DE: 2(x-1)y'=3y. Had to solve it with a power series and an easier alternative method and compare both results. I also had to check whether I can express the power series I found as an elementary question. I'll write the recursive relation, power series, etc in the additional post with the equasion editor.

Question

anonymous · Answer

The solution with the alternative method

$$y=c0(x-1)^{\frac{ 3 }{ 2 }}$$

recursive relationship:

$$c _{n+1}= \frac{ c _{n} (n-\frac{ 3 }{ 2 }) }{ (n+1) }$$

$$ n=0 -> c _{1}=\frac{ -3 }{ 2 } c _{0}$$

anonymous · Answer

$$n=1 -> c _{2}=\frac{ 3 }{ 2^{2}2! } c _{0}$$
$$n=2 -> c _{3}=\frac{ 3 }{ 2^{3}3! } c _{0}$$
$$n=3 -> c _{4}=3 \frac{ 3!! }{ 2^{4}4! } c _{0}$$
$$n=4 -> c _{5}=3 \frac{ 5!! }{ 2^{5}5! } c _{0}$$
$$n=5 -> c _{6}=3 \frac{ 7!! }{ 2^{6}6! } c _{0}$$

and so n, so a double factorial emerges at the top from n=2 on.

$$y=3 c _{0}(-\frac{ 1 }{ 2 } + \frac{ x }{ 2^{2}2! } \sum_{n=2}^{\infty}\sum_{k=1}^{n} \frac{ (2k-1)!! x ^{k+1} }{ 2^{k+2} (k+2)!})$$

Reworking and simplifying gives

$$y=3 c _{0}(-\frac{ 1 }{ 2 } + \sum_{n=1}^{\infty}\frac{ (2(n-1)! x ^{n} }{ 2^{2n} (n-1)!(n+1)!})$$

If I plug in values for n with the latter expression I get the same values as the evaluated forms of the rational expressions for c1, c2, c3, etc. 

But as I said, not sure whether I can turn it into an elementary expression, and it looks quite different from the simple method result. Although I'm wondering whether there's a binomial relationship between the two.

kainui · Answer

Interesting, I don't really have time to work on this right now but I might later today if it hasn't been answered.

anonymous · Answer

haha this is gonna sound stupid but i like how there are exclamations in the equations it makes it sound like someone is either yelling it or way too excited to be doing math

kainui · Answer

I've returned and am now starting. I've already solved it the simpler way and got the same answer as you, now I'm about to do the power series method, give me a moment to work it out on my own.

kainui · Answer

Alright, I haven't been very good since you have the answer already and you're simply just looking for the relationship between the two. 

I am not sure, but it might be that what you have found is the Taylor series for your function.

I wish I could be of more help here, it's an interesting problem.

anonymous · Answer

So, should I try to revert the easy answer into a Taylor series?

anteater · Answer

If you don't mind, I am going to try to look at it too. My son's next class is DE, and I am trying to dust off my brain to see if I remember anything! :)  Got the same answer you did for the simple approach. Can't remember using power series, but am going to review and give it a shot.

anonymous · Answer

@lauren_bisko the exlamation marks are factorials... if you have a multiplication that looks like this 5.4.3.2.1, you can write it down as 5! (in stochastic math, combinations and variations and such you'll start to use it). If it says 5!! it means you only multiply the uneven integers 5.3.1. If it says 8!! you'll only multiply the even number 8.6.4.2.1. Such series increase faster than an exponential function.

anonymous · Answer

Sure @anteater. I'll leave it open. Our university calculus Math 1 ended with DE's and Taylor series and eigenvalues and eigenwaarden for linear algebra. Math 2 was transformational math of vectors. Course Math 3 is about DE with power series, multivariate analysis and complex analysis. It's a new subject because they detected that even the master students still had issues with DEs.

kainui · Answer

@anteater If you want, I have some time and could explain the method of power series for you.

anteater · Answer

It's ok. :)  You don't have to leave it open. I can continue to work on it. I may ask questions of both of you later, but for now I think the best thing to do would be to drag out the books, read, and work a few problems. I will holler if I hit an impasse, though! Thank you both! :)

anonymous · Answer

Well, I haven't got a worked out answer yet to the last part... But per @Kanui's suggestion, I'm going to use y=c0(x-1)^(3/2) into a Taylor series (Maclauclin one) and see whether I'll get something similar.

anteater · Answer

I have a feeling you will.

anteater · Answer

Would you please give me a shout if you are working on further problems? Won't hurt for me to watch as you work them. :)

anonymous · Answer

Yeah, I'm revieuwing the part of the Frobenius method for DEs with singular points and trying exercises on them.

anteater · Answer

I have been away from this stuff for about 30 years now ... serious brain atrophy!!!

anteater · Answer

Happily, the vector calculus came back fairly quickly ... hoping for the same with the rest of it!

anteater · Answer

Should write an ode to ODEs ... ;)

anteater · Answer

Bye for now ... !

anonymous · Answer

Yup it seems to be the Taylor series. I was writing the derivatives of c0(x-1)^(3/2) until the 5th prime, but something blocked all of a sudden and the post was gone, so no. I've tried a Maclauclin series but then I have a complex number i everywhere, although it does seem to trick to make the first term negative and everything else positive when it needs to be. The primes certainly start have the 3 c0 term and the 2^n element in the denominator, and they start the double factorial where it starts in the power series. It doesn't fit completely yet at every detail, but I don't think I'm supposed to figure it out in detail. I can show there is a relation that suggests they are the same answer.