Series soln. with singular point to 2nd order linear differential eqn

[Gridvvk]

So, I'm trying to solve 2nd order linear differential equations (series solutions near a singular point).

(lnx)y" + 0.5y' + y = 0 around the regular singular point x = 1

I got the indicial equation,

r(r-0.5) = 0,

which leads to the roots...

r1 = 0.5, r2 = 0

The problem only asks us to find the first three nonzero terms in the series y1 = Ʃa_n * (x-1)^r+n from n = 0 to infinity. And we only need to find one solution, corresponding to the larger root.

So I took the first and second derivatives of the y1 they gave and plugged it into the differential equation. Now at this point I usually factor out all the x terms. And since the left side has to equal zero for all x, I can divide by that x term to get a recurrence relation (that involves a_n terms). From the relation I can figure out what a_n is. However, in this case, I can't factor out all the x terms because there's a lnx.

Anyone know how to get rid of the lnx?

Thanks!

 

[mighty2000]

Thank you for your post. Solving second order linear differential equations can be a challenging task, especially when dealing with singular points. In this case, the presence of the natural logarithm term in the equation makes it more complicated. However, there are some techniques that can help us solve this type of problem.

Firstly, we can use the method of Frobenius to find a series solution near the singular point. This involves substituting y = Ʃa_n * (x-1)^r+n into the differential equation and solving for the coefficients a_n. In this case, we have to deal with the lnx term, but we can use the fact that lnx = 1 + (x-1) + (x-1)^2/2 + (x-1)^3/3 + ... to rewrite the equation in terms of powers of (x-1). This will allow us to find a recurrence relation for the coefficients and solve for them.

Another approach is to use the method of variation of parameters. This involves finding a particular solution of the differential equation and then using it to find the general solution. In this case, we can use the fact that y = x^0.5 is a solution to the homogeneous equation (with r = 0.5). We can then use the method of variation of parameters to find a particular solution, which will involve a logarithmic term. This can then be used to find the general solution.

I hope these suggestions will help you solve the problem. Good luck!




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