Resolve (x−2)^2 y^(′′) −3(x−2)y′+y=x


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Question Number 168549 by LEKOUMA last updated on 13/Apr/22

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