
\magnification 1400
\parindent 0pt
\parskip 10pt
\pageno=1
\hsize 6.7truein
%\hoffset -0.3truein
\voffset -0.1truein
\vsize 9.3truein
\def\u{\vskip -3pt}
\def\v{\vskip -3pt}
\def\s{\vskip 3pt}
\def\t{\vskip 10pt}
\def\hb{ \hfil \break}
\def\hr{\vskip 5pt \hrule}
%\nopagenumbers

 



\def\u{\vskip -5pt}
\def\v{\vskip 0pt}
\def\vs{\vskip 5pt}

Guess a possible non-homog soln for the  following DEs:
\u

Note homogeneous solution to $y'' - 4y' - 5y = 0$~~ is~~ $y = c_1e^{-t} + c_2e^{5t}$
\vskip 5pt
\centerline{since $r^2 - 4r - 5 = (r-5)(r+1) = 0$}


1.)  
$y'' - 4y' - 5y = 4e^{2t}$
%%\vskip 0.3in
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ Ae^{2t}~~~~}$}

We want to plug in a solution for $y$ into the left-hand side (LHS) of the equation that will give us the right-hand side (RHS) of the equation.  In this case we need the output of the LHS to be a multiple of $e^{2t}$ (in particular the output of the LHS when pluggin in $y$ needs to be the RHS which is $4e^{2t}$).  

Thus we guess $y = Ae^{2t}$.  Plugging  this into the LHS, we can solve for $A$ so that we get the RHS, $4e^{2t}$, thus finding a non-homogeneous solution.

2a.)  $y'' - 4y' - 5y = t^2 - 2t + 1$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ At^2 + Bt + C  ~~~~}$}


2b.)  $y'' - 4y' - 5y = t^2 $
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~  At^2 + Bt + C  ~~~~}$}


2c.)  $y'' - 4y' - 5y = $ a degree 2 polynomial
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~  At^2 + Bt + C   ~~~~}$}

Note that the non-homog solution guess is the same for 2a, 2b, 2c.  In each case, we need to guess a solution such that when we plug it into the LHS we get the RHS, a degree 2 polynomial.  Thus our guess is a degree 2 polynomial (but compare this to example 12).  Note we need $y ~=~  At^2 + Bt + C$ even in case 2b.  Make sure you understand why $y = At^2$ won't work.



3.)  $y'' - 4y' - 5y = 30$
\v
\hskip 0.5in{Guess: $\underline{~~~~ ~~~~~ ~~~~}$}


\eject


3.)  $y'' - 4y' - 5y = 30$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~A   ~~~~}$}


We have a constant on the RHS, so I guess a constant (but again compare to example 12).  If you are observant, you may note that a non-homog solution is $y = -6$)



4a.)  $y'' - 4y' - 5y = 4sin(3t)$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ Asin(3t) + Bcos(3t)  ~~~~}$}

4b.) $y'' - 4y' - 5y =4sin(3t) +  5cos(3t)$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~   Asin(3t) + Bcos(3t) ~~~~}$}



4c.) $y'' - 4y' - 5y = 5cos(3t)$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~   Asin(3t) + Bcos(3t) ~~~~}$}


Note that the non-homog solution guess is the same for 4a, 4b, 4c.   If we plug in $y = Asin(3t)$, the output will contain both $sin(3t)$ and $cos(3t)$ terms.  Thus I need to include both these terms in my guess.  Compare to example 11.


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


\v
\hskip 0.5in{Wrong Guess: $\underline{~~~~~~~~~~~~~~~~~~~~ ~~~~}$}
\v\v
\hskip 0.5in{Correct Guess: $\underline{~~~~~~~~~~~~~~~~~~~~ ~~~~}$}

\eject

5.) $y'' - 4y' - 5y = 4e^{-t}$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ Ate^{-t}  ~~~~}$}

Note $y = Ae^{-t}$ is a homogeneous solution.  Thus if I plug in it, I will get 0.  But I want the RHS, $ 4e^{-t}$.  When a guess doesn't work because it is a homogeneous solution, multiple by $t$.  

Sidenote:  this trick works because when you plug it in, you must use the product rule; the  homogeneous part $e^{-t}$ of $y = Ate^{-t}$ will result in a number of cancellations, but the $t$ part will give you terms that don't cancel out and whose sum is the RHS.

Observe $y =  Ate^{-t}$, ~~$y' =  Ae^{-t} - Ate^{-t}$, ~~
$y'' =  -Ae^{-t} - Ae^{-t} + Ate^{-t} = -2 Ae^{-t} + Ate^{-t}$


Thus $y'' - 4y' - 5y = 
- 2Ae^{-t} + Ate^{-t} - 4( Ae^{-t} - Ate^{-t}) - 5(Ate^{-t})$

$= - 2Ae^{-t}  - 4 Ae^{-t} + At(e^{-t} + 4 e^{-t} - 5e^{-t})$
$= - 6 Ae^{-t} + At(0)$
$= 4e^{-t}$ when $A = -{2 \over 3}$
\vskip 10pt
\centerline{DO NOT FORGET THE PRODUCT RULE!!!!}
\v

6.)   $y'' - 4y' - 5y =(e^t) + (e^{-t}) + (2t^3 + 3t^2) + (4sin(3t) +  5cos(3t))$

