Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : 2015

PI is $y_p=\dfrac 1{f(D)}\\ y_p=\dfrac 1{D^2-4D+4}(e^x+\cos 2x)\\ =\dfrac 1{(D-2)^2}e^x +\dfrac 1{D^2-4D+4}\cos 2x\\ =e^x\dfrac 1{(1-2)^2}+\dfrac 1{-2^2-4D+4}\cos 2x\\ =e^x+\dfrac 1{-4D}\cos 2x\\ =e^x-\dfrac 14\int \cos 2x\space dx\\ =e^x-\dfrac 14\dfrac {\sin 2x}2\\ \therefore PI \space is \space e^x-\dfrac18\sin 2x$