How to find the auxillary equation and the final solution for #(d^2Phi)/(dphi^2) + BPhi = 0# assuming #Phi = e^(im_lphi)#?

1 Answer
Steve M
Feb 4, 2018

How to find the auxiliary equation and the final solution
for #(d^2Phi)/(dphi^2) + BPhi = 0# assuming #Phi = e^(im_lphi)#?

Explanation:

We have:

# (d^2Phi)/(dphi^2) + BPhi = 0 #

We assume that #B in RR, B != 0 #

This is a second order linear Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives.

Complementary Function

The associated Auxiliary equation is:

# m^2+0m+B = 0#
# :. m^2 = B#

The sign of B will determine the possible solution. Then

Case 1: #B gt 0#

Real distinct roots, given by #m=+-B #, so that:

# Phi = C_1e^(Bphi) + C_2Be^(-Bphi) #

Case 2: #B lt 0#

Pure imaginary roots, given by #m=+-Bi #, so that:

# Phi = e^(0phi){C_1cos(Bphi) + C_2sin(Bphi)} #
# \ \ \ = C_1cos(Bphi) + C_2sin(Bphi) #

Here, we are given the form of the solution. Let us consider the given solution:

# Phi = e^(i m_l phi) #

Using Euler's formula , we can write this given solution as:

# Phi = cos(m_l phi) + isin(m_l phi) #

Comparing the given solution with the two possible cases we conclude that #B lt 0#, leading to the solution

# Phi = C_1cos(Bphi) + C_2sin(Bphi) #

We further conclude that:

# C_1=1, C_2=i, B=m_l #

Related questions
Impact of this question
1534 views around the world
You can reuse this answer
Creative Commons License