91Ó°ÊÓ

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$, that is, $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$is constant; i.e., $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Change of angular momentum:

$\mathrm{m}{\mathrm{â„“}}^2\mathrm{θ}=-\mathrm{â„“}\mathrm{³¾²\mu ²õ¾\pm ²Ôθ}$…… (1)

Newton’s rotational law: It is telling that the rate of change of angular momentum is equal to torque.

The mass–spring oscillator equation:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{ext}}=\left[\mathrm{inertia}\right]\mathrm{y}+\left[\mathrm{damping}\right]\mathrm{y}\text{'}+\left[\mathrm{stiffness}\right]\mathrm{y} \\ =\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky} \end{gathered}$$…… (2)

Given that, Legendre’s equation (2) is $\left(1-{\mathrm{t}}^2\right)\mathrm{y}-2\mathrm{ty}\text{'}+2\mathrm{y}=0$…… (3)

To prove: Legendre’s equation (5) is${\mathrm{y}}_2\left(\mathrm{t}\right)=\frac{\mathrm{t}}{2}\ln \left(\frac{1+\mathrm{t}}{1-\mathrm{t}}\right)-1$

Convert the equation (3) into standard form. We get,

$\mathrm{y}-\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}\mathrm{y}\text{'}+\frac{2}{1-{\mathrm{t}}^2}\mathrm{y}=0$

Here,$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}$ is one of the solutions and $\mathrm{p}\left(\mathrm{t}\right)=-\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}$.

Then, find the value of ${\mathrm{y}}_2\left(\mathrm{t}\right)$.

$$\begin{gathered} {\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{y}}_1\left(\mathrm{t}\right)∫\frac{{\mathrm{e}}^{∫\mathrm{p}\left(\mathrm{t}\right)\mathrm{dt}}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt} \\ =\mathrm{t}∫\frac{{\mathrm{e}}^{-∫\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}\mathrm{dt}}}{{\mathrm{t}}^2}\mathrm{dt} \\ =\mathrm{t}∫\frac{{\mathrm{e}}^{\ln {\left(1-{\mathrm{t}}^2\right)}^{-1}}}{{\mathrm{t}}^2}\mathrm{dt} \\ =\mathrm{t}∫\frac{1}{{\mathrm{t}}^2\left(1-{\mathrm{t}}^2\right)}\mathrm{dt} \\ =\mathrm{t}\left(\frac{\ln \left(1+\mathrm{t}\right)}{2}-\frac{\ln \left(1-\mathrm{t}\right)}{2}+\frac{1}{\mathrm{t}}\right) \\ =\frac{\mathrm{t}}{2}\ln \left(\frac{1+\mathrm{t}}{1-\mathrm{t}}\right)-1 \end{gathered}$$

Hence proved