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Quiz 3 ~~~SHOW ALL WORK\hfill  Name:$\underline{~~\hskip 2.4in~~}$


[20]~ 1.)  Solve  $y'' +4y' - 5y =  20 + 12e^t$


Step 1:  Solve homogeneous:   $y'' +4y' - 5y =  0$

$r^2 + 4r - 5 = (r+5)(r-1) = 0$.  Thus $r = -5, 1$.  Thus homogeneous solution is $y = c_1e^{-5t} + c_2e^t$


Step 2a:  Solve  $y'' +4y' - 5y =  20$ 

If $y = A$, then $-5A = 20$ and $A = -4$.  Thus $y = -4$ is a nonhomogeous solution to   $y'' +4y' - 5y =  20$ 


Step 2b:  Solve  $y'' +4y' - 5y =  12e^t$

$y = e^t$ is a homogeneous solution, so multiply standard guess ($y = Ae^t$)  by $t$.  

Let $y = Ate^t$,  then $y' = Ae^t + Ate^t$, and  $y' = Ae^t +  Ae^t + Ate^t = 2Ae^t + Ate^t$


$ 2Ae^t + Ate^t +4( Ae^t + Ate^t) - 5Ate^t =  12e^t$

$ 2Ae^t + Ate^t +4 Ae^t + 4Ate^t - 5Ate^t =  12e^t$

$ 2Ae^t +4 Ae^t  =  12e^t$

$ 6 Ae^t  =  12e^t$ and $A = 2$.   Thus $y = 2te^t$ is a nonhomogeous solution to   $y'' +4y' - 5y =    12e^t$ 

Thus the general non-homogeneous solution is $y =  c_1e^{-5t} + c_2e^t -4 +  2te^t $

Last step:  Solve initial value problem:  $y(0) = -4$, $y'(0) = 8$

 $y =  c_1e^{-5t} + c_2e^t -4 +  2te^t $:    ~~ $-4 =  c_1 + c_2 - 4$.  Thus  $0 =  c_1 + c_2$  and $c_1 = -c_2$

$y =  -5c_1e^{-t} + c_2e^t + 2e^t +  2te^t $:  ~~$8 =  -5c_1 + c_2 + 2  =  5c_2 + c_2 + 2 = 6c_2 + 2 $.  Hence $c_2 = 1, c_1 = -1$



Thus the IVP solution is is $y =  -e^{-5t} + e^t -4 +  2te^t $





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Answer:$\underline{~~\hskip 4.5in~~}$
\end
${dy \over dx} = {1 \over x(2y+3)}$

Separate variables:  $(2y+3)dy = {(5x+4)dx \over x(x+1)}$

${(5x+4) \over x(x+1)} = {A \over x} + {B \over x+1}$

5x + 4 = Bx + A(x+1)




$y^2 + 3y = ln|x| + C$.

$y^2 + 3y - ln|x| - C = 0$.


$y = {-3 \pm \sqrt{9 - 4 (- ln|x| - C)} \over 2}$

$y = {-3 \pm \sqrt{9 + 4 ln|x| + C} \over 2}$

$y = {-3 \pm \sqrt{9  + C)} \over 2}$

$y = {-3 \pm \sqrt{9 -8)} \over 2}$

$y = {-3 - 1 \over 2}$

$y = {-3 \pm \sqrt{9 + 4 ln|x| -8} \over 2}$

$y = {-3 - \sqrt{1 + 4 ln|x| } \over 2}$


Answer:$\underline{~~\hskip 3in~~}$

\end

2.2       14




Solve $sin(t) y' + y cos(t) = 5$ 
 \hfil \break(hint:  this is a short problem if you are observant).

$sin(t) y' + y cos(t) = 5$

 $(sin(t) y)'  = 5$  
 
  $\int (sin(t) y)' dt  = \int 5 dt$  

 $\sin(t) y = 5t + C$  

 $ y  = {5t + C \over sin(t)}$  




Answer:$\underline{~~ y  = {5t + C \over sin(t)}~~}$
\end 
