UNIVERSITY OF GAZIANTEP

Department of Electrical and Electronics Engineering EEE 285 Applied Differential Equations

Name:…………………………………… Date: 25 March 2005 Duration: 15 Minutes

Quiz # 2 Problem 1 Find the general solution of the following differential equation

030133 2

2

3

3

4

4

=++++ ydxdy

dxyd

dxyd

dxyd

if one of the solution is xe x 2sin .

Solution:

This DE is a homogeneous DE with constant coefficients. Hence the solution of the DE equation

is of the form rxey = where r is the root of the characteristic equation of the DE.

Since xe x 2sin is a solution of the DE, xe x 2cos is also another solution. The corresponding

roots of the characteristic equation are complex and they are jr 211 += , jr 212 −= .

The characteristic equation of the DE is

0))()()((30133 4321234 =−−−−=++++ rrrrrrrrrrrr

Then

3 ,2)65())((

))()(52(

))())(21())(21((30133

43

243

432

43234

−=−=⇒++=−−⇒

−−+−=

−−−−+−=++++

rrrrrrrr

rrrrrrrrrrjrjrrrrr

The general solution of the DE is therefore

xxxx ececxecxecxy 34

2321 2cos2sin)( −− +++=