{Guess: }


\eject

6.)   $y'' - 4y' - 5y =(e^t) + (e^{-t}) + (2t^3 + 3t^2) + (4sin(3t) +  5cos(3t))$
\vskip 10pt
\centerline{Guess: $\underline{~~y ~=~ (A_1e^{t}) + (A_2te^{-t}) + (A_3t^3 + B_3t^2 + C_3t + D_3) + (A_4sin(3t) + B_4cos(3t))   ~~~~}$}

Note if we wanted to find a non-homogeneous solution, we would need to determine all our undetermined coefficients.  Note we have 8 undetermined coefficients.  Instead of solving for them all at once (which would require 8 equations for the 8 unknowns), it is easier to divide finding a non-homogeneous solution into 4 simpler parts indicated by the parenthesis and subscripts as described below:  


a.)  Find $A_1$ by plugging $y = A_1e^{t}$ into $y'' - 4y' - 5y =e^t $

b.)  Find $A_2$ by plugging $y = A_2te^{-t}$ into $y'' - 4y' - 5y =e^{-t} $

c.) Find $A_3, B_3, C_3, D_3$ by plugging $y = A_3t^3 + B_3t^2 + C_3t + D_3$ into $y'' - 4y' - 5y = 2t^3 + 3t^2 $

d.) Find $A_4, B_4$ by plugging $y = A_4sin(3t) + B_4cos(3t)$ into 

\rightline{$y'' - 4y' - 5y = 4sin(3t) +  5cos(t)$}


We get the non-homogeneous solution by adding together the non-homogeneous solutions obtained from the above 4 parts since our diff eqn is LINEAR.

We get the general solution by combining the general homogeneous solution with this non-homogeneous solution.



7.)   $y'' - 4y' - 5y =e^t + e^{-t} + 2t^3 + 3t^2 + 4sin(3t) +  5cos(t)$

Guess:

\eject

7.)   $y'' - 4y' - 5y =e^t + e^{-t} + 2t^3 + 3t^2 + 4sin(3t) +  5cos(t)$

\vskip 10pt
\hskip 0.15in{Guess: $\underline{~~y ~=~ (A_1e^{t}) + (A_2te^{-t}) + (A_3t^3 + B_3t^2 + C_3t + D_3)~~ }$}
\vskip 10pt
\rightline{$\underline{~~+ (A_4sin(3t) + B_4cos(3t))   + (A_5sin(t) + B_5cos(t))}$~~}

\vskip 10pt
8.)  
$y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}$

Guess:

\vfill



8.)  
$y'' - 4y' - 5y = 4(t^2 - 2t - 1)e^{2t}$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ (At^2 + Bt + C)e^{2t}   ~~~~}$}

Since the RHS is a product, we guess a product.   \hb
Note I could have guessed
$y ~=~ (At^2 + Bt + C)De^{2t} ~=~ (ADt^2 + BDt + CD)e^{2t}$,
but since $AD, BD, CD$ are just constants, I don't need $D$.


\hr \vskip -5pt \hr \vskip -5pt \hr \vskip 0pt
\eject
\def\o{\vskip -1pt}

Note homogeneous solution to $y'' - 6y' +9y = 0$ ~~is~~ $y = c_1e^{3t} + c_2te^{3t}$
\o
\vskip -5pt
\hskip 1.55in{since $r^2 - 6r +9 = (r-3)(r-3) = 0$}


9.)  $y'' - 6y' +9y = 7e^{3t}$

Guess:

\vfill
9.)  $y'' - 6y' +9y = 7e^{3t}$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ At^2e^{3t}  ~~~~}$}
\o
Note neither $y = Ae^{3t}$ nor $y = Ate^{3t}$ will work since both are homogeneous solutions.  But our trick of multiplying by $t$ until we have a guess that is not a homogeneous solution will work.


10.)  $y'' - 6y' +9y = 7e^{-3t}$

Guess

\vfill

10.)  $y'' - 6y' +9y = 7e^{-3t}$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~ Ae^{-3t}  ~~~~}$}
\o
$y ~=~ Ae^{-3t}$ is not a homogeneous solution (when $A \not= 0$).
\vskip 5pt
\hr \vskip -5pt \hr \vskip -5pt \hr \vskip 0pt

\eject

Some special cases:

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

Best Guess:

\vfill\vfill\vfill

11.)  $y'' - 5y  = 4sin(3t)$
\v
\hskip 0.5in{Best Guess: $\underline{~~~~y ~=~ Asin(3t)  ~~~~}$}
\o
Note, we also could have guessed $y = Asin(3t) + Bcos(3t)$, but since there is no $y'$ term, we don't need the cosine term.  But both guesses will work.  Plugging in $y = Asin(3t) + Bcos(3t)$ will take a little more work, but you will still get the right answer.


12.)  $y'' - 4y'  = t^2 - 2t + 1$

Guess:
\vfill\vfill\vfill

12.)  $y'' - 4y'  = t^2 - 2t + 1$
\v
\hskip 0.5in{Guess: $\underline{~~~~y ~=~At^3 + Bt^2 +Ct   ~~~~}$}

\o
Note there is no $y$ term on the LHS.  Thus to get a $t^2$ term when we plug in our guess, we will need to plug in a $t^3$ term. Hence we guess a degree 3 polynomial.  Note we don't need to include a constant term;  we could have guessed $y ~=~At^3 + Bt^2 +Ct + D$, but any constant $D$ will work (and hence there are an infinite number of solutions for $D$) so we might as well take $D = 0$.

\hr

Don't worry too much about guessing wrong.  You will usually be able to figure out why an incorrect guess doesn't work and use that info to determine a better guess.



\end


