Why Is the Frobenius Method's Second Solution Valid?

[matematikawan]

Assume that x=0 is a regular singular for x²y" + xp(x)y' + q(x)y = 0
and the indicial equation has equal roots \lambda = \lambda_1 = \lambda_2

The first solution is alway known to be of the form y_1(x) = x^{\lambda_1}\sum a_n x^n

Although tedious, I know how to obtain the second linearly independent solution using the Lagrange reduction of order, y₂(x) =u (x)y₁.

I think it is well known that the second solution will be of the form
y_2(x) = y_1(x)lnx + x^{\lambda_1}\sum b_n x^n


The book I'm refering, Schaum Outline Series in Differential Equation (cheap and direct to the method ) gives the following method to compute y₂.

y_2(x) = \frac{\partial y_1(x,\lambda)}{\partial\lambda} |_{\lambda=\lambda_1} \\\ (*) .

As you all know the book never proved any of their theorem/method.

My question is why (*) is a solution for the DE? Any proof for it?

 

[b17m4p]

I don't remember the method well enough to do this for you but I think its sortof similar to how you prove that a linear constant coefficient equation with a repeated root of the characteristic polynomial will have a second solution given by t times the first. I know I have seen a very good exposition on this topic in Hildebrand's Advanced Calculus for Engieneers. I would highly recommend that book for a lot of stuff and that section was very clear - but I didn't work it through well enough to internalise it and be able to answer your question from the top of my head.

 

[matematikawan]

Thanks b17m4p.
There is a copy in our library. I will try to get it. But it looks quite an old edition 1958 ?

QA303 HIL 1949
Advanced calculus for engineers
Hildebrand, Francis Begnaud.
Englewood Cliffs, N.J. : Prentice-Hall, [1958]


If only I have the note with a click of mouse!
Wikipedia ? MIT courses ?