Given differential equation is

\(x^3\frac{d^3y}{dx^3}+3x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=xlog x\) which is

homogeneous linear differential equation.

Let x = e^z

⇒ log x  = z

Then x \(\frac{dy}{dx}=\frac{dy}{dz}=Dy\)

and x²\(\frac{d^2y}{dx^2}=D(D-1)y\)

and x³\(\frac{d^3y}{dx^3}=D(D-1)(D-2)y\)

Then given differential equation reduces to

D(D - 1)(D - 2)y + 3D(D - 1)y + Dy + y = ae^z

⇒ (D³ + 3D² + 2D)y + (3D²- 3D)y + Dy + y = ze^z

⇒ (D³ - 3D² + 3D² + 2D - 3D + D + 1)y = ze^z

⇒ (D³ + 1) y = ze^z

Its auxiliary equation is

m³ + 1 = 0

⇒ (m + 1)(m² - m + 1) = 0

⇒ m = -1 or m = \(\frac{1\pm\sqrt{1-4}}2=\frac{1\pm\sqrt3i}2\)

C.F. = C₁e^-z + e^z/2(C₂ cos(\(\frac{\sqrt3}2\)z) + C₃(\(\frac{\sqrt3}2\)z))

\(=\frac{C_1}x+x^{1/2}(C_2cos(\frac{\sqrt3}2log x)+C_3(\frac{\sqrt3}2log x))\)

(\(\because\) z = log x and e^z = x)

P. I. = \(\frac{1}{D^3+1}ze^z=e^z\frac{1}{(D+1)^3+1}z\)

 = e^z \(\frac{1}{D^3+3D^2+3D+1+1}z\)

 =   e^z \(\frac{1}{D^3+3D^2+3D+2}z\) 

 =  \(\frac{e^z}2\cfrac{1}{1+\frac{D^3+3D^2+3D}2}z\)

=   \(\frac{e^z}2(1+\frac{D^3+3D^2+3D}2)^{-1}\) z

= \(\frac{e^z}2(1-\frac{D^2+3D^2+3D}2+....)z\) (\(\because\) (1 + x)^-1 = 1 - x + x² - x³ + 0)

 = \(\frac{e^z}2(z-\frac32Dz)\)

= \(\frac{e^z}2(z - \frac32)\)

 = \(\frac{x}2(log x-\frac32)\) (\(\because\) z = log x and e^z = x)

\(\therefore\) Complete solution is

y = C.F. + P. I.

 = \(\frac{C_1}x+\sqrt x[C_2cos(\frac{\sqrt3}2log x)C_3sin(\frac{\sqrt3}2log x)]\) + \(\frac{x}2(log x-\frac32)\)