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

The solution for x^2y^'=4x^2y^3-2xy is y=sqrt(3/(x(8+3c_{1)x^3))},y=-sqrt(3/(x(8+3c_{1)x^3))}

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x^2y^'=4x^2y^3-2xy

x^{2} y^{^{'}} = 4 x^{2} y^{3} - 2 x y

y = \frac{3}{x ( 8 + 3 c _{1} x ^{3} )}, y = - \frac{3}{x ( 8 + 3 c _{1} x ^{3} )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{3}{x ( 8 + 3 c _{1} x ^{3} )}, y = - \frac{3}{x ( 8 + 3 c _{1} x ^{3} )}

y = \frac{3}{x ( 8 + 3 c _{1} x ^{3} )}, y = - \frac{3}{x ( 8 + 3 c _{1} x ^{3} )}

n = 1 \sum \infty \frac{( 3 ^{n} )}{n !}

cos (4 t) cos (6 t)

\int 5 sin^{3} (x) cos (x) d x

\frac{d}{d x} ((3 - x) (2 x^{2} - x))

e^{t} cos (t) - e^{t} sin (t)

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