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The boundary value problem $x^2y''-2xy'+2y=0$ subjected to boundary conditions$y(1)+\alpha y'(1)=1 ; y(2)+\beta y'(2)=2$ has unique solution if $$1.\alpha=-1,\beta=2\\2.\alpha=-2,\beta=2\\3.\alpha=-1,\beta=-2\\4.\alpha=-3,\beta=2/3.\\$$ Kindly suggest from where do I prepare to solve such problem and how do I solve them? Question seems very nice to me but I couldn't find any usual ODE technique I could solve it.

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    Is it supposed to be a $2xy'$ instead of $2xy$? If what you have written down is indeed correct, solving it leads to some very complicated modified Bessel Functions. – 2017-01-17
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    @Tom yes you are right . I modified it. – 2017-01-18

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Given the boundary value problem

$$\tag 1 x^2y''-2xy'+2y=0$$

We see that this is an Euler-Cauchy type equation, so we try

$$y = x^m \implies y' = mx^{m-1}, y''=m(m-1)x^{m-2}$$

It is worth noting that as an alternate approach, we can make the substitution $x = e^t$ and then simplify to arrive at the same result.

We substitute this back into $(1)$ and combine like terms

$$x^m(m^2 - 3 m + 2) = 0 \implies m_{1,2} = 1, 2$$

This gives us the solution

$$y(x) = c_1 x^{m_1} + c_2 x^{m_2} = c_1 x + c_2 x^2$$

Now for the BCs.

Case 1:

$$y(1)+\alpha y'(1)=1 , y(2)+\beta y'(2)=2, \alpha=-1,\beta=2$$

This gives

$$y(x) = c_1 x + c_2 x^2 \implies y'(x) = c_1 + 2 c_2 x$$

$$y(1) - y'(1) = c_1 + c_2 - (c_1 + 2 c_2) = -c_2 = 1 \implies c_2 = -1\\y(2) + 2 y'(2) = 2 c_1 + 4 c_2 +2 (c_1 + 4c_2) = 4 c_1 + 12 c_2 = 2 \implies c_1 = \dfrac{7}{2}$$

Thus

$$y(x) = \dfrac{7 }{2}x - x^2$$

You should do the other three cases.

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    the method seems very lengthy – 2017-01-18
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    Actually, do it by hand - it is quick and not hard, I just showed enough details so it makes it easy to follow. In fact, when you try the other three, you'll see what I mean as you only need to do the BCs. – 2017-01-18
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    yes you r right – 2017-01-18