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Find the solution to the initial value problem:
\vskip 10pt
\centerline{$y'' - 6y' + 9y = 8e^{3t} + 27t$,~~ $y(0) = 5$,~~ $y'(0) = 2$.}

\hr

Step 1:  Solve Homogeneous equation $y'' - 6y' + 9y = 0$

Let $y = e^{rt}$.  Then $r^2 - 6r + 9 = 0$.  Thus $(r - 3)^2 = 0$ and $r = 3$.

Thus general homogeneous solution is $y = c_1e^{3t} + c_2te^{3t}$.

\hr

Step 2a:   Solve Non-homogeneous equation $y'' - 6y' + 9y = 8e^{3t} $

Since $y = e^{3t}$ and $y = te^{3t}$ are homogeneous solutions, multiples of these cannot be solutions to the non-homogeneouse equation.  Thus we will try multiplying by another $t$ and try  $y = At^2e^{3t}$.  

$y = At^2e^{3t}$ implies $y' = 2Ate^{3t} +  3At^2e^{3t}$ and   
$y'' = 2Ae^{3t} +  6Ate^{3t} +  6Ate^{3t} +  9At^2e^{3t}$
\vskip 5pt
\rightline{$= 2Ae^{3t} +  12Ate^{3t} +  9At^2e^{3t}$ ~~~~~}

Plugging into $y'' - 6y' + 9y = 8e^{3t} $ and solve for $A$:

$2Ae^{3t} +  12Ate^{3t} +  9At^2e^{3t} - 6(2Ate^{3t} +  3At^2e^{3t}) + 9(At^2e^{3t}) ~=~ 8e^{3t}$

$2Ae^{3t} +  12Ate^{3t} +  9At^2e^{3t} - 12Ate^{3t} -  18At^2e^{3t} + 9At^2e^{3t} ~=~ 8e^{3t}$

$2Ae^{3t} +  (12 - 12)Ate^{3t} +  (9 - 18 + 9)At^2e^{3t}  ~=~ 8e^{3t}$

$2Ae^{3t}   ~=~ 8e^{3t}$ implies $2A ~=~ 8$ and thus $A = 4$.

Thus $y = 4t^2e^{3t}$ is anon-homogeneous solution to  $y'' - 6y' + 9y = 8e^{3t}$. 

Thus  general non-homogeneous solution to  $y'' - 6y' + 9y = 8e^{3t} $ is 
 \vskip 5pt
 \centerline{$y = c_1e^{3t} + c_2te^{3t} + 4t^2e^{3t}$.}
\hr
Step 2b:   Solve Non-homogeneous equation $y'' - 6y' + 9y = 27t$

Guess $y = At + B$.  Then $y' = A$ and $y'' = 0$.

Plugging into $y'' - 6y' + 9y = 27t $ and solve for $A$ and $B$:

 $0 - 6A + 9(At + B) = 27t $

 $9At + 9B - 6A = 27t + 0 $.  Thus $9A = 27$ and  $9B - 6A = 0$.
 
 Hence $A = 3$ and $9B = 6A = 6(3)$ and thus $B = 2$.
 
 Thus $y = 3t + 2$ is a non-homogeneous solution to  $y'' - 6y' + 9y = 27t$. 
 
 Thus  general non-homogeneous solution to  $y'' - 6y' + 9y = 27t $ is 

  \vskip 5pt
 \centerline{$y = c_1e^{3t} + c_2te^{3t} + 3t + 2$.}
 \hr
 
 {\bf Hence  general non-homogeneous solution to  $y'' - 6y' + 9y = 8e^{3t} + 27t$ is }
 \vskip 5pt
 \centerline{$y = c_1e^{3t} + c_2te^{3t} + 4t^2e^{3t} + 3t + 2$.}
  
  or equivalently,
  
   \centerline{$y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2$.}
  

  
\eject
  
  Step 3:  Use initial values to solve for $c_1$ and $c_2$:    

$y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2$
~~implies~~
$y' = 3e^{3t}(4t^2 + c_2t +c_1) +  e^{3t}(8t + c_2) + 3$

 $y(0) = 5$: ~~ $5 =  e^{0} (4(0)^2 + c_2(0) +c_1)  + 3(0) + 2$
 
 \hskip 0.77in $5 =  c_1  + 2$ implies $c_1 = 3$
 
   $y'(0) = 2$:  ~~ $2 = 3e^{0} (4(0)^2 + c_2(0) +c_1) +  e^{0}(8(0) + c_2) + 3$

 \hskip 0.78in $2 =  3c_1  + c_2 + 3$ implies $c_2 = 2 - 3 - 3c_1 = 2 - 3 - 9 = -10$
 
 Thus solution to IVP is $y = e^{3t}(4t^2 -10t +3) + 3t + 2$.
 
