Question

Find the first 3 nonzero terms in each of two solutions (not multiples of each other) about x = 0

Answer 1

\[xy'' + y' - y = 0\]
I have a first solution no problem, but this DE gives a double root, so I have no idea how to obtain the terms of a second solution. The first several terms of the solution I did find are:

\[y_{1}(x) = 1+x+\frac{1}{4}x^{2}+\frac{1}{36}x^{3}+ ...\]

The answer for the 2nd solution, which I cannot find for the life of me, is:

\[y_{2}(x) = y_{1}(x)lnx - 2x-\frac{3}{4}x^{2}-\frac{11}{108}x^{3}+ ...\]

the y2 = y1(x)lnx I know is just a part of the form of a 2nd solution, but those series terms I have no idea. Can anyone show me how to obtain a second series solution and, consequently, those terms that the book has as a solution? x_x

Answer 2

@ikram002p I have a question on x =0. When x =0, P(x) =0 which makes x =0 is not ordinary point, right? so that how to solve it then?

Answer 3

Its not an ordinary point, correct, its a regular singular point.

Answer 4

No, it not to me :(
to check whether it is aregular singular point, I consider:
\[lim_{x\rightarrow 0} x*\dfrac{1}{x}=lim_{x\rightarrow 0}1=1\\lim_{x\rightarrow 0}x^2\dfrac{1}{x}=lim_{x\rightarrow 0}x=0\]
which give me x =0 is not a regular singular point, :(

Answer 5

@Kainui

Answer 6

That's fine, as long as the limits are finite, which they are.

Answer 7

I know. but the characteristic equation from those limits is r^2 +(1-1)r + 0 =0 which give us r =0 , and it's nonsense, right?

Answer 8

That's what its supposed to be o.o You get a double root of r = 0. The one solution I have was derived using r = 0, but theres a way to find a second solution when you have double roots, but Im not sure how to do that. Everything Ive tried has come out wrong.

Answer 9

Ok, ignore it. if I continue your work, the second solution will be \(y-2= y_1ln(x) \)
so, \(y-2 = lnx + xlnx +\dfrac{lnx}{4}x^2+...... \) right?

Answer 10

\(y_2\) not y-2 hihihi

Answer 11

Well, it should come out to whatever I had in the first post:
\[y_{2}(x)= y_{1}(x)lnx - 2x - \frac{3}{4}x^{2}-\frac{11}{108}x^{3}+...\]

Answer 12

why?? is it not that we have formula to find the second solution is \(y_2 =y_1lnx\) ???

Answer 13

if we have double root from C.E?

Answer 14

\[y_{2}(x) = y_{1}(x)lnx+x^{r_{1}}\sum_{n=1}^{\infty}b_{n}x^{n}\] Basically, Im unable to derive the correct coefficients in that series portion of the answer, the bn's

Answer 15

Wonder where is prof @SithAndGiggle ???

Answer 16

Ive tried several different things to get those correct bn's, just not coming out correct.

Answer 17

oh, this is the second type of regular singular point. Since x =0 is not rg. sing. point, we find the value near that point. hihihi. However, your formula for the second solution is not correct. From my book it says \( y_2(x) = y_1(x)ln|x|+|x|^{r_1}\sum_{n=1}^\infty b_n(r_1)x^n\) where \(r_1=r_2~~(double~~root)\)
So that the sum =0 when you have r =0. It turns to my solution \(y_2=y_1ln|x|\)

Answer 18

Your formula is the same as mine, you just typed it more formally. My book has a sort of shortened version like I posted and the same thing you posted. Either way, I only know the solution my textbook has, which is what I listed above x_x So you're saying the answer in the back of my book is incorrect?

Answer 19

this looks like a power series solution to me

Answer 20

let:
\[y=\sum_0a_nx^n\]\[y'=\sum_1a_n~nx^{n-1}\]\[y''=\sum_2a_n~n(n-1)x^{n-2}\]
this gives us:

\[x\sum_2a_n~n(n-1)x^{n-2} + \sum_1a_n~nx^{n-1} - \sum_0a_nx^n = 0\]
even out the powers on x by shifting indexes:
\[\sum_2a_n~n(n-1)x^{n-1} + \sum_1a_n~nx^{n-1} - \sum_1a_{n-1}x^{n-1} = 0\]

now line up the starting summations by pulling out an n:
\[a_1 -a_{0}+\sum_2a_n~n(n-1)x^{n-1} + \sum_2a_n~nx^{n-1} - \sum_2a_{n-1}x^{n-1} = 0\]
and combine the summations by factoring out the x power

\[a_1 -a_{0}+\sum_2~[a_n~n(n-1)+ a_n~n -a_{n-1}]~x^{n-1} = 0\]

this is identically 0 when all the coefficients are 0, and we can devise a recurrsion equation now.
\[a_n~n(n-1)+ a_n~n -a_{n-1}=0\]
\[a_n~(n(n-1)+n) -a_{n-1}=0\]
\[a_n~(n(n-1)+n) =a_{n-1}\]
\[a_n =\frac{a_{n-1}}{n(n-1)+n}~:~n\ge2\]

Answer 21

that denom looks like it can simplify

n(n-1)+n

n(n-1+1) = n^2

Answer 22

x2y"+xy'-xy=0
r2+(1-1)r+0=0...so r2=0...r=0,0
then
y1=sigma(an.x^n)
y2=y1lnx+sigma(bnx^n)

Answer 23

x2y"--->n(n-1)an
xy'---->nan
xy---->a(n-1)...n-1 is index
so recursive equation is going to be...suppose a0=1
n(n-1)an+nan-a(n-1)=0----->an/a(n-1)=-1/n2------>an=(-1)^n/n!^2
then
y1=sigma((-1)^n/n2.x^n)

Answer 24

@amistre64 How do I get a second solution out of that, though? :(

Answer 25

@mahmit2012 That gives me y1, which I already have been already able to get. How about y2? I don't know how to get the bn coefficients in that 2nd solution.

Answer 26

a0 and a1 are your .... what are they called ..... constants for determining solutions.

as is, you get a set of coefficients in terms of a0, and another in terms of a1

Answer 27

or it may very well be that a0 is a constant, and the rest of the terms are in a1 .... havent worked it out too far yet