Question Number 168549 by LEKOUMA last updated on 13/Apr/22
$${Resolve} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} {y}^{''} −\mathrm{3}\left({x}−\mathrm{2}\right){y}'+{y}={x} \\ $$
Answered by mindispower last updated on 13/Apr/22
$${y}={ax}+{b} \\ $$$$\Leftrightarrow−\mathrm{3}{a}\left({x}−\mathrm{2}\right)+{ax}+{b}={x}\Rightarrow\begin{cases}{−\mathrm{2}{a}=\mathrm{1}}\\{\mathrm{6}{a}+{b}=\mathrm{0}}\end{cases} \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}},{b}=\mathrm{3} \\ $$$${y}=−\frac{{x}}{\mathrm{2}}+\mathrm{3}, \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} {y}''−\mathrm{3}\left({x}−\mathrm{2}\right){y}'+{y}=\mathrm{0} \\ $$$${x}−\mathrm{2}={t}\Rightarrow{y}\left({x}\right)={y}\left({t}+\mathrm{2}\right)={z}\left({t}\right) \\ $$$$\frac{{dy}}{{dx}}={z}' \\ $$$$\Leftrightarrow{t}^{\mathrm{2}} {z}''−\mathrm{3}{tz}'+{z}=\mathrm{0} \\ $$$${t}={e}^{{s}} \Rightarrow{z}\left({t}\right)={z}\left({e}^{{s}} \right)={f}\left({s}\right) \\ $$$${z}'\left({t}\right)={f}'\left({s}\right).\frac{{ds}}{{dt}}=\frac{\mathrm{1}}{{t}}{f}'\left({s}\right) \\ $$$${z}''\left({t}\right)=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{f}'\left({s}\right)+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{f}''\left({s}\right) \\ $$$${tz}'\left({t}\right)={f}'\left({s}\right) \\ $$$${t}^{\mathrm{2}} {z}''\left({t}\right)={f}''\left({s}\right)−{f}'\left({s}\right) \\ $$$$\Leftrightarrow{f}''\left({s}\right)−{f}'\left({s}\right)−\mathrm{3}{f}'\left({s}\right)+{f}\left({s}\right)=\mathrm{0} \\ $$$${f}''\left({s}\right)−\mathrm{4}{f}'\left({s}\right)+{f}\left({s}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left({s}\right)={ae}^{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){s}} +{be}^{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){s}} \\ $$$${z}\left({t}\right)={at}^{\mathrm{2}+\sqrt{\mathrm{3}}} +{bt}^{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${y}\left({x}\right)={a}\left({x}−\mathrm{2}\right)^{\mathrm{2}+\sqrt{\mathrm{3}}} +{b}\left({x}−\mathrm{2}\right)^{\mathrm{2}−\sqrt{\mathrm{3}}} −\frac{{x}}{\mathrm{2}}+\mathrm{3} \\ $$$$ \\ $$$$ \\ $$