Answer
$y = \frac{x~ln~x}{2}-\frac{x}{4}+\frac{1}{4x}$
Work Step by Step
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation:
$xy' + y = x~ln~x$
$y' + \frac{y}{x} = ln~x$
Let $P(x) = \frac{1}{x}$
Let $Q(x) = ln~x$
We can find the integrating factor:
$I(x) = e^{\int \frac{1}{x} ~dx} = e^{ln~x} = x$
We can multiply by the integrating factor:
$xy'+y = x~ln~x$
$\frac{d}{dx}(x~y) = x~ln~x$
We can integrate both sides of the equation:
$x~y = \int x~ln~x~dx$
Let $u = ln~x$
$\frac{du}{dx} = \frac{1}{x}$
Let $dv = x~dx$
$v = \frac{x^2}{2}$
$\int u~dv = uv - \int v~du$
$\int u~dv = \frac{x^2~ln~x}{2} - \int \frac{x}{2}~dx$
$\int u~dv = \frac{x^2~ln~x}{2} - \frac{x^2}{4}+C$
Then:
$x~y = \frac{x^2~ln~x}{2} - \frac{x^2}{4}+C$
$y = \frac{x~ln~x}{2}-\frac{x}{4}+\frac{C}{x}$
We can use the initial condition to find $C$:
$y(1) = 0$
$\frac{(1)~ln~1}{2}-\frac{1}{4}+\frac{C}{1} = 0$
$0-\frac{1}{4}+C = 0$
$C = \frac{1}{4}$
We can write the solution:
$y = \frac{x~ln~x}{2}-\frac{x}{4}+\frac{1}{4x}$
