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

Question

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

Christopher Agresta · Accepted Answer

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

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 #

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