What is the solution for (x^2-1)y^'=xyln(y) ?

The solution for (x^2-1)y^'=xyln(y) is y=e^{c_{1}sqrt(x^2-1)}

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(x^2-1)y^'=xyln(y)

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

y = e^{c_{1} x^{2} - 1}

Solution steps

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

Solve separable ODE:

y = e^{c_{1} x^{2} - 1}

y = e^{c_{1} x^{2} - 1}

\int_{1}^{\infty} x^{- \frac{4}{3}} d x

\frac{\partial}{\partial x} (e^{- x^{2} - 2 y^{2}})

\int 8 sec (x) (sec (x) - 4 tan (x)) d x

y^{^{''}} + 8 y^{^{'}} + 4 y = t

\frac{\partial}{\partial y} (sin (z))

What is the solution for (x^2-1)y^'=xyln(y) ?
The solution for (x^2-1)y^'=xyln(y) is y=e^{c_{1}sqrt(x^2-1)}