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Quiz 2 \hfill  Show your work \break
Feb 19, 2016 \hfill  Circle your answer.

[10]~ 1.)  Given that $y(t) = {1 \over t}$ and $y(t) = t^{3 \over 2}$ are solutions to $2t^2y'' + ty' - 3y = 0$, state the general solution to this 2nd order homogeneous linear differential equation:

Given two linearly independent solutions to a 2nd order homogeneous linear differential equation, one can create the general solution by taking their linear combination.  Thus the answer is
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\centerline{$y(t) = {c_1  \over t} + c_2 t^{3 \over 2} $}

In other words, $\{ {1 \over t},  t^{3 \over 2} \}$ forms a basis for the solution set (since every solution can be written uniquely as a linear combination of these two functions).

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

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






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


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

Separate variables:  ${dy \over dx} = ycos(x)$.

${dy \over y} = cos(x)dx$

$\int {dy \over y} = \int cos(x)dx$


$ln|y| = sin(x) + C$

$ln|y| = sin(x) +  C$



$|y| = e^{sin(x) + C} = e^Ce^{sin(x)} $

Thus $y = Ce^{sin(x)}$

\centerline{Answer:$\underline{~~y = Ce^{sin(x)}~~}$}


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