What is the solution for y^'+1/(1+x)y= 2/(1+x) ?

The solution for y^'+1/(1+x)y= 2/(1+x) is y=-(c_{1})/(1+x)+2

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y^'+1/(1+x)y= 2/(1+x)

y^{^{'}} + \frac{1}{1 + x} y = \frac{2}{1 + x}

y = - \frac{c _{1}}{1 + x} + 2

Solution steps

y^{'} (x) + \frac{1}{1 + x} y = \frac{2}{1 + x}

Solve separable ODE:

y = - \frac{c _{1}}{1 + x} + 2

y = - \frac{c _{1}}{1 + x} + 2

11 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 3

y^{^{'}} = 1 - x + y - x y, y (0) = 0

5 y^{^{'}} = y cos (x), y (0) = 2

\frac{d v}{d t} = 9.81 - 3 v

y^{^{'}} = x y^{2} sin (x^{2})

What is the solution for y^'+1/(1+x)y= 2/(1+x) ?
The solution for y^'+1/(1+x)y= 2/(1+x) is y=-(c_{1})/(1+x)+2