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Resolve-x-2-2-y-3-x-2-y-y-x-




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

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