Question

Solve using power series.
y" + 4y = 0

Answer 1

No dude. would u help me with this then?

Answer 2

yes I know the power series for e^x. i bet its 1 + x + x^2/2! + x^3/3! + ....
but I don't know how to execute it. ;(

Answer 3

with a bullet in the head

Answer 4

let \[\Large y=\sum_{n=0}^{\infty} a_n x^n=a_0+a_1x+a_2x^2+\cdots\] for some constants \(a_n\)
\[\Large y''=2(1)a_2+3(2)a_3 x+4(3)a_4 x^2+\cdots\]
now for the equation \(\Large y''+4y=0\)
\[\large 2(1)a_2+3(2)a_3x+4(3)a_4x^2+\cdots +4(a_0+a_1x+a_2x^2+\cdots) = 0\]
for this equation to be true for all x, the coefficients should all equal zero, that is
\[\Large{2a_2+4a_0=0\\3(2)a_3+4a_1=0\\4(3)a_4+4a_2=0\\\vdots}\]

Answer 5

bullet to the stomach is more painful

Answer 6

@kropot72 . thanks dear. So I have to find then how it arrives to the general solution.

Answer 7

\[\Large{2a_2+4a_0=0\Rightarrow a_2=-(4/2)a_0\\6a_3+4a_1=0\Rightarrow a_3=-(4/6)a_1\\12a_4+4a_2=0\Rightarrow a_4=-(4/12)a_2\\\vdots\\a_{n}=-(4/n(n-1))a_{n-2}\qquad,n=2,3,4,\dots}\]

Answer 8

@kropot72 , \[\Large y=Ae^{2x}+Be^{2x}\] is not the correct solution.

Answer 9

use the power series

\[\sum_{i=0}^\infty a_nx^n\]

derivative of this will be step up one

\[\frac{dy}{dx}=\sum_{i=1}^\infty na_nx^{n-1}\]

Answer 10

so do that again and then what you want to do is get everything to the \[x^n\]

Answer 11

once you do that you should get something of the such

\[(n-1)(n-2)(n+1)+a_{n+1}+a_{n-2}=0\\]

Answer 12

\[(n-1)(n-2)(n+1)+a_{n+1}+a_{n-2}=0\]

Answer 13

isolate one of the a terms and then you'll get something of the suh

\[a_{n+1}=\frac{a_{n-2}}{(n-1)(n-2)(n+1)}\]

Answer 14

then you can plug in values for n to find the power series

Answer 15

This question has a small trick, but i will let others finish, because they started before me, if you want i can put the final answer as i solved on a piece of paper...

Answer 16

oh also you want the n=0 to be the smae for all in each

in other words... you want to take the derivatives of that sum and then you'll get 2 different sums...

those sums can then be added up to one sum

Answer 17

All help is much appreciated. thanks alot.
Still solving here, using all your ideas. Thanks ^_^

Answer 18

Well as the guys mentioned and they have done a great job, you need to differentiate the sums....

Answer 19

\[y = \sum_{n=0}^{\infty} a_nx^n\]

Answer 20

http://tutorial.math.lamar.edu/Classes/DE/HOSeries.aspx

Answer 21

\[y \prime = \sum_{n=1}^{\infty}n a_nx^{n-1}\]

Answer 22

\[y'' =\sum_{n=2}^{\infty}n(n-1 )a_nx^{n-2}\]

Answer 23

\[\sum_{n=2}^{\infty}n(n-1 )a_nx^{n-2} + 4\sum_{n=0}^{\infty} a_nx^n = 0\]

Answer 24

this is done by plugging the series in y''+4y=0

Answer 25

now the only trick in such questions is when you have different indexes in summations, and to solve such a problem we need to shift the sum in order to have the same starting point...

Answer 26

as you can see we have n =0 in one summation and we have n =2 in the other one...

Answer 27

so we need to shift n=2 to be n=0

Answer 28

\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n+4\sum_{n=0}^\infty a_x x^n=0\]

Answer 29

\[\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2}x^n\]

Answer 30

can you see what happened? because we started from 2, but we need to start from 0, then we shift the sum by 2 units at every component in the series

Answer 31

\[\sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2}x^n+4\sum_{n=0}^{\infty} a_nx^n = 0\]

Answer 32

am i disconnected my pc is going mad

Answer 33

now what we can do, we factorize the x_n and rearrange the sums

Answer 34

\[\large \sum_{n=0}^{\infty}\left\{ (n+2)(n+1)a_{n+2} +4a_n\right\}x^n=0\]
this means \[\large {(n+2)(n+1)a_{n+2}+4a_n=0\qquad n=0,1,2,\cdots\\a_{n+2}=-\frac{4}{(n+2)(n+1)}a_n}\]

Answer 35

yes that is the right answer saved me time...

Answer 36

but there is still more for this question... to come

Answer 37

now from the power series, we will get some sort of relation, most likely similar to functions expanded using Taylor series...

Answer 38

the philosophy behind such a system its a bridge between explicit ODE solution and Numerical Analysis for ODE

Answer 39

So sirm3d, I solved down to \[a _{n+2}= \frac{ -4 a _{n} }{ (n+2)(n+1) }\] and got the series \[\sum_{n=0}^{\infty} \frac{ (-1)(2x)^{2n} }{ (2n)! }\] for the c0 terms and \[\sum_{n=0}^{\infty} \frac{ (-1)x ^{2n+1} }{ (2n+1)! }\] for the c1 terms. Is this correct?