 
\hr \vskip -5pt \hr
\vskip -5pt
Quiz 3  \hfil  
Oct. 14, 2016 

\vskip -5pt


1.)  Suppose $y = c_1 e^{3t} + c_2te^{3t} + 4t^2e^{3t}$
is a solution to $y'' - 6y' + 9y = 8e^{3t}$.
Find the solution to the initial value problem:
\vskip 5pt
\centerline{$y'' - 6y' + 9y = 8e^{3t} + 27t$,~~ $y(0) = 5$,~~ $y'(0) = 2$.}

Note:  Solving this IVP is a 4 part problem, but I have already done the first two parts for you.

{\bf ANSWER:}  Since  $y = c_1 e^{3t} + c_2te^{3t} + 4t^2e^{3t}$
is a solution to $y'' - 6y' + 9y = 8e^{3t}$, 
\vskip -5pt
we know $y = c_1 e^{3t} + c_2te^{3t}$ is the general solution to the homogeneous equation
is a solution to $y'' - 6y' + 9y = 0$ ~~and~~
$y = 4t^2e^{3t}$
is a solution to $y'' - 6y' + 9y = 8e^{3t}$.

Thus parts 1 and 2a are already completed.  Repeating the remaining 2 parts:

{\bf Step 2b:}   Solve Non-homogeneous equation $y'' - 6y' + 9y = 27t$:~~
Guess $y = At + B$.  Then $y' = A$ and $y'' = 0$.

Plugging into $y'' - 6y' + 9y = 27t $ and solve for $A$ and $B$:
~~~~
 $0 - 6A + 9(At + B) = 27t $

 $9At + 9B - 6A = 27t + 0 $.  Thus $9A = 27$ and  $9B - 6A = 0$.
 
 Hence $A = 3$ and $9B = 6A = 6(3)$ and thus $B = 2$.
 
 Thus $y = 3t + 2$ is a non-homogeneous solution to  $y'' - 6y' + 9y = 27t$. 
 
 Thus  general non-homogeneous solution to  $y'' - 6y' + 9y = 27t $ ~~is~~ 
{$y = c_1e^{3t} + c_2te^{3t} + 3t + 2$.}
 \hr
 \vskip -5pt
  Thus  general non-homog. soln to  $y'' - 6y' + 9y = 8e^{3t} + 27t$ is
 %\vskip 5pt
 %\centerline
 {$y = c_1e^{3t} + c_2te^{3t} + 4t^2e^{3t} + 3t + 2$.}
  
  or equivalently,
  %\centerline
   {$y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2$.}
  \hr

  
  
 {\bf Step 3:}  Use initial values to solve for $c_1$ and $c_2$:    

$y = e^{3t}(4t^2 + c_2t +c_1) + 3t + 2$
~~implies~~
$y' = 3e^{3t}(4t^2 + c_2t +c_1) +  e^{3t}(8t + c_2) + 3$

 $y(0) = 5$: ~~ $5 =  e^{0} (4(0)^2 + c_2(0) +c_1)  + 3(0) + 2$
 %\hskip 0.77in
 implies
   $5 =  c_1  + 2$ implies $c_1 = 3$
 
   $y'(0) = 2$:  ~~ $2 = 3e^{0} (4(0)^2 + c_2(0) +c_1) +  e^{0}(8(0) + c_2) + 3$

 \hskip 0.78in 
$2 =  3c_1  + c_2 + 3$ implies $c_2 = 2 - 3 - 3c_1 = 2 - 3 - 9 = -10$
 
 Thus solution to IVP is $y = e^{3t}(4t^2 -10t +3) + 3t + 2$.
 
\vfill

\centerline{Answer:  \underbar{~~~ $y = e^{3t}(4t^2 -10t +3) + 3t + 2$~~~}}


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