A differential equation is given asx2d2y- 2xdy+ 2y = 4dx2dxThe

A differential equation is given as
x² d²y - 2x dy + 2y = 4
dx² dx

The solution of the differential equation in terms of arbitrary constants C₁ and C₂ is

x² =	d²y	- 2x	dy	+ 2y = 4
	dx²		d

Let x = e^z

CF :	d²y	= θ(θ - 1)
	dx²

⇒ θ(θ - 1) - 2θ + 2 = 0
⇒ θ² - 3θ + 2 = 0
(θ - 1)(θ - 2) = 0
θ = 1,2
θ = C₁e^z + C₁e^2z
y = C₁x² + C₂x

PI : -	4	+	4
	(θ - 1)		(θ - 2)

= + 4(1 -θ)^-1 -	4			1 -	θ		^-1
	2				2

= + 4(1 - θ + ...) - 2		1 +	θ	+ .....
			2

= + 4 – 2 = 2 (Neglecting higher order term)
y = CF + PI
y = C₁ x² C₂ x + 2