Find the solutions of a differential equation Let $u:\mathbb{R^+}\to\mathbb{R}$. Find the solutions of the following differential equation
$$u''(x)+\displaystyle\frac{u'(x)}{x}=c\log x $$. 
 A: You can set $\frac{du(x)}{dx} = v(x)$ to start and solve for the linear differential equation
$$v'(x)+\dfrac{v(x)}{x} = c\log(x).$$
set $\mu (x) = e^{\int\frac{1}{x}dx} = x.$
Multiply both sides of the equation by $\mu (x)$ to get 
$$x v'(x) +v(x) = cx\log(x)$$ substitute $\frac{d(x)}{dx} =1$ to get
$$xv'(x)+\frac{d}{dx}(x v(x)) = cx\log(x).$$ Then you can use the reverse product rule to get 
$$xv'(x)+\frac{d}{dx}(x v(x)) = \frac{d}{dx}(x v(x)) = cx\log(x).$$ So now your equation is 
$$\frac{d}{dx}(x v(x)) = cx\log(x).$$ Integrate both sides. From here you can evaluate the integrals and substitute $v(x) = \frac{du(x)}{dx}$.
The solution is 
$$u(x) = \dfrac{-cx^2}{4}+\dfrac{1}{4}cx^2\log(x)+a_1\log(x)+a_2$$
where $a_1$ and $a_2$ are arbitrary constants. 
A: The trick is to rewrite the equation as
$xu''(x) + u'(x) = c x \log x$,
and then to recognize that the left hand side is the derivative of $xu'(x)$ by the product rule. Then integrating, we have
$xu'(x) = c \left( \frac{1}{2} x^2 \log x - \frac{1}{4}x^2 \right) + A$,
where is a constant of integration, and the integral was computed using integration by parts. Then divide the whole equation by $x$ and integrate again:
$u'(x) = c \left( \frac{1}{2} x \log x - \frac{1}{4}x \right) + \frac{A}{x}$
$u(x) = c \left( \frac{1}{4} x^2 \log x - \frac{1}{4} x^2 \right) + A \log x + B$.
