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

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

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y^'=((x^2))/y

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

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

Solution steps

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

Solve separable ODE:

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

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

\int (2 - \frac{1}{x})^{2} d x

\int_{3}^{0} (- x^{3} - 2 x^{2} + x - 1) d x

y = \frac{x}{x + 3}, (1, 0.25)

\frac{d}{d x} (\frac{1}{( 3 x ^{2} + x - 3 ) ^{9}})

f (x) = x^{x^{2}}

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