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

Feb 4, 2018

How to find the auxiliary equation and the final solution
for $\frac{{d}^{2} \Phi}{\mathrm{dp} h {i}^{2}} + B \Phi = 0$ assuming $\Phi = {e}^{i {m}_{l} \phi}$?

Explanation:

We have:

$\frac{{d}^{2} \Phi}{\mathrm{dp} h {i}^{2}} + B \Phi = 0$

We assume that $B \in \mathbb{R} , B \ne 0$

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:

${m}^{2} + 0 m + B = 0$
$\therefore {m}^{2} = B$

The sign of B will determine the possible solution. Then

Case 1: $B > 0$

Real distinct roots, given by $m = \pm B$, so that:

$\Phi = {C}_{1} {e}^{B \phi} + {C}_{2} B {e}^{- B \phi}$

Case 2: $B < 0$

Pure imaginary roots, given by $m = \pm B i$, so that:

$\Phi = {e}^{0 \phi} \left\{{C}_{1} \cos \left(B \phi\right) + {C}_{2} \sin \left(B \phi\right)\right\}$
$\setminus \setminus \setminus = {C}_{1} \cos \left(B \phi\right) + {C}_{2} \sin \left(B \phi\right)$

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:

$\Phi = \cos \left({m}_{l} \phi\right) + i \sin \left({m}_{l} \phi\right)$

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

$\Phi = {C}_{1} \cos \left(B \phi\right) + {C}_{2} \sin \left(B \phi\right)$

We further conclude that:

${C}_{1} = 1 , {C}_{2} = i , B = {m}_{l}$