What is the solution for y^'=-2/4 x^3y^8 ?

The solution for y^'=-2/4 x^3y^8 is y=-\sqrt[7]{8/(7(-x^4+c_{1))}}

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y^'=-2/4 x^3y^8

y^{^{'}} = - \frac{2}{4} x^{3} y^{8}

Solution steps

y^{'} (x) = - \frac{2}{4} x^{3} y^{8}

Solve separable ODE:

y^{^{'}} = - t y^{3}

\frac{d y}{d x} y = (3 + x)^{\frac{5}{x}}

f^{^{'}} (x) 4 f (x) = 2 x - 6

y y^{^{'}} = (1 - y^{2}) sin (x)

H (x) x d x - x (- ln (H (x))) (H (x) d x + x d H (x)) = 0

What is the solution for y^'=-2/4 x^3y^8 ?
The solution for y^'=-2/4 x^3y^8 is y=-\sqrt[7]{8/(7(-x^4+c_{1))}}