Question

I will give a medal..
Find the two independent solution of d^2y/dx^2 + x.dy/dx + y=0 when the series expansion is about the origin

Answer 1

suppose the solution is

$$
\Large y(x) = a_0 + a_1x + a_2x^2 + a_3 x^3 + ...
\\
\text{we want to solve } \\
\Large y ' ' + x y' + y = 0


$$

Answer 2

$$
\large y(x) = a_0 + a_1x + a_2x^2 + a_3 x^3 + a_4x^4... \\
\large y'(x) = a_1 + 2a_2x + 3a_3x^2 + 4a_3x^3... \\
\large y ' '(x) = 2a_2 +6a_3x + 12a_4x^2...\\
\large \text{plug these solutions into } \\
\large y ' ' + x y' + y = 0
\large \\
\large \text {we have} \\
( 2a_2 +6a_3x + 12a_4x^2... )+ x(a_1 + 2a_2x + 3a_3x^2 + 4a_3x^3... ) \\
+ ( a_0 + a_1x + a_2x^2 + a_3 x^3 + a_4x^4...)=0

$$

Answer 3

$$( 2a_2 +6a_3x + 12a_4x^2... )+ x(a_1 + 2a_2x + 3a_3x^2 + 4a_3x^3... ) \\
+ ( a_0 + a_1x + a_2x^2 + a_3 x^3 + a_4x^4...)=0\\
\iff \\
(2a_2 + a_0) + (6a_3 + a_1+ a_1)x + (12a_4 +2a_2 + a_2)x^2 + ...=0
\\\iff\\
(2a_2 + a_0) + (6a_3 +2a_1)x + (12a_4 +3a_2)x^2 + ...=0
\\ \iff \\
\text {since the right side is equal to } \\

0 + 0x +0x^2 + 0x^3 + ...\\
\text {equating coefficients we have}\\

2a_2 + a_0 =0\\
6a_3 +2a_1 = 0\\
12a_4 + 3a_2 = 0

$$

Answer 4

it is easier if we use sum notation

Answer 5

Thanks prof. Am learning more. Please ride on sir

Answer 6

Ok sir.Sir from the first method you used, what are then the independent solutions

Answer 7

$$
\text{we want to solve } \\

\Large y ' ' + x y' + y = 0 \\
\text {Assume your solution is } \\
\Large y(x)=\sum_{n=0}^{\infty}a_n x^n\\
\text {then it follows that} \\
\Large y'(x) = \sum_{n=1}^{\infty}n \cdot a_n x^{n-1} \\
\Large y''(x) = \sum_{n=2}^{\infty}n(n-1) \cdot a_n x^{n-2}\\

\text {Substitute: }\Large y ' ' + x y' + y = 0 \\
\large \sum_{n=2}^{\infty}n(n-2) \cdot a_n x^{n-2} + x\sum_{n=1}^{\infty}n \cdot a_n x^{n-1} + \sum_{n=0}^{\infty}a_n x^n=0\\
\large \iff \\
\large \sum_{n=2}^{\infty}n(n-2) \cdot a_n x^{n-2} + \sum_{n=1}^{\infty}n \cdot a_n x^{n} + \sum_{n=0}^{\infty}a_n x^n=0\\
\large \iff \\
\large \sum_{n=0}^{\infty}(n+2)(n+1) \cdot a_{n+2} x^{n} + \sum_{n=1}^{\infty}n \cdot a_n x^{n} + \sum_{n=0}^{\infty}a_{n+2} x^n=0\\
\large \iff\\

\large \sum_{n=0}^{\infty}(n+2)(n+1) \cdot a_{n+2} x^{n} + \sum_{n=1}^{\infty}n \cdot a_n x^{n} + \sum_{n=0}^{\infty}a_n x^n=0\\


$$

Answer 8

Ok.i understand sir but I instead of using

Answer 9

make sure there are no errors, we need this for the next step

Answer 10

using the sum notationn can you deduce the two linear independent soluctions from the first method

Answer 11

$$

\iff \\
\ (2)(1)a_2x^0 + \sum_{n=1}^{\infty}(n+2)(n+1)a_{n+2}x^n +

\sum_{n=1}^{\infty}n \cdot a_n x^{n} + a_0 x^0 + \sum_{n=1}^{\infty}a_n x^n=0 \\

\iff \\

\large (2a_2 +a_0) + \sum_{n=1}^{\infty}[(n+2)(n+1)a_{n+2} + n\cdot a_n + a_n] x^n =0\\

\iff \\
\large (2a_2 +a_0) + \sum_{n=1}^{\infty}[(n+2)(n+1)a_{n+2} + (n+1)\cdot a_n] x^n =0

\\ \large \text{Equating coefficients this gives us: } \\
\large (2a_2+a_0)= 0\\
\large (n+2)(n+1)a_{n+2} + (n+1)\cdot a_n=0 \\
\text{Therefore}\\
\large a_0 = -2a_2 \\
\large a_{n+2} = \frac{-a_n(n+1)}{(n+1)(n+2)} = \frac{-a_n}{n+2} \\

$$

Answer 12

is that the answer sir?

