What is the solution for yy^'=x^2 ?

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

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yy^'=x^2

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

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

Solution steps

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

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 a^{x} d x

x \to 1 lim (x^{2} + 2 x)

\frac{d}{d x} ((sin (3 x))^{4})

\int \frac{1}{2 - x} d x

\frac{5 + x}{1 - 5 x}

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