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

The solution for (dy)/(dx)= 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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(dy)/(dx)= 1/(e^y-x)

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

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

Solution steps

\frac{d y}{d 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})

\frac{d y}{d x}, \frac{6 x}{y ^{7}} = 4

\frac{\partial}{\partial t} (e^{t} cos (x))

20 t e^{- 2 t} \cdot sin (4 t)

ln (ln (x))

x \to 0 + lim (\frac{x}{- 3 x ^{3} - 10 x})

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