What is the solution for y^'= 1/(e^y-x) ?

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

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y^'= 1/(e^y-x)

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

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

Solution steps

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

Solve exact ODE:

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

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

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