Paper 2, Section II, 6 A6 \mathrm{~A}

Differential Equations | Part IA, 2010

(a) By using a power series of the form

y(x)=k=0akxky(x)=\sum_{k=0}^{\infty} a_{k} x^{k}

or otherwise, find the general solution of the differential equation

xy(1x)yy=0.x y^{\prime \prime}-(1-x) y^{\prime}-y=0 .

(b) Define the Wronskian W(x)W(x) for a second order linear differential equation

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

and show that W+p(x)W=0W^{\prime}+p(x) W=0. Given a non-trivial solution y1(x)y_{1}(x) of (2)(2) show that W(x)W(x) can be used to find a second solution y2(x)y_{2}(x) of (2)(2) and give an expression for y2(x)y_{2}(x) in the form of an integral.

(c) Consider the equation (2) with

p(x)=P(x)x and q(x)=Q(x)xp(x)=-\frac{P(x)}{x} \quad \text { and } \quad q(x)=-\frac{Q(x)}{x}

where PP and QQ have Taylor expansions

P(x)=P0+P1x+,Q(x)=Q0+Q1x+P(x)=P_{0}+P_{1} x+\ldots, \quad Q(x)=Q_{0}+Q_{1} x+\ldots

with P0P_{0} a positive integer. Find the roots of the indicial equation for (2) with these assumptions. If y1(x)=1+βx+y_{1}(x)=1+\beta x+\ldots is a solution, use the method of part (b) to find the first two terms in a power series expansion of a linearly independent solution y2(x)y_{2}(x), expressing the coefficients in terms of P0,P1P_{0}, P_{1} and β\beta.

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