What is the solution for e^y(1+x^2)dy-2x(1+e^y)dx=0 ?

The solution for e^y(1+x^2)dy-2x(1+e^y)dx=0 is y=ln(e^{c_{1}}+e^{c_{1}}x^2-1)

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e^y(1+x^2)dy-2x(1+e^y)dx=0

e^{y} (1 + x^{2}) d y - 2 x (1 + e^{y}) d x = 0

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

Solution steps

e^{y} (1 + x^{2}) d y - 2 x (1 + e^{y}) d x = 0

Solve separable ODE:

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

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

\int 2 e x + 8 x^{3} + 6 x d x

\int_{0}^{1} \frac{61}{4 x - 1} d x

\frac{d}{d x} (\frac{1}{1 - \frac{2}{x}})

\frac{d}{d t} (cos (8 t))

y^{^{'}} = \frac{1}{( t \cdot y + y + t + 1 )}, y (0) = - 1

What is the solution for e^y(1+x^2)dy-2x(1+e^y)dx=0 ?
The solution for e^y(1+x^2)dy-2x(1+e^y)dx=0 is y=ln(e^{c_{1}}+e^{c_{1}}x^2-1)