What is the solution for sqrt((x-4)(x+4))y^'-xy^2=0 ?

The solution for sqrt((x-4)(x+4))y^'-xy^2=0 is y=-(sqrt(x^2-16)-c_{1})/(x^2-16-c_{1)^2}

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sqrt((x-4)(x+4))y^'-xy^2=0

(x - 4) (x + 4) y^{^{'}} - x y^{2} = 0

y = - \frac{x ^{2} - 16 - c _{1}}{x ^{2} - 16 - c _{1}^{2}}

Solution steps

(x - 4) (x + 4) y^{'} (x) - x y^{2} = 0

Solve separable ODE:

y = - \frac{x ^{2} - 16 - c _{1}}{x ^{2} - 16 - c _{1}^{2}}

y = - \frac{x ^{2} - 16 - c _{1}}{x ^{2} - 16 - c _{1}^{2}}

x \frac{d y}{d x} + 3 y = 0

\frac{d y}{d x} = 90 y x^{9}

x^{3} y + 7 x^{2} y^{^{'}} + 6 y = 0

\frac{d y}{d x} x^{4} + x^{2} y^{2} = y^{2}

2 y y^{^{'}} = 12 t^{2}, y (0) = 5

What is the solution for sqrt((x-4)(x+4))y^'-xy^2=0 ?
The solution for sqrt((x-4)(x+4))y^'-xy^2=0 is y=-(sqrt(x^2-16)-c_{1})/(x^2-16-c_{1)^2}