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

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

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(x+1)y^'+y=ln(x)

(x + 1) y^{^{'}} + y = ln (x)

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

Solution steps

(x + 1) y^{'} (x) + y = ln (x)

Solve linear ODE:

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

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

\frac{1 + x}{1 - x}

x \to 0 + lim ((1 + 2 x)^{\frac{1}{x}})

x \to 1.5 lim (\frac{1}{x - 2})

x y^{^{'}} + y = 4 x y^{2}, y (1) = 9

\int - 8 e^{x + 4} - \frac{5}{x + 7} - 5 sec^{2} (x + 6) d x

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