Question

So I was looking at an ordinary differential equation xy"+2y'+2y=0 using power series but I could only obtain one independent solution: sinx/x. However, solving it using a substitution of y(x)=v(x)/x, I was able to obtain the full solution of cosx/x and sinx/x. Where is the cosx/x in the power series solution? I am only getting one recursive formula and constant, which I used to find the sinx/x portion of the solution.

Answer 1

I've already graduated from college and this is purely for my own interest and to freshen up on some of the things I've learned.

Answer 2

xy"+2y'+2y=0

let
y = sum[0] an x^n
y' = sum[1] an n x^(n-1)
y'' = sum[2] an n(n-1) x^(n-2)


x sum[2] an n(n-1) x^(n-2) + 2sum[1] an n x^(n-1) + 2sum[0] an x^n=0

sum[2] an n(n-1) x^(n-1) + sum[1] 2an n x^(n-1) + sum[0] 2an x^n=0


we need to get the exponents the same by shifting the index of sum[0] by +1

sum[2] an n(n-1) x^(n-1) + sum[1] 2an n x^(n-1) + sum[1] 2an x^(n-1) =0


we need to get the indexes lined up to sum[2]

sum[2] an n(n-1) x^(n-1)
+ 2a_1 + sum[2] 2an n x^(n-1)
+ 2a_1 + sum[2] 2an x^(n-1) =0


4a_1
+ sum[2] an n(n-1) x^(n-1) + sum[2] 2an n x^(n-1) + sum[2] 2an x^(n-1) =0



now we can combine the summations

4a_1 + sum[2] (an n(n-1) + 2an n + 2an x^(n-1)) =0


this is identically 0 when all the coefficients a_n are 0 right?

Answer 3

Here is my work if that helps anyone to help me. Keep me honest. XD

Answer 4

i missed some n parts in that last run ... i only did the exponent and not the subindexes

Answer 5

\[\sum_2 a_n~n(n-1) x^{n-1} + \sum_1 2a_n~ n x^{n-1} + \sum_0 2a_n~ x^n=0\]
\[... + \sum_0 2a_n~ x^n=0\]
\[... + \sum_{0+1} 2a_{n-1}~ x^{n-1}=0\]
\[\sum_2 a_n~n(n-1) x^{n-1} + \sum_1 2a_n~ n x^{n-1} + \sum_{1} 2a_{n-1}~ x^{n-1}=0\]
\[...+ 2a_1+\sum_2 2a_n~ n x^{n-1} +2a_{0}+ \sum_{2} 2a_{n-1}~ x^{n-1}=0\]
\[2a_0+2a_1+\sum_2 a_n~n(n-1) x^{n-1} + \sum_2 2a_n~ n x^{n-1} + \sum_{2} 2a_{n-1}~ x^{n-1}=0\]
\[2a_0+2a_1+\sum_2 \left[a_n~n(n-1) + 2a_n~ n + 2a_{n-1}\right]~ x^{n-1}=0\]

Answer 6

does that all look in order now?

Answer 7

ive got about 10 minutes before my classes start :/

Answer 8

The last summation term should be multiplied by x not 2.

Answer 9

I didn't type it correctly. My failure..

Answer 10

So the correct problem is: xy"+2y'+xy=0

Answer 11

So sorry for the inconvenience.

Answer 12

lets use this as an example then :)

taking the rule for the coefficients to be:

\[a_n~n(n-1) + 2a_n~ n + 2a_{n-1}=0\]
solving for a_n gives us a formula to assess

\[a_n(n(n-1) + 2n) =-2a_{n-1}\]
\[a_n =-\frac{2a_{n-1}}{n(n-1) + 2n}~:~n\ge2\]
this would develop 2 sets to play with, one starting with n=0, and another starting with n=1

if the solutions are to be in sin and cos, you would see a pattern in the even and odd ns

\[a_{2n+1}=sin\]
\[a_{2n}=cos\]

in some fashion

Answer 13

Yeah I'm expecting I'm missing something and that the \[a_{1}=0\] is wrong somewhere which would give me a second recursion for the cos.

