x^2 (d^2 y/dx^2 ) −3x (dy/dx) + 4y = ℓn^2 (x)−ℓn(x)


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Question Number 123207 by benjo_mathlover last updated on 24/Nov/20

$x^{2} \frac{d^{2} y}{{d x}^{2}} - 3 x \frac{d y}{d x} + 4 y = ℓ n^{2} (x) - ℓ n (x)$
Commented by mnjuly1970 last updated on 24/Nov/20

$c a u c h y d i f f e r e n t i a l e q u a t i o n .$
Commented by mohammad17 last updated on 24/Nov/20
$put:x=e^u ⇒u=lnx let:xD=D_1 ⇒x^2 D^2 =D_1 ^2 −D_1 ⇒(D_1 ^2 −4D_1 +4)y=u^2 −u (m−2)^2 =0⇒m_1 =m_2 =2 Yh=(c_1 +c_2 x)e^(2u) Yp=(1/((D_1 −2)^2 ))(u^2 −u) Yp=(1/4)×(1/((1−(D_1 /2))^2 ))(u^2 −u) Yp=(1/4)[1+D_1 +3(D_1 ^2 /4)+4(D_1 ^3 /8)+....](u^2 −u) Yp=(1/4){u^2 −u+2u−1+(3/2)} Yp=(1/4)u^2 +(1/4)u+(1/8) put:u=lnx ∴Yh=(C_1 +C_2 lnx)x^2 ∴Yp=(1/4){ln^2 x+lnx+(1/2)} Y=Yh+Yp=(C_1 +C_2 lnx)x^2 +(1/4){ln^2 x+lnx+(1/2)} by:⟨m.o⟩$
$p u t : x = e^{u} \Rightarrow u = l n x$ $l e t : x D = D_{1} \Rightarrow x^{2} D^{2} = D_{1}^{2} - D_{1}$ $\Rightarrow (D_{1}^{2} - 4 D_{1} + 4) y = u^{2} - u$ ${(m - 2)}^{2} = 0 \Rightarrow m_{1} = m_{2} = 2$ $Y h = (c_{1} + c_{2} x) e^{2 u}$ $Y p = \frac{1}{{(D_{1} - 2)}^{2}} (u^{2} - u)$ $Y p = \frac{1}{4} \times \frac{1}{{(1 - \frac{D_{1}}{2})}^{2}} (u^{2} - u)$ $Y p = \frac{1}{4} [1 + D_{1} + 3 \frac{D_{1}^{2}}{4} + 4 \frac{D_{1}^{3}}{8} + . . . .] (u^{2} - u)$ $Y p = \frac{1}{4} {u^{2} - u + 2 u - 1 + \frac{3}{2}}$ $Y p = \frac{1}{4} u^{2} + \frac{1}{4} u + \frac{1}{8}$ $p u t : u = l n x$ $∴ Y h = (C_{1} + C_{2} l n x) x^{2}$ $∴ Y p = \frac{1}{4} {{l n}^{2} x + l n x + \frac{1}{2}}$ $Y = Y h + Y p = (C_{1} + C_{2} l n x) x^{2} + \frac{1}{4} {{l n}^{2} x + l n x + \frac{1}{2}}$ $b y : ⟨ m . o ⟩$
	Answered by Lordose last updated on 24/Nov/20

	$I can solve but {can}^{'} t type$
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