## Calculus (3rd Edition)

$$y=\frac{e^{x}}{x}+\frac{3-e}{x}$$
Given $$x y^{\prime}+y=e^{x}, \quad y(1)=3$$ Then $$y^{\prime}+\frac{1}{x}y=\frac{1}{x}e^{x}$$ This a linear equation with $p(x) =\frac{1}{x}$ and $q(x) =\frac{1}{x}e^{x}$, so \begin{align*} \mu(x) &= e^{\int p(x)dx}\\ &=e^{\int\frac{1}{x} dx}\\ &= e^{\ln x}\\ &=x \end{align*} Hence the general solution is \begin{align*} \mu(x) y &=\int \mu (x)xq(x)dx\\ x y&=\int \frac{x}{x}e^{x}dx \\ &=\int e^{x}dx\\ &=e^{x}+C \end{align*} Then $$y= \frac{1}{x}e^{x}+C \frac{1}{x}$$ Since $y(1)= 3$, then $C= 3-e$, and hence $$y=\frac{e^{x}}{x}+\frac{3-e}{x}$$