Question

(3D^2-2D^2+D-1)z=4e^(x+y).cos(x+y)

Answer 1

Given equation is-

$(3D^2-2D^2+D-1)z=4e^{(x+y)}.cos(x+y)\\[9pt](D^2+D-1)z=4e^{(x+y)}.cos(x+y)$

Auxiliary equation is-

$m^2+m-1=0$

$m=\dfrac{-1\pm \sqrt{1-4}}{2}=\dfrac{-1\pm \sqrt{3}i}{2}$

Complimentary function is-

$CF=e^{-\frac{x}{2}}(c_1cos \dfrac{\sqrt{3}}{2}x+c_2sin\dfrac{\sqrt{3}}{2}x)$

Particular integral is-

$PI=\dfrac{4e^{x+y}cos(x+y)}{D^2+D-1}$

$=4e^{x+y}\dfrac{cos(x+y)}{(D+1)^2+(D+1)-1}\\[9pt]=4e^{x+y}\dfrac{cos(x+y)}{D^2+3D+1} \\[9pt]=4e^{x+y}\dfrac{cos(x+y)}{-1+3D+1}\\[9pt]=4e^{x+y}\dfrac{cos(x+y)}{3D}\\[9pt]=-4e^{(x+y)}\times \dfrac{sin(x+y)}{3}$

Complete solution is-

y=CF+PI

$y = e^{-\frac{x}{2}}(c_1cos \dfrac{\sqrt{3}}{2}x+c_2sin\dfrac{\sqrt{3}}{2}x)-4e^{(x+y)}\times \dfrac{sin(x+y)}{3}$