Answer 14

ill be able to focus better on this in about 5 hours :) in the meanitme im sure someone else will respond. Make sure youve got the correct information posted.

Answer 15

Yeah I'm so sorry about that and thanks for your help.

Answer 16

\[xy"+2y'+xy=0\]
\[x\sum_2a_n~n(n-1)x^{n-2}+2\sum_1a_n~nx^{n-1}+x\sum_0a_n~x^n=0\]
\[\sum_2a_n~n(n-1)x^{n-1}+\sum_12a_n~nx^{n-1}+\sum_0a_n~x^{n+1}=0\]
\[\sum_2a_n~n(n-1)x^{n-1}+\sum_12a_n~nx^{n-1}+\sum_{0+2}a_{n-2}~x^{n-2+1}=0\]
\[\sum_2a_n~n(n-1)x^{n-1}+\sum_12a_n~nx^{n-1}+\sum_{2}a_{n-2}~x^{n-1}=0\]
\[\sum_2a_n~n(n-1)x^{n-1}+2a_1+\sum_22a_n~nx^{n-1}+\sum_{2}a_{n-2}~x^{n-1}=0\]
\[2a_1+\sum_2[a_n~n(n-1)+2a_n~n+a_{n-2}]~x^{n-1}=0\]
\[a_n~n(n-1)+2a_n~n+a_{n-2}=0\]
\[a_n(n(n-1)+2n)=-a_{n-2}\]
\[a_n=-\frac{a_{n-2}}{n^2+n}~:~n\ge2 \]
\[a_2=-\frac{a_0}{2^2+2}\]
\[a_3=-\frac{a_1}{3^2+3}\]
\[a_4=-\frac{a_2}{4^2+4}=\frac{a_0}{(2^2+2)(4^2+4) }\]
\[a_5=-\frac{a_3}{5^2+5}=\frac{a_1}{(3^2+3)(5^2+5) }\]

Answer 17

\[a_{2n}=a_0\frac{(-1)^n}{(2n+3)!}\to a_0\sum_2\frac{(-1)^n}{(2n+3)!}x^{2n} \]
\[a_{2n+1}=a_1\frac{(-1)^n}{(2n+4)!}\to a_1\sum_2\frac{(-1)^n}{(2n+4)!}x^{2n+1} \]

Answer 18

dividing each by x gets you into a more normalized sin/cos structure

Answer 19

if i see that correctly

Answer 20

But what about the \[2a _{1}\] that was left outside the summation? This is where my issue lies. It is the only constant term and since the coefficients of each power of x should equal zero when summed, don't you end up with \[a _{1}=0\] which kills off the other possible solution?

Answer 21

Also the denominators should be (2n)! and (2n+1)! which follow the form for the power series for sin and cos. I generally leave things like n(n+1) factored so you can see all the factors and follow the pattern easier.

Answer 22

However, looking at the a1 I am definitely seeing where the other solution could be. I just don't see how a1 could be nonzero.

Answer 23

a0 and a1 become the general placeholders in the solution - they are just unknowns with respect to the rest of the an parts.

We know that for n >= 2 that the formula holds (assuming i doint mess it up along the way :) :
\[a_n~n(n-1)+2a_n~n+a_{n-2}=0\]
solving for an gets us:
\[a_n=-\frac{a_{n-2}}{n(n+1)}~:~n\ge2\]
lets use n=4
\[a_4~4(3)+2a_4~4+a_{4-2}=0\]
is this true for some general a0? since a0 is the basis for the even ns
\[a_0\frac{4.3}{2.3.4.5}\\
+2a_0\frac{4}{2.3.4.5}\\
-a_0\frac{1}{2.3}=0\]
by factoring out the a0, we get: either a0 = 0 or the rest of it does
\[\frac{4.3}{2.3.4.5}+\frac{4.2}{2.3.4.5}-\frac{1}{2.3}=0\]
\[\frac{1}{2.5}+\frac{1}{3.5}-\frac{1}{2.3}=0\]
\[\frac{3}{1}+\frac{2}{1}-\frac{5}{1}=0\]
so, regardless of the value of a0, or a1, we can zero out the summation.
Now, the question is, what value does a0 and a1 have to be to get the whole ting equal to zero in this case since the solution is for =0? Thats where we can then define the unknown "constants".

