What is the solution for e^xdx-(1+e^x)ydy=0 ?

The solution for e^xdx-(1+e^x)ydy=0 is y=sqrt(2ln(1+e^x)+c_{1)},y=-sqrt(2ln(1+e^x)+c_{1)}

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e^xdx-(1+e^x)ydy=0

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{x ^{2}}{y ^{2} - 3} \frac{d y}{d x} = \frac{1}{2 y}

sin (x) cos (y) d x + cos (x) sin (y) d y = 0

3 x^{2} y^{2} \frac{d y}{d x} = 2 x - 1

\frac{d y}{d x} = \frac{2 x sin ^{2} ( y )}{( x ^{2} + 10 ) cos ( y )}

y^{2} d x + d y = 0

What is the solution for e^xdx-(1+e^x)ydy=0 ?
The solution for e^xdx-(1+e^x)ydy=0 is y=sqrt(2ln(1+e^x)+c_{1)},y=-sqrt(2ln(1+e^x)+c_{1)}