What is the solution for (x^2y^2-x^4)y^'=x^2y^2-y^4 ?

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

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(x^2y^2-x^4)y^'=x^2y^2-y^4

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

y = - \frac{x}{1 + c _{1} x}

Solution steps

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

Solve separable ODE:

y = - \frac{x}{1 + c _{1} x}

y = - \frac{x}{1 + c _{1} x}

x \to - 1 lim (ln (1 + x))

\frac{d}{d x} (ln (4 x - 5))

\frac{1}{1 - x - x ^{2}}

t - 2

f (x) = x^{2} + 4, at x = - 1

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