How do I use Leibnitz's theorem to prove relations like (1−x²)y_(n+2) − (2n+1)x y_(n+1) − (n²+m²)y_n = 0?

In one line

First get a simple differential equation connecting \(y,y_1,y_2\), then differentiate it \(n\) times using Leibnitz's theorem on each product term.

Explained simply

Finding \(y_n\) for \(y=e^{m\sin^{-1}x}\) directly is impossible to write neatly. But the function satisfies a simple second-order equation. Differentiating that equation \(n\) times gives a recurrence linking \(y_{n+2},y_{n+1},y_n\). It is like knowing the rule of a sequence instead of every term. These recurrences are used to compute values like \(y_n(0)\) and to build power series.

Technically, each term in the equation is a product of a simple polynomial (\(1-x^2\), \(x\)) and a derivative of \(y\). Leibnitz's theorem handles each product, and because the polynomials vanish after two derivatives, only a few terms survive.

The same three-step recipe (differentiate once, clear the root or fraction, differentiate again) works for all the standard functions in these questions: \(\sin^{-1}x\), \(e^{m\sin^{-1}x}\), \(\cos(m\log x)\), \((\sin^{-1}x)^2\) and \(\tan^{-1}x\).

How it works

1. Differentiate \(y\) once and clear roots or fractions by squaring or cross-multiplying.
2. Differentiate again and cancel a common factor (often \(2y_1\)) to reach an equation in \(y,y_1,y_2\).
3. Apply Leibnitz to each product: \(D^n[(1-x^2)y_2]=(1-x^2)y_{n+2}-2nxy_{n+1}-n(n-1)y_n\) and \(D^n[xy_1]=xy_{n+1}+ny_n\).
4. Collect terms in \(y_{n+2},y_{n+1},y_n\).

Key formula

Two results to remember: \(D^n[(1-x^2)y_2]=(1-x^2)y_{n+2}-2nx\,y_{n+1}-n(n-1)y_n\); \(\ D^n[xy_1]=xy_{n+1}+ny_n\).

Quick example

\(y=e^{m\sin^{-1}x}\). Then \(y_1=\dfrac{my}{\sqrt{1-x^2}}\), so \((1-x^2)y_1^2=m^2y^2\).

Differentiate: \((1-x^2)2y_1y_2-2xy_1^2=2m^2yy_1\). Divide by \(2y_1\): \((1-x^2)y_2-xy_1-m^2y=0\).

Differentiate \(n\) times: \((1-x^2)y_{n+2}-2nxy_{n+1}-n(n-1)y_n-xy_{n+1}-ny_n-m^2y_n=0\).

Collect: \((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\). Proved.

Putting \(x=0\): \(y_{n+2}(0)=(n^2+m^2)y_n(0)\), which gives every derivative at 0 from \(y(0)=1\), \(y_1(0)=m\).

Common mistakes

• Not squaring to remove the square root, which makes the next derivative messy.
• Forgetting the \(-n(n-1)y_n\) term from \(D^n[(1-x^2)y_2]\) (it comes from \(v_2=-2\) with \({}^nC_2\)).
• Sign errors when collecting; write each Leibnitz expansion on its own line.

How MU asks it

Classic 6-mark question: "If \(y=\sin^{-1}x\), prove \((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-n^2y_n=0\)" or "If \(y=\cos(m\log x)\), prove \(x^2y_{n+2}+(2n+1)xy_{n+1}+(n^2+m^2)y_n=0\)". Show \(y_1\), the squared form, \(y_2\) equation, the two Leibnitz expansions separately, and the collected result.