Answer 24

a0 and a1 become the general placeholders in the solution - they are just unknowns with respect to the rest of the an parts.

We know that for n >= 2 that the formula holds (assuming i doint mess it up along the way :) :
\[a_n~n(n-1)+2a_n~n+a_{n-2}=0\]
solving for an gets us:
\[a_n=-\frac{a_{n-2}}{n(n+1)}~:~n\ge2\]
lets use n=4
\[a_4~4(3)+2a_4~4+a_{4-2}=0\]
is this true for some general a0? since a0 is the basis for the even ns
\[a_0\frac{4.3}{2.3.4.5}\\
+2a_0\frac{4}{2.3.4.5}\\
-a_0\frac{1}{2.3}=0\]
by factoring out the a0, we get: either a0 = 0 or the rest of it does
\[\frac{4.3}{2.3.4.5}+\frac{4.2}{2.3.4.5}-\frac{1}{2.3}=0\]
\[\frac{1}{2.5}+\frac{1}{3.5}-\frac{1}{2.3}=0\]
\[\frac{3}{1}+\frac{2}{1}-\frac{5}{1}=0\]
so, regardless of the value of a0, or a1, we can zero out the summation.
Now, the question is, what value does a0 and a1 have to be to get the whole ting equal to zero in this case since the solution is for =0? Thats where we can then define the unknown "constants".

Answer 25

Let me zero in on the step that is causing me trouble:\[2a _{1}+\sum_{2}^{}[a _{n}n(n-1)+2a _{n}n+a _{n-2}]x ^{n-1}=0\]
I can't see how this does not imply that a1=0 since a1 is the only x^0 term. In just a second, I'm going to work out the recursion and confirm that it does yield the cosine solution when I assume a1 is not 0.

Answer 26

So here is what I got for the recursion formula:\[a _{2n-1}=\frac{ 2(-1)^{n-1}a _{1} }{ (2n)! }\] which works for all the odd an's except for a1. Also not that it does not work for n=0. Plugging this into the power series:\[y _{2}(x)=\sum_{1}^{\infty}\frac{ 2(-1)^{(n-1)}a _{1}x ^{2n-1} }{ (2n)! }\]

Answer 27

Doing some tidying:\[y _{2}(x)=\frac{ -2a _{1} }{ x }\sum_{1}^{\infty}\frac{ (-1)^{n}x ^{2n} }{ (2n)! }\] This sum is soooooo close to being cos(x) but cos(x) starts at n=0. So we basically get that the sum is cos(x)-1.\[y _{2}(x)=\frac{ a _{1} }{ x }(\cos x -1)\]

Answer 28

Plugging this solution into the differential equations shows it does not satisfy the equation, which is expected because of the (cosx-1) instead of cosx.

Answer 29

Also I forgot to mention that I absorbed the -2 into the a1 since it's just a constant anyway.

Answer 30

those are some good intermediary steps; i had been working with it some last night and came to the same basic conclusions.

but I like to solve for an
\[a _{n}=-a_{n-2}\frac{1}{n(n+1)}\]
since \(y=\sum~a_n~x^n\) this produces

\[a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+ ....\]

\[a_0+2a_1\frac1{2!}x-a_0\frac 1{3!}x^2-2a_1\frac{1}{4!}x^3+a_0\frac{1}{5!}x^4+2a_1\frac{1}{6!}x^5+....\]
divvying this up into a0 and a1 parts:

\[a_0(1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\frac{1}{9!}x^8-...)\]
\[2a_1(\frac{1}{2!}x-\frac{1}{4!}x^3+\frac{1}{6!}x^5-\frac{1}{8!}x^7+\frac{1}{10!}x^9-...)\]

multiply and divide by x

\[a_0\frac1x(x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\frac{1}{9!}x^9-...)\]
\[2a_1\frac1x(\frac{1}{2!}x^2-\frac{1}{4!}x^4+\frac{1}{6!}x^6-\frac{1}{8!}x^8+\frac{1}{10!}x^{10}-...)\]

