What is the solution for 2yy^'-xy^'=y-2x ?

The solution for 2yy^'-xy^'=y-2x is y=(x+sqrt(-3x^2+c_{1)})/2 ,y=(x-sqrt(-3x^2+c_{1)})/2

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2yy^'-xy^'=y-2x

2 y y^{^{'}} - x y^{^{'}} = y - 2 x

y = \frac{x + - 3 x ^{2} + c _{1}}{2}, y = \frac{x - - 3 x ^{2} + c _{1}}{2}

Solution steps

2 y y^{'} (x) - x y^{'} (x) = y - 2 x

Solve exact ODE:

y = \frac{x + - 3 x ^{2} + c _{1}}{2}, y = \frac{x - - 3 x ^{2} + c _{1}}{2}

y = \frac{x + - 3 x ^{2} + c _{1}}{2}, y = \frac{x - - 3 x ^{2} + c _{1}}{2}

\int (2 y sin (x) - x^{3} + ln (y)) d y

y^{^{''}} - 9 y^{^{'}} = 81 t

\frac{d}{d x} (\frac{arccos ( x )}{x})

\frac{d}{d x} (tan (x) + \frac{1}{x ^{2}})

\int 4 - x d x

What is the solution for 2yy^'-xy^'=y-2x ?
The solution for 2yy^'-xy^'=y-2x is y=(x+sqrt(-3x^2+c_{1)})/2 ,y=(x-sqrt(-3x^2+c_{1)})/2