What is the solution for xyy^'-y^2=y^2sqrt(y/x) ?

The solution for xyy^'-y^2=y^2sqrt(y/x) is y=(4x)/(ln^2(x)+2c_{1)ln(x)+c_{1}^2}

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xyy^'-y^2=y^2sqrt(y/x)

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

y = \frac{4 x}{ln ^{2} ( x ) + 2 c _{1} ln ( x ) + c _{1}^{2}}

Solution steps

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

Solve using substitution:

y = \frac{4 x}{ln ^{2} ( x ) + 2 c _{1} ln ( x ) + c _{1}^{2}}

y = \frac{4 x}{ln ^{2} ( x ) + 2 c _{1} ln ( x ) + c _{1}^{2}}

\int \frac{x ^{2} - 81}{x ^{4}} d x

f (x) = 4 x^{4} - 5 x - 3

\int_{0}^{2} e^{- t^{2}} d t

\int \frac{2 x + 1}{x ^{2} + x + 1} d x

x = - y - 5, x = 5 y - 11, y = 0

What is the solution for xyy^'-y^2=y^2sqrt(y/x) ?
The solution for xyy^'-y^2=y^2sqrt(y/x) is y=(4x)/(ln^2(x)+2c_{1)ln(x)+c_{1}^2}