Let y=arcsin(x)/sqrt(1-x^2). Show that (1-x^2) y'-xy-1=0, and prove that, for all integers n>=0, (1-x^2)y^{n+2}-(2n+3)xy^{n+1} -(n+1)^2 y^{n}=0. (Superscripts denote repeated differentiation)

This is the first part of a STEP question (STEP 3, 2013, Q1), and is an example of a recurring pattern - "Induction Differential Equation".

The first part is a computation, combining the chain and quotient rules. We know that the derivative of f(x)=arcsin(x) is 1/sqrt(1-x²), and we need to differentiate the denominator. We use the chain rule: let u=1-x^2, and v=sqrt(u). Then du/dx=-2x, and dv/du = 1/(2 sqrt(u)). Combining, dv/dx=-x/sqrt(1-x²). Now, we put everything together using the quotient rules: y' = (f'v-v'f)/v². Multiplying by (1-x²)=v², we obtain (1-x²)y'= sqrt(1-x²)/sqrt(1-x²) +xarcsin(x)/sqrt(1-x²), which we can rewrite as 1+xy, as desired. Let's call this equation ().

We prove the second part by induction on n. To do this, we need two things: a base case, that is, proving the case n=0, and an inductive step, where we prove case n+1 from case n. To do the base case, we differentiate both sides of equation (*) that we proved above: d/dx ((1-x² )y') = d/dx (xy+1 ). On the left-hand side, we get -2xy'+(1-x² )y'', using the product rule, and on the right-hand side, we get xy'+y, similarly. Rearranging, we've proven the case n=0 of the desired statement. The inductive step is similar: we differentiate the equation from case n using the chain rule. Each of the first two terms splits into two: respectively, -2xy⁽ⁿ⁺²⁾+(1-x²)y⁽ⁿ⁺³⁾, and (2n+3)y⁽ⁿ⁺¹⁾+(2n+3)xy⁽ⁿ⁺²⁾. If we combine the terms with the same numbers of derivatives on y, we get (1-x²)y⁽ⁿ⁺³⁾-(2n+5)xy⁽ⁿ⁺²⁾-((n+1)²+2n+3)y⁽ⁿ⁺¹⁾=0 -which is what we wanted to prove, since ((n+1)²+2n+3) simplifies to (n+2)².