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

The solution for y^'=x^2sqrt(y) is y=(x^6}{36}+\frac{c_{1}x^3)/6+(c_{1}^2)/4

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y^'=x^2sqrt(y)

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

y = \frac{x ^{6}}{36} + \frac{c _{1} x ^{3}}{6} + \frac{c _{1}^{2}}{4}

Solution steps

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

Solve separable ODE:

y = \frac{x ^{6}}{36} + \frac{c _{1} x ^{3}}{6} + \frac{c _{1}^{2}}{4}

y = \frac{x ^{6}}{36} + \frac{c _{1} x ^{3}}{6} + \frac{c _{1}^{2}}{4}

y^{^{'}} - \frac{x}{5 y ^{4}} = 0, y (0) = 2

\frac{d y}{d x} = y - a

3 t^{5} \frac{d x}{d t} + 6 x - 5 = 0

\frac{d y}{d x} = 3 - 0.04 y

\frac{d y}{d x} = \frac{y cos ( x )}{1 + y ^{2}}, y (0) = 1

What is the solution for y^'=x^2sqrt(y) ?
The solution for y^'=x^2sqrt(y) is y=(x^6}{36}+\frac{c_{1}x^3)/6+(c_{1}^2)/4