equation in D'(\R) we consider in $\mathcal{D}'(\mathbb{R})$ the equation 
 $$2 x T'' - T' =\delta$$.
I begin to solve the homogeneous equation $2x T'' - T'=0$
i search an solution $T=x^r$ with $r \in \mathbb{R}$, then i found that:
$T_h(x)= C_1 x^{1/2} + C_2 x$ 
my question is: can you give me please an indication to found an particular solution for the on homogeneous equation?
Thank's for the hel
Ma question est: pouvez vous me donner une indication pour trouver une solution particulière à l'équation non homogène? S'il vous plaît.
Merci d'avance.

Let's pose $S = T'$. Then:
$$2 x T'' - T' =\delta \Rightarrow 2x S' - S = \delta \Rightarrow \\
\Rightarrow 2xS' = \delta + S \Rightarrow 2x \frac{dS}{dx} = \delta + S \Rightarrow \\
\frac{dS}{\delta + S} = \frac{dx}{2x}.$$
Integrating both side...
$$\int\frac{dS}{\delta + S} = \int \frac{dx}{2x} \Rightarrow \log(\delta+S) = \frac{1}{2}\log(x) + c.$$
Imposing $c = \frac{1}{2}\log(a)$ then:
$$\int\frac{dS}{\delta + S} = \int \frac{dx}{2x} \Rightarrow \log(\delta+S) = \frac{1}{2}\log(ax) \Rightarrow \delta + S = \sqrt{ax} \Rightarrow \\
S(x) = \sqrt{ax} + \delta.$$
Let's  find $T(x)$:
$$T(x) = \int \left(\sqrt{ax} + \delta\right)dx = \frac{2}{3}\sqrt{a}x^{\frac{3}{2}} + \delta x + b.$$
Now, we can check if this work:
$$T'(x) = \sqrt{ax} + \delta, T''(x) = \frac{1}{2}\sqrt{\frac{a}{x}},$$
and hence:
$$2 x T'' - T' =\delta \Rightarrow 2x\frac{1}{2}\sqrt{\frac{a}{x}} - \sqrt{ax} + \delta = \delta \Rightarrow \\
\Rightarrow \sqrt{ax} - \sqrt{ax} + \delta = \delta \Rightarrow 0 = 0.$$
Finally, we said that the solution is:
$$T(x) = \frac{2}{3}\sqrt{a}x^{\frac{3}{2}} + \delta x + b.$$
Since $a$ and $b$ are constants, then we can pose:
$$\begin{cases}
C_1 = \frac{2}{3}\sqrt{a}\\
C_2 = b
\end{cases},$$
thus obtaining
$$T(x) = C_1 x^{\frac{3}{2}} + \delta x + C_2.$$
