1997 Paper 3 Q6

Year: 1997
Paper: 3
Question Number: 6

Course: UFM Pure
Section: Second order differential equations

Difficulty: 1700.0 Banger: 1516.0

differential equation second order substitution Chebyshev polynomials recurrence relation trigonometric change of variable SHM

Problem

Suppose that \(y_n\) satisfies the equations \[(1-x^2)\frac{{\rm d}^2y_n}{{\rm d}x^2}-x\frac{{\rm d}y_n}{{\rm d}x}+n^2y_n=0,\] \[y_n(1)=1,\quad y_n(x)=(-1)^ny_n(-x).\] If \(x=\cos\theta\), show that \[\frac{{\rm d}^2y_n}{{\rm d}\theta^2}+n^2y_n=0,\] and hence obtain \(y_n\) as a function of \(\theta\). Deduce that for \(|x|\leqslant1\) \[y_0=1,\quad y_1=x,\] \[y_{n+1}-2xy_n+y_{n-1}=0.\]

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Difficulty Rating: 1700.0

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Banger Rating: 1516.0

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Problem source

Suppose that $y_n$ satisfies the equations
\[(1-x^2)\frac{{\rm d}^2y_n}{{\rm d}x^2}-x\frac{{\rm
d}y_n}{{\rm d}x}+n^2y_n=0,\]
\[y_n(1)=1,\quad y_n(x)=(-1)^ny_n(-x).\]
If 
$x=\cos\theta$, show that
\[\frac{{\rm d}^2y_n}{{\rm d}\theta^2}+n^2y_n=0,\] and hence
obtain
$y_n$ as a function of
$\theta$. Deduce that for $|x|\leqslant1$
\[y_0=1,\quad y_1=x,\]
\[y_{n+1}-2xy_n+y_{n-1}=0.\]

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