What is the solution for y^'-xy=2xy^{(-1)/2} ?

The solution for y^'-xy=2xy^{(-1)/2} is y=\sqrt[3]{e^{(3x^2+6c_{1})/2}-4e^{(3x^2)/4+(3c_{1})/2}+4}

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y^'-xy=2xy^{(-1)/2}

y^{^{'}} - x y = 2 x y^{\frac{- 1}{2}}

y = 3 e^{\frac{3 x ^{2} + 6 c _{1}}{2}} - 4 e^{\frac{3 x ^{2}}{4} + \frac{3 c _{1}}{2}} + 4

Solution steps

y^{'} (x) - x y = 2 x y^{\frac{- 1}{2}}

Solve separable ODE:

y = 3 e^{\frac{3 x ^{2} + 6 c _{1}}{2}} - 4 e^{\frac{3 x ^{2}}{4} + \frac{3 c _{1}}{2}} + 4

y = 3 e^{\frac{3 x ^{2} + 6 c _{1}}{2}} - 4 e^{\frac{3 x ^{2}}{4} + \frac{3 c _{1}}{2}} + 4

3 y^{2} (x - 3) \frac{d y}{d x} = k

\frac{d x}{d y} = tan (x)

y^{^{'}} = \frac{e ^{- x}}{sin ( y )}, y (0) = 1

\frac{d y}{d x} = \frac{x ( y - 1 )}{( y - 2 ) ( x ^{2} + 1 )}

y^{^{'}} = x^{2} y^{- 3}, y (0) = 4

What is the solution for y^'-xy=2xy^{(-1)/2} ?
The solution for y^'-xy=2xy^{(-1)/2} is y=\sqrt[3]{e^{(3x^2+6c_{1})/2}-4e^{(3x^2)/4+(3c_{1})/2}+4}