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}}

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)