5pt
~~~~~~~~$y' = 3A cos(3t) - 3B sin(3t)$
\vskip -5pt
~~~~~~~~$y'' = -9A sin(3t)  - 9B cos(3t)$

$y'' - 4y' - 5y = 4sin(3t)$

$\matrix{  -9A sin(3t)& -& 9B cos(3t)&& \cr
12B sin(3t) &- &12A cos(3t) &&\cr
 -5A sin(3t)  &- &5 cos(3t) &&\cr
% && & \cr
   \noalign{\hrule}
&&& \cr
    (12B - 14A) sin(3t)  &- &(-14B - 12A) cos(3t)&=& 4sin(3t)
}$

Since $\{sin(3t), cos(3t)\}$ is a linearly independent set:

$12B - 14A = 4$ and $-14B - 12A = 0$

%Thus $B = -{12 \over 14}A = -{6 \over 7}A$ and 
%$12(-{6 \over 7}A) - 14A  = {-72 - 98 \over  7}A= {-170 \over 7}A= 4$

Thus $A = -{14 \over 12}B = -{7 \over 6}B$ and 

$12B - 14(-{7 \over 6}B) = 12B + 7({7 \over 3}B) =  {36 + 49 \over  3}B= {85 \over 3}B= 4$

Thus $B = 4({3 \over 85 }) = {12 \over 85}$ ~~and~~ $A  = -{7 \over 6}B
 = -{7 \over 6}( {12 \over 85}) = -{14 \over 85}$

Thus $y = (-{14 \over 85})sin(3t) + {12 \over 85}cos(3t)$ is one solution to the nonhomogeneous equation.



Thus the general solution to the 2nd order linear  nonhomogeneous equation is 

\centerline{$y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)$}



3.) {\bf If initial value problem:}
\vskip -7pt
{Once general solution is known, can solve initial value 
problem (i.e., use initial conditions to find $c_1, c_2$).}



NOTE:  you must know the GENERAL solution to the ODE BEFORE you can solve for the initial values.  The homogeneous solution and the one nonhomogeneous solution found in steps 1 and 2 above do NOT need to separately satisfy the initial values.
\vskip -4pt
Solve $y'' - 4y' - 5y = 4sin(3t)$, ~$y(0) = 6$, $y'(0) = 7$.

General solution:
$y = c_1e^{-t} + c_2e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)$

Thus $y' =  -c_1e^{-t} + 5c_2e^{5t} -({42 \over 85})cos(3t) - {36 \over 85}sin(3t)$

$y(0) = 6$: ~~~~ $6 =  c_1 + c_2  + {12 \over 85}$
~~~~~~~~~ $ {498 \over 85} =  c_1 + c_2 $

$y'(0) = 7$:  ~~~~$7 =  -c_1 + 5c_2 -{42 \over 85}$
 ~~~~~~~${637 \over 85} =  -c_1 + 5c_2 $


$6c_2 = {498 + 637 \over 85} = {1135 \over 85} = {227 \over 17}$.
Thus $c_2 = {227 \over 102}$.

$c_1 =  {498 \over 85}  - c_2 =  {498 \over 85}  -  {227 \over 102}
= {2988  -  1135 \over 510} = {1853 \over 510} = {109 \over 30}$

\vskip 5pt

Thus $y = ( {109 \over 30})e^{-t} + ({227 \over 102})e^{5t} -({14 \over 85})sin(3t) + {12 \over 85}cos(3t)$.

