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

The solution for yy^'=x-2x^3 is y=sqrt(x^2-x^4+c_{1)},y=-sqrt(x^2-x^4+c_{1)}

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yy^'=x-2x^3

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

y = x^{2} - x^{4} + c_{1}, y = - x^{2} - x^{4} + c_{1}

Solution steps

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

Solve separable ODE:

y = x^{2} - x^{4} + c_{1}, y = - x^{2} - x^{4} + c_{1}

y = x^{2} - x^{4} + c_{1}, y = - x^{2} - x^{4} + c_{1}

\frac{d}{d x} (a^{7} + (cos (x))^{7})

x = y^{2} + 4 y - 22, x = 3 y + 8

x - tan (x)

x (x + y) \frac{d y}{d x} = y (x - y)

4 x + 9

What is the solution for yy^'=x-2x^3 ?
The solution for yy^'=x-2x^3 is y=sqrt(x^2-x^4+c_{1)},y=-sqrt(x^2-x^4+c_{1)}