Tricks to Solving Higher Order Differential Equations -3 (Auxiliary Eqns.with Complex roots)

We now consider the situation where the auxiliary equation with real coefficients contains a complex root, say a + bi, which means it must also contain its conjugate, a – bi. The complementary function for the given differential equation would then contain the terms:

 

c1 exp(a +bi)x + c2(a- bi)x

 

which may be written, alternatively,

 

c1exp (ax)[ cos bx + i sin bx]  +  c2exp (ax)[ cos bx -  sin bx] 

 

The terms can then be re-arranged to get:

exp (ax) [(c1 + c2) cos bx + i(c1 – c2) sin bx]

 

If we let c3 = c1 + c2 and i(c1 – c2) = c4, we have:

 

y= exp(ax) [c3 cos bx + c4 sin bx]

 

so that part of the solution corresponding to the two complex roots, m1 = a +bi and m2 = a – bi, has real numbers c3 and c4.  (if we choose c1 and c2 as complex conjugate numbers).

 

Note that a 2^nd part of the solution can be obtained by choosing:

 

c = [c3² + c4 ²]^1/2  and  tan a = c3/ c4

Then we have:

sin  a = c3/ c  and cos a = c4/ c

 

Multiply the y solution in c3, c4 by c/c = 1 and one gets:

 

y= exp(ax) c  [c3/ c cos bx + c4/c sin bx]

 

or: 

 

y= c exp(ax)   [sin  a cos bx + cos a  sin bx]

 

=  c exp(ax)   sin ( bx +  a)

 

In this last expression c and a are two arbitrary constants.

 

Example:

 

Solve the differential equation: y” + y’ + y = 0

 We write out the auxiliary equation:

m² + m + 1 = 0

Solve using the quadratic formula to obtain:

 

m1 =  - ½  + Ö3 i / 2

and:

m2 =  - ½  -  Ö3  i/ 2

 Then the general solution is:

y= exp (-x/2)[ c1 cos Ö3 x/ 2   + c2 sin Ö3  x/ 2 ]

 

Problems for the Math Maven:

 

1)     Find the general solution of: y”’ – 3y” + 7y’ – 5y = 0

2)     Find the general solution of: y^iv + 18y” + 81 = 0

In each case put the solution in a form analogous to:

y= exp(ax) [c3 cos bx + c4 sin bx]