What is the solution for y^'+2x^{-1}y=6x^4y^2 ?

The solution for y^'+2x^{-1}y=6x^4y^2 is y= 1/(x^2(-2x^3+c_{1))}

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y^'+2x^{-1}y=6x^4y^2

y^{^{'}} + 2 x^{- 1} y = 6 x^{4} y^{2}

y = \frac{1}{x ^{2} ( - 2 x ^{3} + c _{1} )}

Solution steps

y^{'} (x) + 2 x^{- 1} y = 6 x^{4} y^{2}

Solve Bernoulli ODE:

y = \frac{1}{x ^{2} ( - 2 x ^{3} + c _{1} )}

y = \frac{1}{x ^{2} ( - 2 x ^{3} + c _{1} )}

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What is the solution for y^'+2x^{-1}y=6x^4y^2 ?
The solution for y^'+2x^{-1}y=6x^4y^2 is y= 1/(x^2(-2x^3+c_{1))}