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

[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:


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
\vskip 10pt
\centerline{$y(x) = c_1 x^{3 \over 2} +  {c_2  \over x}$}

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

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

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


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

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


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


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


Separate variables:  ${dy \over dx} = y(sinx + 1)$.

${dy \over y} = (sinx + 1)dx$

$\int {dy \over y} = \int (sinx + 1)dx$


$ln|y| = -cos(x) + x + C$

$ln|y| = -cos(x) + x + C$



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

Thus $y = Ce^{x-cos(x)}$

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