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the Green function $G(x,t)$ of the boundary value problem
$\frac{d^2y}{dx^2}-\frac{1}{x}\frac{dy}{dx} = 1$ , $y(0)=y(1)=0$ is
$G(x,t)= f_1(x,t)$ if $x≤t$ and $G(x,t)= f_2(x,t)$ if $t≤x$ where

(a)$f_1(x,t)=-\frac{1}{2}t(1-x^2)$ ; $f_2(x,t)=-\frac{1}{2t}x^2(1-t^2)$
(b)$f_1(x,t)=-\frac{1}{2x}t^2(1-x^2)$ ; $f_2(x,t)=-\frac{1}{2t}x^2(1-t^2)$
(c)$f_1(x,t)=-\frac{1}{2t}x^2(1-t^2)$ ; $f_2(x,t)=-\frac{1}{2}(1-x^2)$
(d)$f_1(x,t)=-\frac{1}{2t}x^2(1-x^2)$ ; $f_2(x,t)=-\frac{1}{2x}t^2(1-x^2)$
then which are correct. i am not getting my calculation right. my answer was ver similar to them but does not match completely to first i multiply the equation by (1/x) both side and convert it to a S-L equation but not getting my answer right.any help from you please.

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can anyone have solution of the above problem.i am waiting eagerly.thanks – dumm Dec 18 '12 at 4:33

Green's function is symmetric, so answer can be (b) and (d).

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For this question answer is a and c.Because Green's function is symmetric w.r.t to the variables.

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