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Question Number 168549 by LEKOUMA last updated on 13/Apr/22
Resolve (x−2)^2 y^(′′) −3(x−2)y′+y=x
Resolve(x−2)2y″−3(x−2)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⇔−3a(x−2)+ax+b=x⇒{−2a=16a+b=0a=−12,b=3y=−x2+3,(x−2)2y″−3(x−2)y′+y=0x−2=t⇒y(x)=y(t+2)=z(t)dydx=z′⇔t2z″−3tz′+z=0t=es⇒z(t)=z(es)=f(s)z′(t)=f′(s).dsdt=1tf′(s)z″(t)=−1t2f′(s)+1t2f″(s)tz′(t)=f′(s)t2z″(t)=f″(s)−f′(s)⇔f″(s)−f′(s)−3f′(s)+f(s)=0f″(s)−4f′(s)+f(s)=0⇒f(s)=ae(2+3)s+be(2−3)sz(t)=at2+3+bt2−3y(x)=a(x−2)2+3+b(x−2)2−3−x2+3

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