Question: Solve differential equation $(y'')^2-5y'+6=0$

For there is no "x" in equation, I made substitution $y'=p(y)$ and $y''=p'p$

$(p'p)^2-5p+6=0$

What can I do with the squared member?

Answer: 
You have to use
$$ L\bigg\{\frac{f(t)}{t}\bigg\}=\int\_s^\infty F(\tau)d\tau $$
twice, where $F(s)=L\{f(t)\}$.
Let $f(t)=\frac{1-\cos t}{t}$ and then
\begin{eqnarray}
F(s)=L\{f(t)\}&=&\int\_s^\infty L\{1-\cos t\}(\tau)d\tau=\int\_s^\infty\bigg(\frac1\tau-\frac{\tau}{\tau^2+1}\bigg)d\tau\\
&=&\frac12\ln\bigg(\frac{\tau^2}{\tau^2+1}\bigg)\bigg|\_s^\infty=-\frac12\ln\bigg(\frac{s^2}{s^2+1}\bigg). 
\end{eqnarray}
So
\begin{eqnarray}
L\bigg\{\frac{1-\cos t}{t^2}\bigg\}&=&L\bigg\{\frac{f(t)}{t}\bigg\}=\int\_s^\infty F(\tau)d\tau\\
&=&-\frac12\int\_s^\infty \ln\bigg(\frac{\tau^2}{\tau^2+1}\bigg)d\tau\\
&=&\cdots
\end{eqnarray}
You finish the rest.
Well, first thing I thaught of was to use the fact that : $$ \left(\forall x\in\mathbb{R}^{\*}\right),\ \frac{1-\cos{x}}{x^{2}}=\int\_{0}^{1}{\left(1-y\right)\cos{\left(xy\right)}\,\mathrm{d}y} $$

Then, with Fubini's theorem will, I shall find the result.

Let $ s>0 $, we have : \begin{aligned}\int\_{0}^{+\infty}{\mathrm{e}^{-sx}\left(\frac{1-\cos{x}}{x^{2}}\right)\mathrm{d}x}&=\int\_{0}^{+\infty}{\mathrm{e}^{-sx}\int\_{0}^{1}{\left(1-y\right)\cos{\left(xy\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int\_{0}^{1}{\left(1-y\right)\int\_{0}^{+\infty}{\mathrm{e}^{-sx}\cos{\left(xy\right)}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=s\int\_{0}^{1}{\frac{1-y}{s^{2}+y^{2}}\,\mathrm{d}y}\\ &=\int\_{0}^{1}{\frac{\frac{\mathrm{d}y}{s}}{1+\left(\frac{y}{s}\right)^{2}}}-\frac{s}{2}\int\_{0}^{1}{\frac{\frac{2y\,\mathrm{d}y}{s^{2}}}{1+\left(\frac{y}{s}\right)^{2}}}\\ &=\int\_{0}^{\frac{1}{s}}{\frac{\mathrm{d}x}{1+x^{2}}}-\frac{s}{2}\int\_{0}^{\frac{1}{s^{2}}}{\frac{\mathrm{d}x}{1+x}}\\ \int\_{0}^{+\infty}{\mathrm{e}^{-sx}\left(\frac{1-\cos{x}}{x^{2}}\right)\mathrm{d}x}&=\arctan{\left(\frac{1}{s}\right)}-\frac{s}{2}\ln{\left(1+\frac{1}{s^{2}}\right)}\end{aligned}