Solve by annihilator method y 2y y ex xSolutionFirst we

Solve by annihilator method y\" + 2y\' + y = e^-x + x

Solution

First we look at annihilator of associated homogenous equation

y\'\'+2y\'+y=0

y\'\'=D^2y, 2y\'=2Dy

SO we have

(D^2+2D+1)y=0

SO annihilatro is

D^2+2D+1=(D+1)^2

Annihilator of e^{-x} is?

De^{-x}=-e^{-x}

So, (D+1)e^{-x} =0

HEnce annihilator of e^{-x}=D+1

For x it is D^2

So annihilator of this equation is

D^2(D+1)(D+1)^2=D^2(D+1)^3

General solution corrspondong to D^2=A+Bx

Corresponding to (D+1)^3=e^{-x}(C+Dx+Ex^2)

Hence general solution is

y=A+Bx+e^{-x}(C+Dx+Ex^2)