1 problem found
Show that \(\sin(k\sin^{-1} x)\), where \(k\) is a constant, satisfies the differential equation $$(1-x^{2})\frac {\d^2 y}{\d x^2} -x\frac{\d y}{\d x} +k^{2}y=0. \tag{*}$$ In the particular case when \(k=3\), find the solution of equation \((*)\) of the form \[ y=Ax^{3}+Bx^{2}+Cx+D, \] that satisfies \(y=0\) and \(\displaystyle \frac{\d y}{\d x}=3\) at \(x=0\). Use this result to express \(\sin 3\theta\) in terms of powers of \(\sin\theta\).
Solution: \begin{align*} && y &= \sin(k \sin^{-1} x ) \\ &&y' &= \cos (k \sin^{-1} x) \cdot k \frac{1}{\sqrt{1-x^2}} \\ && y'' &= -\sin (k \sin^{-1} x) \cdot k^2 \frac{1}{(1-x^2)} - \cos(k \sin^{-1} x) \cdot k \frac{x}{(1-x^2)\sqrt{1-x^2}} \\ && (1-x^2)y'' &= -k^2y -xy' \\ \Rightarrow && 0 &= (1-x^2)y''+xy' + k^2y \end{align*} \begin{align*} && y &= Ax^3 + Bx^2 + Cx + D \\ && y' &= 3Ax^2 + 2Bx + C \\ && y'' &= 6Ax+2B \\ && 0 &= (1-x^2)(6Ax+2B) - x( 3Ax^2 + 2Bx + C) + 9(Ax^3 + Bx^2 + Cx + D ) \\ &&&= x^3(-6A-3A+9A) + x^2(-2B-2B+9B) + x(6A-C+9C) + (2B +9D) \\ \Rightarrow && B &= 0 \\ \Rightarrow && D &= 0 \\ \Rightarrow && C &= -\frac34 A \\ \\ x = 0, y = 0, y' = 0: && y &= 3x-4x^3 \\ \end{align*} And so \(\sin 3 x = 3 \sin x - 4\sin^3 x\)