## Calculus (3rd Edition)

$$y(x)=\frac{1}{x+1}\left(5+\ln x-\frac{1}{x}\right)$$
Given $$y^{\prime}+\frac{1}{x+1} y=x^{-2}, \quad y(1)=2$$ This a linear equation with $p(x) =\frac{1}{1+x}$ and $q(x) =x^{-2}$, so \begin{align*} \mu(x) &= e^{\int p(x)dx}\\ &=e^{\int\frac{1}{1+x} dx}\\ &= e^{\ln (1+x)}\\ &=1+x \end{align*} Hence the general solution is \begin{align*} \mu(x) y &=\int \mu (x)xq(x)dx\\ (1+x) y&=\int (1+x)x^{-2}dx \\ &=\int \frac{1}{x}+ x^{-2}dx\\ &=\ln x- x^{-1}+C \end{align*} Then $$y(x)=\frac{1}{x+1}\left(C+\ln x-\frac{1}{x}\right)$$ Since $y(1)= 2$, then $C=5$, and hence $$y(x)=\frac{1}{x+1}\left(5+\ln x-\frac{1}{x}\right)$$