What is the solution for yy^'-3x^2-x=0 ?

The solution for yy^'-3x^2-x=0 is y=sqrt(2x^3+x^2+c_{1)},y=-sqrt(2x^3+x^2+c_{1)}

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yy^'-3x^2-x=0

y y^{^{'}} - 3 x^{2} - x = 0

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

Solution steps

y y^{'} (x) - 3 x^{2} - x = 0

Solve separable ODE:

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

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

e^{x} cos (y) d x + (1 + e^{x}) sin (y) d y = 0

y^{^{'}} = 6 (x y)^{2}

\frac{d y}{d x} = e^{x + y}, y (0) = 0

x (H (x) x)^{2} d x + x^{2} (H (x) x) (H (x) d x + x d H (x)) = 0

(x y^{2} - x) d x + (2 x^{2} y + 8 y) d y = 0

What is the solution for yy^'-3x^2-x=0 ?
The solution for yy^'-3x^2-x=0 is y=sqrt(2x^3+x^2+c_{1)},y=-sqrt(2x^3+x^2+c_{1)}