What is the solution for (dy)/(dx)=(1+y^2)/(2yx) ?

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

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(dy)/(dx)=(1+y^2)/(2yx)

\frac{d y}{d x} = \frac{1 + y ^{2}}{2 y x}

y = c_{1} x - 1, y = - c_{1} x - 1

Solution steps

\frac{d y}{d x} = \frac{1 + y ^{2}}{2 y x}

Solve separable ODE:

y = c_{1} x - 1, y = - c_{1} x - 1

y = c_{1} x - 1, y = - c_{1} x - 1

y = x y^{^{'}} - \frac{1}{y}

y^{^{'}} = \frac{2 e ^{x}}{e ^{y}}

(y + 1) cos (x) d x - d y = 0

\frac{d}{d x} (x^{4} + y^{4}) = 4, x^{3} + 4 y^{3} y^{^{'}} = 0

3 (1 + q^{2}) \frac{d p}{d q} = 2 pq (p^{3} - 1)

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