\vskip 5pt
\hrule
%\vfill
\vskip -5pt
Partial Check:  $y(0) = ( {109 \over 30}) + ({227 \over 102}) + {12 \over 85} =  6$.

~~~~~~~~~~~~~~~~~~~~$y'(0) =  -{109 \over 30} + 5({227 \over 102}) -{42 \over 85} = 7$.

$(e^{-t})'' - 4(e^{-t})' - 5(e^{-t}) = 0$ and 
$(e^{5t})'' - 4(e^{5t})' - 5(e^{5t}) = 0$ 


\end
Potential proofs for exam 1:


Proof by (counter) example:


\item\item 1. Prove a function is not 1:1, not onto, not a bijection, not linear.

\item\item 2. Prove that a differential equation can have multiple solutions.

\vskip 10pt
\hrule

Prove convergence of a series using ratio test.

Induction proof.


Prove a function is linear. 



%Theorem 2.4.1: Existence of a unique solution for a 1st order Linear ODE IVP


Theorem 3.2.2:  If $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to the 2nd order linear ODE, $ay'' + by' + cy = 0$, then their linear combination $y = c_1\phi_1(t) + c_2\phi_2(t)$ is also a solution for constants $c_1$ and $c_2$.


%Theorem 3.5.2: The general solution of the nonhomogeneous equation can be written as a linear combination of the solutions to the homogeneous equation %and a specific solution of the nonhomogeneous equation.

 Note you may use what you know from both pre-calculus and calculus (e.g., integration and derivatives are linear).

\end
\eject

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3.6   Variation of Parameters~~~~~~\hfill
%\vskip 5pt
%\centerline
{Solve $y'' - 2y' + y = e^t ln(t)$}~~
\vskip -8pt
{\bf 1)  Find homogeneous solutions:  Solve $y'' - 2y' + y = 0$}
\u
Guess:  $y = e^{rt}$, then $y' = re^{rt}$, $y'' = r^2e^{rt}$, and
\vskip 5pt
\centerline{$r^2e^{rt} - 2re^{rt} + e^{rt} = 0$
%\vskip 10pt
%\centerline
implies {$r^2 - 2r + 1 = 0$}}
\u
$(r - 1)^2 = 0$, and hence $r = 1$
\s\s\s\s
General homogeneous solution:  $y = c_1e^{t} + c_2 te^{t}$
\v
since have two linearly independent solutions:  $\{e^{t}, te^{t}\}$ 
\v

{\bf 2.)  Find a non-homogeneous solution:}
\u
Sect. 3.5 method:  Educated guess
\u
Sect. 3.6:  {\bf Guess $y = u_1(t)e^t + u_2(t) te^t$ and solve for $u_1$ and 
$u_2$}

\u


$u_1(t) = \int { \left|\matrix{0 & \phi_2 \cr 1 & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 $= -\int {\phi_2(t) g(t) \over W(\phi_1, \phi_2)} dt =  -\int {(te^t)(e^t ln(t)) \over e^{2t}} dt $

\rightline{$=  -\int tln(t) = -[{t^2 ln(t) \over 2} - \int {t \over 2}] = -{t^2 ln(t) \over 2} + {t^2 \over 4}  $}


$u_2(t) =   \int { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & 1}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} g(t)dt$
 $= \int {\phi_1(t) g(t) \over W(\phi_1, \phi_2)} dt =  \int {(e^t)(e^t ln(t)) \over e^{2t}} dt $
 
\rightline{$ =  \int ln(t) = 
% [tln(t) - \int dt] = 
 tln(t) - t$ }


$ W(\phi_1, \phi_2) = \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|
 = \left|\matrix{e^t & te^t \cr e^t & e^t + te^t}\right|$

$\matrix{u = ln(t) & dv = t dt\cr du = {dt \over t} & v = {t^2 \over 2}}$
\hfill
$\matrix{u = ln(t) & dv = dt\cr du = {dt \over t} & v = {t}}$
\s\s\s\s\s
General solution :
%\vskip 5pt
%\centerline
{ $y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}$ }
\v
which simplifies to  $y = c_1e^{t} + c_2 te^{t} +  ({ln(t) \over 2}  -  {3\over 4} )t^2 e^t $


\eject
