An MCQ on Greens function

Question

$$G(x,t) =\begin{cases} a+ b\log t & \text{if $0<x<t$ } \\[2ex] c+ d\log t & \text{if $t<x<1$ } \end{cases}$$

is a Greens function for $xy''+y'=0$ subject to $y$ being bounded as $x$ tends to $0$ and $y(1)=y'(1)$ if

The options are

1) $a=b=c=d=1$

2) $a=c=1$, $b=d=0$

3) $a=c=0$, $b=d=1$

4) $a=b=c=d=0$

I don't know how to tackle the problem. I've tried to find the Green's function. And I've got

$$ G(x,t) =\begin{cases} a+ b(\log x) & \text{if $0<x<t$ } \\[2ex] a+ b(\log x)-\log t+\log x & \text{if $t<x<1$ } \end{cases} $$

but this is not in the above given form. Also how to apply the boundary conditions? Please help me to solve the problem.

Answer 1

I hope you meant $$G(x,t) =\begin{cases} a+ b\log t & \text{if $0<x\le t$ } \\[2ex] c+ d\log x & \text{if $t\le x\le1$ } \end{cases}$$

Now, using:

1. $G(x,t)$ satisfies the boundary condition.

$$\implies G(1,t)=G'(1,t)\implies c+d\log 1=\frac d 1\implies c=d$$ That narrows our options to a and d.

2. Derivative, $G'(x,t)$ jumps at $t$ by $1\over p(t)$=$1\over t$ $$\implies d=1$$

Hence 1.