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**A Beautiful Mind** Linear: $\displaystyle x^2y'-x(x^2-2y), y(1) = 5/4

$

$\displaystyle x^2y'-x(x^2-2y)= 0$

$\displaystyle x^2y' = x^3-2xy$

$\displaystyle y' = \frac{x^3-2xy}{x^2}$

$\displaystyle y' = x - \frac{2y}{x}$

$\displaystyle y' + (\frac{2}{x})(y) = x $

$\displaystyle e^\int (\frac{2}{x}dx) = ....x^2$

$\displaystyle x^2y'+x^2(\frac{2}{x})y = x^3$

$\displaystyle \int(\frac{d}{dx} x^2y) dx = \int x^3 dx$

$\displaystyle x^2y = \frac{x^4}{4}$ Mr F adds: + C

$\displaystyle y = \frac{x^2}{2}$ Mr F says: That 2 should be a 4.

...lost, don't know what I'm doing wrong.