What is the solution for y^'=sqrt(yx) ?

The solution for y^'=sqrt(yx) is y=(x^3)/9+(c_{1}x^{3/2})/3+(c_{1}^2)/4

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y^'=sqrt(yx)

y^{^{'}} = y x

y = \frac{x ^{3}}{9} + \frac{c _{1} x ^{\frac{3}{2}}}{3} + \frac{c _{1}^{2}}{4}

Solution steps

y^{'} (x) = y x

Solve separable ODE:

y = \frac{x ^{3}}{9} + \frac{c _{1} x ^{\frac{3}{2}}}{3} + \frac{c _{1}^{2}}{4}

y = \frac{x ^{3}}{9} + \frac{c _{1} x ^{\frac{3}{2}}}{3} + \frac{c _{1}^{2}}{4}

\frac{d y}{d x} y = 1

\frac{d v}{d t} + \frac{- k v ( t )}{0.018} = - 9.81

\frac{d y}{d t} = - 0.06 y

y^{5} \frac{d y}{d x} = 2 x^{4} y^{6} - 6 x^{4}

\frac{y ^{2} d y}{d x} = x

What is the solution for y^'=sqrt(yx) ?
The solution for y^'=sqrt(yx) is y=(x^3)/9+(c_{1}x^{3/2})/3+(c_{1}^2)/4