What is the solution for y(dy)/(dx)=x^2,y(0)=1 ?

The solution for y(dy)/(dx)=x^2,y(0)=1 is y=sqrt((2x^3)/3+1)

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y(dy)/(dx)=x^2,y(0)=1

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (f (x)^{- 1}) x^{2} - 8 x - 9 x = - 21

y^{^{'}} = \frac{y - 1}{x}

\frac{d y}{d x} = - 2 (y - 40)

3 + 2 q + (4 + 2 p) \frac{d q}{d p} = 0

sin (3 x) d x + 2 y cos (33 x) d y = 0

What is the solution for y(dy)/(dx)=x^2,y(0)=1 ?
The solution for y(dy)/(dx)=x^2,y(0)=1 is y=sqrt((2x^3)/3+1)