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

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

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x(dy)/(dx)=y^2-4

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

y = \frac{2 e ^{4 c_{1}}}{x ^{4} - e ^{4 c_{1}}} + \frac{2 x ^{4}}{x ^{4} - e ^{4 c_{1}}}

Solution steps

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

Solve separable ODE:

y = \frac{2 e ^{4 c_{1}}}{x ^{4} - e ^{4 c_{1}}} + \frac{2 x ^{4}}{x ^{4} - e ^{4 c_{1}}}

y = \frac{2 e ^{4 c_{1}}}{x ^{4} - e ^{4 c_{1}}} + \frac{2 x ^{4}}{x ^{4} - e ^{4 c_{1}}}

x y^{^{'}} + y = 5 x y^{2}, y (1) = 3

\frac{( x - 1 ) ^{3}}{x ^{2}}

\int \frac{4}{4 + e ^{x}} d x

\frac{d}{d t} ((\frac{2 t + 5}{t ^{2} + 1})^{4})

\frac{\partial}{\partial u} (\frac{1}{u ^{2} + v ^{2}})

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