What is the solution for 2xy+4x+(x^2-4)(dy)/(dx)=0 ?

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

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2xy+4x+(x^2-4)(dy)/(dx)=0

2 x y + 4 x + (x^{2} - 4) \frac{d y}{d x} = 0

y = \frac{c _{1}}{x ^{2} - 4} - 2

Solution steps

2 x y + 4 x + (x^{2} - 4) \frac{d y}{d x} = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{2} - 4} - 2

y = \frac{c _{1}}{x ^{2} - 4} - 2

(1 - x^{2}) \frac{d y}{d x} - 2 x \frac{d y}{d x} + 3 y = 0

\frac{d y}{d t} = \frac{y ^{2}}{t + 2}

\frac{d x}{d y} = 10 - x

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

\frac{d y}{d t} = (0.2 + sin (t)) (\frac{y}{2})

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