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

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

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y^'=(ln(x)(y^2+1))/(xy)

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

y = e^{l n^{2} (x) + c_{1}} - 1, y = - e^{l n^{2} (x) + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{l n^{2} (x) + c_{1}} - 1, y = - e^{l n^{2} (x) + c_{1}} - 1

y = e^{l n^{2} (x) + c_{1}} - 1, y = - e^{l n^{2} (x) + c_{1}} - 1

\frac{d y}{d x} = x y, y (- 1) = \frac{25}{16}

(1 + e^{x}) e^{y} y^{^{'}} = y^{- 1}

(2 y + x^{2} y) d y + (3 x + x y^{2}) d x = 0

y^{^{'}} - y^{2} - y = 0

y^{^{'}} = - x y + \frac{4 x}{y}

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