Question

How do you find the auxiliary equation and the final solution for \[\dfrac{{{d}^{2}}\phi }{d{{\phi }^{2}}}+B\phi =0\] assuming \[\phi ={{e}^{i{{m}_{l}}\phi }}\] ?

Answer 1

Hint: In this problem, we have to find the auxiliary equation and the final solution for the given derivative. We can assume that \[B\in \mathbb{R},B\ne 0\], where it is a second order linear homogeneous differentiation equation with constant coefficient. Here we have to find the solution of the homogeneous equation by looking at the auxiliary equation, which is the polynomial equation with the coefficient of the derivative.

Complete step-by-step solution:
We know that the given derivative is,
\[\dfrac{{{d}^{2}}\phi }{d{{\phi }^{2}}}+B\phi =0\]
It is a second order linear homogeneous differentiation equation with constant coefficient.
We can now assume \[B\in \mathbb{R},B\ne 0\]
In complementary function, the associative auxiliary equation can be written as,
\[\begin{align}
  & \Rightarrow {{m}^{2}}+0m+B=0 \\
 & \Rightarrow {{m}^{2}}=-B.......(1) \\
\end{align}\]
We know that where the sign of B will determine the possible solution.
We can now find the solution from the possible cases below.
Case:1
For, \[B>0\]
We will get real distinct roots from the auxiliary equation, which is given by
\[m=\pm B\]
So that,
\[\Rightarrow \phi ={{C}_{1}}{{e}^{B\phi }}+{{C}_{2}}{{e}^{-B\phi }}\]
Case:2
For, \[B<0\]
We will get pure imaginary roots from the auxiliary equation, which is given by
\[m=\pm iB\]
So that,
\[\begin{align}
  & \Rightarrow \phi ={{e}^{0\phi }}\left\{ {{C}_{1}}\cos \left( B\phi \right)+{{C}_{2}}\sin \left( B\phi \right) \right\} \\
 & \Rightarrow \phi ={{C}_{1}}\cos \left( B\phi \right)+{{C}_{2}}\sin \left( B\phi \right) \\
\end{align}\]
We can see that from (1), the sign of B is negative,
So that from case:2, we can write as
\[\begin{align}
  & \Rightarrow \phi ={{e}^{0\phi }}\left\{ {{C}_{1}}\cos \left( B\phi \right)+{{C}_{2}}\sin \left( B\phi \right) \right\} \\
 & \Rightarrow \phi ={{C}_{1}}\cos \left( B\phi \right)+{{C}_{2}}\sin \left( B\phi \right) \\
\end{align}\]
We are given a form of the solution, let us consider the given solution.
\[\phi ={{e}^{i{{m}_{l}}\phi }}\]
Now we can use the Euler’s formula.
We can write the given solution from the above step, we get
\[\Rightarrow \phi =\cos \left( {{m}_{l}}\phi \right)+i\sin \left( {{m}_{l}}\phi \right)\]
We can now compare the given solution, with the two possible cases, we conclude that \[B<0\], leading to the solutions.
\[\Rightarrow \phi ={{C}_{1}}\cos \left( B\phi \right)+{{C}_{2}}\sin \left( B\phi \right)\]
We can further conclude that,
\[{{C}_{1}}=1,{{C}_{2}}=i,B={{m}_{l}}\]
As the sign of B is negative which gives complex values.
Therefore, the solution is \[\phi =\cos \left( {{m}_{l}}\phi \right)+i\sin \left( {{m}_{l}}\phi \right)\] whose auxiliary equation is \[{{m}^{2}}+0m+B=0\].

Note: Students make mistakes while using the Euler’s formula and comparing the solutions with the possible cases. We should remember that we can find the solutions for a homogeneous equation by looking at the auxiliary equation, which is the polynomial equation with the coefficient of the derivatives.