Question

d^2-6dd'+9d'^2=12x^2+36xy

Answer 1

Given differential equation is $(D^2-6DD'+9D'^2)z=12x^2+36xy$

where $D = \frac{d}{dx}, D' = \frac{d}{dy}$ .

$\implies (D-3D')^2 z = 12 x^2 + 36 xy$

So, Auxiliary equation is $(D-3D')^2 = 0 \implies D-3D' = 0 , D - 3D' = 0$ .

Hence, Complementary function (C.F.) = $\phi_1(y+3x) + x \phi_2(y+3x)$ .

Particular Integral (P.I.) = $\frac{1}{(D-3D')^2} (12x^2 + 36 xy)$

$= \frac{1}{D^2} \left(1-\frac{3D'}D{}\right)^{-2} (12x^2+36xy)$

$= \left(1+2\frac{3D'}{D}+3\right(\frac{3D'}{D}\left)^2+4\right(\frac{3D'}{D}\left)^3+…\right) \int\int(12x^2+36xy) dxdx$

$= \left(1+6\frac{D'}{D}+27\right(\frac{D'}{D}\left)^2+108\right(\frac{D'}{D}\left)^3+…\right) \left[ x^4 + 6 x^3 y \right]$

$= x^4 + 6 x^3 y + \frac{6}{D} (6x^3)$

$= x^4 + 6x^3 y + 9 x^4 \\ = 10 x^4 + 6x^3 y$

Hence, solution of given differential equation is

$z$ = C.F. + P.I.

$\implies z = \phi_1(y+3x)+x\phi_2(y+3x) + 10 x^4 + 6x^3 y$