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Quiz 3 Math 3600\hfill  Show your work \break
Oct. 14, 2016 

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 10pt
\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.

\vskip 5in
\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill\vfill
\centerline{Answer:  \underbar{\hskip 3in}}


\end
Quiz 2 Math 3600 Sect 2\hfill  Show your work \break
Sept. 23, 2016

[20]~ 1.)  Transform the given initial value problem into an equivalent problem with initial point at the origin:

\centerline{${dy \over dt} = t^2 + y^2$, ~~~~$y(1) = 2$}
\vskip 2.45in

\centerline{Answer:  \underbar{\hskip 3in}}
\vskip 10pt
Quiz 2 Math 3600 Sect 2\hfill  Show your work \break
Sept. 23, 2016

[20]~ 1.)  Transform the given initial value problem into an equivalent problem with initial point at the origin:

\centerline{${dy \over dt} = t^2 + y^2$, ~~~~$y(1) = 2$}
\vskip 2.45in

\centerline{Answer:  \underbar{\hskip 3in}}

\end

[10]~ 1.)  Given that $y(x) = x^{3 \over 2}$ and $y(x) = {1 \over x}$  are solutions to $2x^2y'' + xy' - 3y = 0$, state the general solution to this 2nd order homogeneous linear differential equation:

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%$2t^2(2t^{-3}) + t(-t^{-2}) - 3(t^{-1}) =4(t^{-1}) - t^{-1} - 3(t^{-1}) = 0$

%$2t^2({3 \over 4}t^{-1 \over 2}) + t({3 \over 2} t^{1 \over 2}) - 3( t^{3 \over 2}) =({3 \over 2}t^{3 \over 2}) + ({3 \over 2} t^{3 \over 2}) - 3( t^{3 \over 2})= 0$


%$t^2(6t) + t(3t^{2}) - 9(t^{3}) =4(t^{-1}) - t^{-1} - 3(t^{-1}) = 0$

%$t^2(2) + t(2t) - 4(t^{2}) =4(t^{-1}) - t^{-1} - 3(t^{-1}) = 0$


%Answer:$\underline{~~\hskip 2in~~}$


[10]~ 2.)  Solve: $y' = y~sin(x) + y$.

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%Answer:$\underline{~~\hskip 2in~~}$
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