Question

Find the Wronskian of two solutions of the given differential equation without solving the equation. \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0, \quad\) Legendre's equationn

Short answer

#Answer# The Wronskian of two solutions of Legendre's equation is a constant, and it can be determined by the initial conditions of the problem. That is, \(W(u,v) = C\), where \(C\) is a constant.

Step-by-step solution

Differentiate the Wronskian with respect to \(x\)

The Wronskian \(W(u,v)\) is defined by \(W(u,v) = u(x) v'(x) - u'(x) v(x)\). First, differentiate both sides with respect to \(x\): $$W'(u,v) = u(x) v''(x) + u'(x) v'(x) - u''(x) v(x) - u'(x) v'(x)$$

Substitute Legendre's equation in place of \(u''\) and \(v''\)

Now, we substitute the expressions of \(u''\) and \(v''\) from Legendre's equation into the expression for \(W'(u,v)\). First of all, we rewrite Legendre's equation in terms of \(y''\): $$y'' = \frac{1}{(1-x^2)}\left[2xy' - \alpha(\alpha+1)y\right]$$ Applying the above expression for \(u\) and \(v\): $$u''=\frac{1}{(1-x^2)}\left[2xu'-\alpha(\alpha+1)u\right],\qquad v''=\frac{1}{(1-x^2)}\left[2xv'-\alpha(\alpha+1)v\right]$$ Now substitute these expressions into the expression for \(W'(u,v)\): $$W'(u,v) = u(x) \left(\frac{1}{(1-x^2)}\left[2xv'-\alpha(\alpha+1)v\right]\right) + u'(x) v'(x) - v(x) \left(\frac{1}{(1-x^2)}\left[2xu'-\alpha(\alpha+1)u\right]\right) - u'(x) v'(x)$$

Simplify the expression for \(W'(u,v)\)

Now we simplify the expression for \(W'(u,v)\): $$W'(u,v) =u'(x)v'(x) -u'(x)v'(x) + \frac{u(x) v(x)}{(1-x^2)}\alpha(\alpha+1) - \frac{u(x) v(x)}{(1-x^2)}\alpha(\alpha+1)$$ We observe that all terms cancel out except the last ones. So, $$W'(u,v) = 0$$

Integrate to find the Wronskian

Since \(W'(u,v) = 0\), the Wronskian must be a constant. To find this constant, we integrate both sides with respect to \(x\): $$\int W'(u,v) \, dx = \int 0 \, dx$$ Which gives us, $$W(u,v) = C$$ where \(C\) is a constant. So, the Wronskian of two solutions of Legendre's equation is a constant, and it can be determined by the initial conditions of the problem.