Solve $y'' + p(t)y' + q(t)y = g(t)$ where $y = c_1\phi_1(t) + c_2\phi_2(t)$ is solution to homogeneous equation  $y'' + p(t)y' + q(t)y = 0$

\def\p{\phi}

Guess  $y = u_1(t)\phi_1(t) + u_2(t) \phi_2(t)$

$y = u_1 \p_1 + u_2 \p_2$
implies
$y' = u_1 \p_1' + u_1' \p_1 + u_2 \p_2' + u_2' \p_2$


Two unknown functions, $u_1$ and $u_2$, but only one equation ($y'' + p(t)y' + q(t)y = g(t)$).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in $y''$: ~~Choose $u_1'\p_1 + u_2' \p_2 = 0$}

%$y = u_1 \p_1 + u_2 \p_2$ 

\v
$y' = u_1 \p_1' + u_2 \p_2' $
implies
$y'' =  u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2'$


Plug into $y'' + p(t)y' + q(t)y = g(t)$:

$u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p( u_1 \p_1' + u_2 \p_2') + q(u_1 \p_1 + u_2 \p_2) = g$

$u_1 \p_1'' +  u_1' \p_1' + u_2 \p_2''+ u_2' \p_2' + p u_1 \p_1' + pu_2 \p_2') + qu_1 \p_1 + qu_2 \p_2 = g$

$u_1 \p_1'' +  p u_1 \p_1' + qu_1 \p_1 + u_1' \p_1' + u_2 \p_2'' + pu_2 \p_2' + qu_2 \p_2 + u_2' \p_2'= g$

$u_1( \p_1'' +  p  \p_1' + q \p_1) + u_1' \p_1' + u_2 (\p_2'' + p \p_2' +  q\p_2) + u_2' \p_2'= g$

$\p_1$, $\p_2$ are homogeneous solutions.  Thus $\p_i'' + p \p_i' +  q\p_i = 0$.


Hence $u_1( 0) + u_1' \p_1' + u_2 (0) + u_2' \p_2'= g$

Thus we have 2 eqns to find  2 unknowns, the functions $u_1$ and $u_2$:
\v
%$u_1'\p_1 + u_2' \p_2 = 0$
%\v
%$ u_1' \p_1' + u_2' \p_2'= g$

$\matrix{
u_1'\p_1 + u_2' \p_2 = 0 \cr
\ u_1' \p_1' +  u_2' \p_2'= g
}$
implies $\left[\matrix{\p_1 &  \p_2  \cr
 \p_1' &  \p_2'}\right]
 \left[\matrix{ u_1'\cr u_2' }\right]
\left[ \matrix{ 0\cr g }\right]$



Cramer's rule: $u_1'(t) =  { \left|\matrix{0 & \phi_2 \cr g & \phi_2'}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|}$ and  $u_2'(t) =   { \left|\matrix{\phi_1 & 0 \cr  \phi_1' & g}\right|
\over  \left|\matrix{\phi_1 & \phi_2 \cr \phi_1' & \phi_2'}\right|} $

\eject

Sect.3.6:  {\bf Guess $y = u_1(t)e^t + u_2(t) te^t$ and solve for $u_1$ and 
$u_2$}

$y' = u_1'e^t +u_1e^t + u_2'te^t + u_2(e^t + te^t) = e^{2t} + te^{2t} - te^{2t} - e^{2t}$.

Two unknown functions, $u_1$ and $u_2$, but only one equation ($y'' - 2y' + y = e^t ln(t)$).  Thus might be 
OK to choose 2nd eq'n.

{\bf Avoid 2nd derivative in $y''$: ~~Choose $u_1'e^t + u_2' te^t = 0$}

Hence $y' = u_1e^t +  u_2(e^t + te^t)$.

and 
$y'' = u_1'e^t +u_1e^t + u_2'(e^t + te^t) + u_2(e^t + e^t + te^t)$.

\v
~~~~~~~~~$ = u_1'e^t +u_1e^t + u_2'e^t + u_2'te^t + u_2(2e^t + te^t)$.
\v
~~~~~~~~~$ = u_1e^t + u_2'e^t + u_2(2e^t + te^t)$.

\vskip 10pt

\centerline{Solve $y'' - 2y' + y = e^t ln(t)$}


$  u_1e^t + u_2'e^t + u_2(2e^t + te^t) - 2[ u_1e^t +  u_2(e^t + te^t)] +  u_1e^t + 
u_2te^t = e^t ln(t)$

$  u_2'e^t + 2u_2e^t + u_2te^t - 2u_2 e^t  -2u_2 te^t +  u_2te^t = e^t ln(t)$

$  u_2'    = ln(t)$
or in other words,
${du_2 \over dt} = ln(t)$

Thus $\int du_2 = \int ln(t) dt$

$u_2 = t ln(t) - t$.   Note only need one solution, so don't need $+C$.


 $y = u_1(t)e^t + [ t ln(t) - t ] te^t$

$u_1'e^t + u_2' te^t = 0$.  Thus $u_1' + u_2' t = 0$.  Hence $u_1' =  -u_2' t = -tln(t) $

Thus $u_1 =  -\int t ln(t) dt =  -{t^2 ln(t) \over 2} + {t^2 \over 4}  $

Thus the general solution is 

\vskip 5pt
\centerline { $y = c_1e^{t} + c_2 te^{t} +  ( -{t^2 ln(t) \over 2} + {t^2 \over 4} )e^{t} + ( tln(t) - t)te^{t}$ }

\eject



\end


\end

\end