0
22kviews
Solve the $[D^4-4D^3+8D^2-8D+4]y=0$

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

Marks : 3

Year : 2014

1

Please explain this question to me


1 Answer
2
5.4kviews

Auxiliary equation is $[D^4-4D^3+8D^2-8D+4]y=0 \\ \therefore D^4+4D^2+4-4D^3-8D+4D^2=0 \\ \therefore (D^2)^2 + (-2D)^2(2)^2 +2D^2(-2D)+2(-2D)(2)+2D^2(2)=0\\ =(D^2-2d+2)^2=0\\ \therefore D=\dfrac {2\pm\sqrt{2^2-4(1)(2)}}2 \\ =\dfrac {2\pm\sqrt{-4}}2 \\ =\dfrac {2\pm2i}2\\ \therefore D=1\pm i,1\pm i \\$ $C.F \space is\space y_c=e^x[(c_1+xc_2)\cos x+(c_3+xc_4)\sin x]$

Please log in to add an answer.