Question

d^3y/dx^3-3 d^2y/dx^3+ 4 dy/dx -2y=e^x+cosx

Answer 1

Given equation is-

$\dfrac{d^3y}{dx^3}-3\dfrac{ d^2y}{dx^3}+ 4 \dfrac{dy}{dx} -2y=e^x+cosx$

It's auxilary equation is:-

$m^3-3m^2+4m-2=0$

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

The roots of the above equation are $1,1\pm i$

Complimentary function (C.F.) is given by

= $C_1e^x+(C_2cosx+C_3sinx)e^x$

Particular Integral-

$=\dfrac{e^x}{D^3-3D^2+4D-2}+\dfrac{cosx}{D^3-3D^2+4D-2}$

$=\dfrac{xe^x}{3D^2-6D+4}+\dfrac{cosx}{D.(-1^2)-3(-1^2)+4D-2}$

$=\dfrac{xe^x}{3-6+4}+\dfrac{cosx}{3D+1}$

$=\dfrac{xe^x}{1}+(1+3D)^{-1}cosx$

$=xe^x+(1-3D)cosx$

=$xe^x+cosx+3sinx$

Therefore Complete solution is-

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

$y=C_1e^x+(C_2cosx+C_3sinx)e^x+xe^x+cosx+3sinx$