Question

((D)^2 - (D')^2 - 3D + 3D')z = sin(x-2y)

Answer 1

((D)²-(D')²-3D+3D')z=Sin(x-2y)

Auxiliary equation of(D²-D'²-3D+3D')z=0 is;

m²-1²-3m+3*1=0

m²-3m+2=0

m²-2m-m+2=0

m(m-2)-1(m-2)=0

(m-2)(m-1)=0

m=2 or 1

Complementary function is $\phi_1(y+x)+\phi_2(y+2x)$

P.I=$\frac{1}{D^{2}-D'^{2}-3D+3D'}Sin(x-2y)$

Let f(D,D')=D²-D'²-3D+3D'

Now co-eff of x is 1=a(say)

And that of y is -2=b(say)

so that f(a,b)=f(1,-2)

=1²-(-2)²-3(1)+3(-2)

=1-4-3-6

=-12 $\not=$ 0

P.I=$\frac{1}{f(a,b)}\iint Sin \ vdvdv$ let x-2y=v

=$\frac{-1}{12}\int -Cot \ vdv=\frac{1}{12}Sin\ v$

=$\frac{1}{12}Sin(x-2y)$

Therefore the general solution is :

C.F+P.I

$=\phi_1(y+x)+\phi_2(y+2x)+\frac{1}{12}Sin(x-2y)$