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Quiz 2 ~~~SHOW ALL WORK\hfil \break
Sept 20, 2018

[10]~ 1.) Solve $ty' + (t+1) y = t$

$y' + {(t+1)\over t} y = 1$

$u(t) = e^{\int  {(t+1)\over t} dt } = e^{\int  (1+{1\over t}) dt } = 
e^{t+ln| t|} = e^te^{ln| t|} = |t|e^{t}$.

Let $u(t) = te^t$

$ te^ty' + { e^t(t+1)} y =  te^t$


$ (te^ty)' =  te^t$ \hfill {\bf Check:}  $(te^ty)' =   te^ty' +  (e^t + te^t)y $

$ \int (te^ty)' =  \int te^t$ 

$ te^ty =  te^t - e^t + C$  by integration by parts

Thus $y =  1 - t^{-1}e^t + Ct^{-1}e^{-t}$


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Answer:$\underline{~~y =  1 - t^{-1}e^t + Ct^{-1}e^{-t}~~}$
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[10]~ 2.)  Suppose ${\phi_n}$ is defined by sucessive approximation where  $y' = {2t \over y+1}$, ~$y(0) = 0$.
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If $\phi_1(t) = t^2$, then $\phi_2(t) = \underline{~~ ln|t^2 + 1|~~}$

$\phi_1(t) = \int_0^t f(s, 0)ds = \int_0^t 2sds  = t^2$ 

$\phi_2(t) = \int_0^t f(s, \phi(s))ds= \int_0^t {2s \over s^2 + 1} = ln|s^2 + 1| |_0^t  = ln|t^2 + 1| -  ln|1| =  ln|t^2 + 1| $



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${dy \over dx} = {1 \over x(2y+3)}$

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

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

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


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

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

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

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

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

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

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

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