Assume that x=0 is a regular singular for x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}y" + xp(x)y' + q(x)y = 0

and the indicial equation has equal roots [tex]\lambda = \lambda_1 = \lambda_2[/tex]

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

Although tedious, I know how to obtain the second linearly independent solution using the Lagrange reduction of order, y_{2}(x) =u (x)y_{1}.

I think it is well known that the second solution will be of the form

[tex]y_2(x) = y_1(x)lnx + x^{\lambda_1}\sum b_n x^n [/tex]

The book I'm refering, Schaum Outline Series in Differential Equation (cheap and direct to the method ) gives the following method to compute y_{2}.

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

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?

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# Frobenius Method again!

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