Question

Show that the differential equation $sin\theta \frac{d^{2}y}{d{\theta }^2}+cos\theta \frac{dy}{d\theta }+n(n+1)\left(sin\theta \right)y=0$can be transformed into Legendre’s equation by means of the substitution.$x=cos\theta$

Short answer

Hence the differential equation is $sin\theta \frac{d^{2}y}{d{\theta }^2}+cos\theta \frac{dy}{d\theta }+n(n+1)\left(sin\theta \right)y=0$verified.

Step-by-step solution

Step 1:DefineLegendre equation

The Legendre differential equation is the second order ordinary differential equation (ODE) which can be written as:

$(1-x^2)\frac{d^{2}y}{d{x}^2}-2x\frac{dy}{dx}+l(l+1)y=0$

Derive thederivative of the function

Let the given equation be $\sin \theta \frac{d^{2}y}{d{\theta }^2}+\cos \theta \frac{dy}{d\theta }+n(n+1)\left(\sin \theta \right)y=0$… (1), and the function be$x=\cos \theta$.

By using the chain rule, the derivatives are,

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{\mathrm{dy}}{\mathrm{dx}}\cdot \frac{\mathrm{dx}}{\mathrm{d\theta }}\\ =\frac{\mathrm{dy}}{\mathrm{dx}}\frac{\mathrm{d}}{\mathrm{d\theta }}\left[\mathrm{cos\theta }\right]\end{array}$

$\frac{dy}{d\theta }=-\sin \theta \frac{dy}{dx}$… (2)

$\begin{array}{c}\frac{d^{2}y}{d{\theta }^2}=\frac{d}{d\theta }\left[\frac{dy}{dx}\right]\frac{dx}{d\theta }+\frac{dy}{dx}\frac{d}{d\theta }\left[\frac{dx}{d\theta }\right]\\ =\frac{dx}{d\theta }\frac{d}{dx}\left[\frac{dy}{dx}\right]\frac{dx}{d\theta }+\frac{dy}{dx}\frac{d^{2}x}{d{\theta }^2}\\ =\frac{d^{2}y}{d{x}^2}{\left[\frac{dx}{d\theta }\right]}^2+\frac{dy}{dx}\frac{d^{2}x}{d{\theta }^2}\\ =\frac{d^{2}y}{d{x}^2}{\left[-\sin \theta \right]}^2+\frac{dy}{dx}\frac{d}{d\theta }[-\sin \theta ]\end{array}$

$\frac{d^{2}y}{d{\theta }^2}={\sin }^2\theta \frac{d^{2}y}{d{x}^2}-\cos \theta \frac{dy}{dx}$… (3)

Derive the required solution.

Substitute the equation (2) and (3) into (1) yields,$\begin{array}{rcl}\theta \left[{\sin }^2\theta \frac{d^{2}y}{d{x}^2}-\cos \theta \frac{dy}{dx}\right]+\cos \theta \left[-\sin \theta \frac{dy}{dx}\right]+n(n+1)\left(\sin \theta \right)y& =& 0\\ {\sin }^3\theta \frac{d^{2}y}{d{x}^2}-\sin \theta \cos \theta \frac{dy}{dx}-\cos \theta \sin \theta \frac{dy}{dx}+n(n+1)\left(\sin \theta \right)y& =& 0\\ {\sin }^3\theta \frac{d^{2}y}{d{x}^2}-2\sin \theta \cos \theta \frac{dy}{dx}+n(n+1)\left(\sin \theta \right)y& =& 0\\ {\sin }^2\theta \frac{d^{2}y}{d{x}^2}-2\cos \theta \frac{dy}{dx}+n(n+1)y& =& 0\\ \left[1-{\cos }^2\theta \right]\frac{d^{2}y}{d{x}^2}-2\cos \theta \frac{dy}{dx}+n(n+1)y& =& 0\\ \left[1-x^2\right]\frac{d^{2}y}{d{x}^2}-2x\frac{dy}{dx}+n(n+1)y& =& 0\\ & & \end{array}$

Hence verified.