Answer 31

the wolf is showing an imaginary solution so i have to wonder if trying to conform it to someting else is making this go haywire

Answer 32

\[y = a_0+2a_1\frac1{2!}x-a_0\frac 1{3!}x^2-2a_1\frac{1}{4!}x^3+a_0\frac{1}{5!}x^4+2a_1\frac{1}{6!}x^5-....\]
\[y' = 2a_1\frac1{2!}-a_0\frac 2{3!}x-2a_1\frac{3}{4!}x^2+a_0\frac{4}{5!}x^3+2a_1\frac{5}{6!}x^4-a_0\frac{6}{7!}x^5....\]
\[y'' = -a_0\frac 2{3!}-2a_1\frac{3.2}{4!}x+a_0\frac{4.3}{5!}x^2+2a_1\frac{5.4}{6!}x^3-a_0\frac{6.5}{7!}x^4-2a_1\frac{7.6}{8!}x^5....\]
----------------------------
\[xy = a_0x+2a_1\frac1{2!}x^2-a_0\frac 1{3!}x^3-2a_1\frac{1}{4!}x^4+a_0\frac{1}{5!}x^5+2a_1\frac{1}{6!}x^6-....\]
\[2y' = 2a_1-a_0\frac {2.2}{3!}x-2a_1\frac{3.2}{4!}x^2+a_0\frac{4.2}{5!}x^3+2a_1\frac{5.2}{6!}x^4-a_0\frac{6.2}{7!}x^5....\]
\[xy'' = -a_0\frac 2{3!}x-2a_1\frac{3.2}{4!}x^2+a_0\frac{4.3}{5!}x^3+2a_1\frac{5.4}{6!}x^4-a_0\frac{6.5}{7!}x^5-2a_1\frac{7.6}{8!}x^6....\]
-----------------------------------
\[xy = a_0x+a_1x^2-a_0\frac 1{3!}x^3-a_1\frac{2}{4!}x^4+a_0\frac{1}{5!}x^5+a_1\frac{2}{6!}x^6-....\]
\[2y' = 2a_1-a_0\frac {2.2}{3!}x-2a_1\frac{3.2}{4!}x^2+a_0\frac{4.2}{5!}x^3+2a_1\frac{5.2}{6!}x^4-a_0\frac{6.2}{7!}x^5....\]
\[xy'' = -a_0\frac 2{3!}x-2a_1\frac{3.2}{4!}x^2+a_0\frac{4.3}{5!}x^3+2a_1\frac{5.4}{6!}x^4-a_0\frac{6.5}{7!}x^5-2a_1\frac{7.6}{8!}x^6....\]
everything is zeroing out except that 2a1 .... might be a slight error in the process getting to y

Answer 33

of course since its an arbitrary constant, a1=0 would be fine

Answer 34

Yeah it works but it just doens't feel complete without the general solution being of the form\[y(x)=c _{1} y _{1}(x)+c _{2}y _{2}(x)\] where the constants are arbritrary and only determined after initial conditions are applied.

Answer 35

Also the solution given by wolfram alpha with imaginary parts could be transformed into the same form using cos and sin using Euler's Identity. I'll look for the thing I found on wolfram alpha where it showed the cos sin solution and the transformation and all of that.

Answer 36

Check this out:
http://www.wolframalpha.com/widgets/view.jsp?id=e602dcdecb1843943960b5197efd3f2a

Answer 37

Then look at the step by step solution they give.

Answer 38

It shows how they solve it using a transformation and it yields the cos/sin solution. This is primarily how I knew what to be looking for. It's always easier working out problems when you have the correct answer to guide you. XD