Question

(D3 − 3D2 − 6D + 8)y = xe−3x

Answer 1

Given,

$(D^3 − 3D^2 − 6D + 8)y = xe^{−3x}$

Its auxilary equation is-

$m^3-3m^2-6m+8=0$

$m^3-m^2-2m^2+2m-8m+8=0\\ m^2(m-1)-2m(m-1)-8(m-1)=0\\ (m-1)(m^2-2m-8)=0$

The roots of the equation is-

$m=1,4,-2$

Complimentary function is-

$C.F.=c_1e^x+c_2e^{4x}+c_3e^{-2x}$

Particular integral

$P.I.=\dfrac{e^{-3x}x}{D^3-3D^2-6D+8}$

$=\dfrac{e^{-3x}x}{(D-3)^2-3(D-2)^2-6(D-1)+8}$

$=\dfrac{e^{-3x}x}{D^3-12D^2+33D-25}$

$=\dfrac{e^{-3x}x}{-25(1-(\dfrac{D^3-12D^2+33D)}{25})}$

$=\dfrac{e^{-3x}}{-25}(1-\dfrac{D^3-12D^2+33D}{25})x$

$=\dfrac{-e^{-3x}}{25}(x-\dfrac{33}{25})$

$=\dfrac{-e^{-3x}}{25}(x-\dfrac{33}{25})$

Complete solution is-

$y=c_1e^x+c_2e^{4x}+c_3e^{-2x}-\dfrac{e^{-3x}}{25}(x-\dfrac{33}{25})$