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

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

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

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

y = \frac{ln ( 1 + e ^{x} )}{e ^{x}} + \frac{c _{1}}{e ^{x}}

Solution steps

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

Solve linear ODE:

y = \frac{ln ( 1 + e ^{x} )}{e ^{x}} + \frac{c _{1}}{e ^{x}}

y = \frac{ln ( 1 + e ^{x} )}{e ^{x}} + \frac{c _{1}}{e ^{x}}

\frac{d}{d x} (sin (x^{0}))

\int_{0}^{t} 5 e^{s} sin (t - s) d s

\frac{d y}{d x} + y = - 1, x > 1

\int \frac{cos ( x )}{( sin ^{2} ( x ) ) ^{\frac{1}{3}}} d x

x \to 0 lim (\frac{sin ( 14 x )}{x})

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