Show that the generalized functions displayed on the right are generalized solutions in $\mathscr{D'}(\mathbb{R}^{1})$
$$x u' = p.v.(\frac{1}{x}), \quad u = c_{1} + c_{2} \theta(x) - p.v.(\frac{1}{x})$$
$$x^{2} u' = 1, \quad u = c_{1} + c_{2} \theta(x) + c_{3} \delta(x) - p.v.(\frac{1}{x}),$$
where $\theta(x)$ is Heaviside's function such that $\theta(x) = 1, x \geq 0$ and $\theta(x) = 0, x < 0$ and $\theta'(x) = \delta(x)$; where $\delta(x)$ is Dirac's delta function.
I am confused about what precisely the problem asks and especially by the notations $x u'$ and $x^{2} u'$.
As I understand it is just derive $u$ and set into the respective equations i.e. for the first one it would be $u' = \frac{d}{dx} u = \theta'(x) + p.v.(\frac{1}{x^{2}})$.
And for the second we have $u' = \theta'(x) + \delta'(x) + p.v.(\frac{1}{x^{2}})$. But then what?
The first one: $$ u = c_{1} + c_{2} \theta(x) - \text{pv}\frac{1}{x} \\ u' = c_{2} \delta(x) + \text{fp}\frac{1}{x^2} \\ xu' = c_{2} x \,\delta(x) + x \,\text{fp}\frac{1}{x^2} = \text{pv}\frac{1}{x} \\ $$
The second one: $$ u = c_{1} + c_{2} \theta(x) + c_{3} \delta(x) - \text{pv}\frac{1}{x} \\ u' = c_{2} \delta(x) + c_{3} \delta'(x) + \text{fp}\frac{1}{x^2} \\ xu' = c_{2} x\,\delta(x) + c_{3} x\,\delta'(x) + x\,\text{fp}\frac{1}{x^2} = - c_{3} \delta(x) + \text{pv}\frac{1}{x} \\ x^2 u' = - c_{3} x\, \delta(x) + x \, \text{pv}\frac{1}{x} = 1 \\ $$