What is the solution for y^'=((y^2-1))/(xy) ?

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

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y^'=((y^2-1))/(xy)

y^{^{'}} = \frac{( y ^{2} - 1 )}{x y}

y = c_{1} x^{2} + 1, y = - c_{1} x^{2} + 1

Solution steps

y^{'} (x) = \frac{( y ^{2} - 1 )}{x y}

Solve separable ODE:

y = c_{1} x^{2} + 1, y = - c_{1} x^{2} + 1

y = c_{1} x^{2} + 1, y = - c_{1} x^{2} + 1

\frac{d}{d x} (x e^{x^{4}})

\frac{d}{d z} (x^{2} y z)

\int x^{5} \cdot ln (x) d x

y^{^{''}} + 9 y = 19

f (x) = \frac{x ^{2}}{8 - 8 x}

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