Question

(D^3-7DD'^2-6D'^3)z=cos(2x-3y)

Answer 1

$(D^3-7DD'^2-6D'^3)z=cos(2x-3y)$

Auxillary equation is:

$m^3-7m-6=0$

$(m+2)(m+1)(m-3)=0$

$m = -1, -2, 3$

C.F.=$f_1(y-x)+f_2(y-2x)+f_3(y+3x)$

P.I. $=\frac 1 {D^3-7DD'^2-6D'^3}cos(2x-3y)$

Put $D^2 = -4, D'^2 = -9$

P.I. $= \frac 1 {-4D+63D+54D'}cos(2x-3y) =$ $\frac 1 {59D+54D'}cos(2x-3y)$

$=\frac {59D-54D'} {3481D^2-2916D'^2}cos(2x-3y)$ $= \frac {59D-54D'} {=13924+26244}cos(2x-3y)$

$=\frac 1 {12320}(59Dcos(2x-3y)-54D'cos(2x-3y))$

$=\frac 1 {12320}(-118sin(2x-3y)-162sin(2x-3y))$

$=-\frac {280}{12320}sin(2x-3y)=-\frac{7}{176}sin(2x-3y)$

The complete solution $z=$ C.F.+P.I.$=f_1(y-x)+f_2(y-2x)+f_3(y+3x)$ $-\frac{7}{176}sin(2x-3y)$

Answer: $z=f_1(y-x)+f_2(y-2x)+f_3(y+3x)-\frac{7}{176}sin(2x-3y)$