31A (Asymptotic Methods)

y^{''} + p (x) y^{'} + q (x) y = 0

, infinity is a regular singular point only when

p (x) = O (x^{- 1})

and

q (x) = O (x^{- 2})

. Here

p = - 1

and

q = - 2/ x - 2/ x^{2}

, so infinity is an irregular singular point.

ii

 0  0

Solution

 0  0

Use the standard elimination of the first derivative by setting

y (x) = e^{x /2} u (x) .

(1)

Then

y^{''} - y^{'} = e^{x /2} (u^{''} - \frac{1}{4} u),

(2)

so the transformed equation is

u^{''} + q (x) u = 0, q (x) = - \frac{1}{4} - \frac{2}{x} - \frac{2}{x ^{2}} .

(3)

iii

 0  0

Solution

 0  0

Put

Q (x) = - q (x) = \frac{1}{4} + \frac{2}{x} + \frac{2}{x ^{2}} > 0

(1)

for sufficiently large

x

. The Liouville–Green approximation gives

u_{\pm} (x) \sim Q (x)^{- 1/4} exp (\pm \int^{x} Q (t) d t) .

(2)

Since

Q (x) = \frac{1}{2} + \frac{2}{x} + O (x^{- 2}), Q (x)^{- 1/4} = 2 [1 + O (x^{- 1})],

(3)

we obtain, after absorbing nonzero constants into the two solutions,

u_{+} (x) \sim x^{2} e^{x /2}, u_{-} (x) \sim x^{- 2} e^{- x /2} .

(4)

iv

 0  0

Solution

 0  0

Multiplying the two Liouville–Green forms by

e^{x /2}

gives

y_{+} (x) \sim x^{2} e^{x}, y_{-} (x) \sim x^{- 2} .

(1)

The first expression is in fact exact:

y = x^{2} e^{x}, y^{'} = e^{x} (x^{2} + 2 x), y^{''} = e^{x} (x^{2} + 4 x + 2),

(2)

and direct substitution gives

y^{''} - y^{'} - \frac{2 ( x + 1 )}{x ^{2}} y = 0.

(3)

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 31A (Asymptotic Methods)

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