Question

(x^4D^4+6x^3D^3+9x^2D^2+3xD+1)y=(1+lnx)^2

Answer 1

Given differential equation is:

$(x^4D^4+6x^3D^3+9x^2D^2+3xD+1)y=(1+lnx)^2$

Let us put, $x = e^z \implies z = ln(x), \frac{d}{dx} = D$

$(D(D-1)(D-2)(D-3)+6D(D-1)(D-2)+9D(D-1)+3D+1 )y = (1+z)^2$

$(D^4+2D^2+1)y = (1+z)^2$

The auxiliary equation will be,

$D^4+2D^2+1 =0$

$D = -i,-i,i,i$

Then,

$y = c_1(1+z)cos(z)+c_2(1+z)sin(z)$

P.I. $\frac{1}{D^4+2D^2+1}(1+z)^2 = \frac{1}{D^4+2D^2+1}(1+2z+z^2)$

$= \frac{1}{1+(D^4+2D^2)}(1+2z+z^2) = (1+(D^4+2D^2))^{-1}(1+2z+z^2)$

Expanding $(1+(D^4+2D^2))^{-1}$ binomially.

$= (1-(D^4+2D^2)+\frac{-1(-1-1)}{2!}(D^4+2D^2)^2 + .........)(1+2z+z^2)$

$= (1+2z+z^2 - 2(2)) = (z^2+2z-3)$

The solution to the equation is,

$y = c_1(1+z)cos(z)+c_2(1+z)sin(z) + (z^2+2z-3)$

$y = c_1(1+lnx)cos(lnx) + c_2 (1+lnx)sin(lnx) + (lnx)^2+3lnx - 3$