Answer 13

made a typo

Answer 14

$$


\text {look at the even terms }\\
\Large a_2 = \frac{-a_0}{2} \\
\Large a_4 = \frac{-a_2}{(2+2)}= \frac{-(\frac{-a_0}{2})}{4}= \frac {a_0}{8} \\

\Large a_6 = \frac{-a_4}{(4+2)}= \frac{-(\frac{a_0}{8})}{6}= -\frac {a_0}{48} \\

... \\
\Large a_{2n}= \frac{a_0}{2^n~n!}(-1)^n \\
~~~\\
\text {look at odd terms }\\
\Large a_3 = \frac{-a_1}{1+2}= \frac{-a_1}{3}\\
\Large a_5 = \frac{-a_3}{3+2}= \frac{-(\frac{-a_1}{3})}{5}= \frac {a_1}{15} \\

\Large a_6 = \frac{-a_4}{5+2}= \frac{-(\frac{a_1}{15})}{7}= -\frac {a_1}{105} \\

... \\
\Large a_{2n+1}=\frac{a_1(-1)^n (2n+1)!}{2^n n!}

$$

Answer 15

OK SIR BUT CAN I ASK ANOTHER QUESTION HERE?

Answer 16

better put a new post

Answer 17

BUT PLEASE HELP ME ON THAT ......
THANKS NEW POST COMING

Answer 18

to get y1 , let a_0 = a_1 = 1 to get y2 , let a_0 = -a_1 = 1 PLEASE COMPLETE IT SIR

Answer 19

$$\large \text{plug in the a_n that we found}\\
\large y(x)=\sum_{n=0}^{\infty}a_n x^n\\
\large = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4...\\
\large = a_0 + a_1x + \frac{-a_0}{2}x^2 + \frac{-a1}{3}x^3 + \frac{a_0}{8}x^4 +\frac{a_0}{15}x^4...
=



$$

Answer 20

now we can factor

Answer 21

SIR CAN YOU WORK ON THIS
FIND THE FIRST FIVE TERM IN THE SOLUCTION DY/DX + (1+X^2)Y= SINX WITH Y(0)= a

Answer 22

?

Answer 23

notice that there are two independent constants there $$ a_0, a_1 $$

Answer 24

$$

\large \text{plug in the a_n that we found}\\
\large y(x)=\sum_{n=0}^{\infty}a_n x^n\\
\large = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4...\\
\large = a_0 + a_1x + \frac{-a_0}{2}x^2 + \frac{-a1}{3}x^3 + \frac{a_0}{8}x^4 +\frac{a_0}{15}x^4...\\
= a_0 + a_1x + \frac{-a_0}{2}x^2 + ... +\frac{a_0(-1)^n}{2^n n!} +\frac{a_0(-1)^{n+1}(2n+1)!}{2^n n!}... \\

= a_0 \Big\{1 - \frac{x^2}{2} +\frac{x^4}{8} - ... \Big\}
+ a_1\Big\{x - \frac{x^3}{3} +\frac{x^5}{15} - ... \Big\} \\
\large = a_0 \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^n n!}+ a_1 \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}(2n+1)!}{2^n n!}

$$

Answer 25

so sir what will be my conclusion

Answer 26

if you set a_0 = 0, and a_1 = 1, you will have one solution

if you set a_0 = 1 and a_0 = 0 you will have another solution

Answer 27

ok sir

Answer 28

thank you sir

Answer 29

$$

\large \text{If we substitute in the} ~a_n \text{that we found recursively above}\\
\large y(x)=\sum_{n=0}^{\infty}a_n x^n\\
\large = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4...\\
\large = a_0 + a_1x + \frac{-a_0}{2}x^2 + \frac{-a1}{3}x^3 + \frac{a_0}{8}x^4 +\frac{a_0}{15}x^4...\\
= a_0 + a_1x + \frac{-a_0}{2}x^2 + ... +\frac{a_0(-1)^n}{2^n n!} +\frac{a_0(-1)^{n+1}(2n+1)!}{2^n n!}... \\

= a_0 \Big\{1 - \frac{x^2}{2} +\frac{x^4}{8} - ... \Big\}
+ a_1\Big\{x - \frac{x^3}{3} +\frac{x^5}{15} - ... \Big\} \\
\large = a_0 \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^n n!}+ a_1 \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}(2n+1)!}{2^n n!} \\


\text {Therefore} \\
\large y(x) =a_0 \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^n n!}+ a_1 \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}(2n+1)!}{2^n n!} \\

\large \text{if } a_0 = 1 , a_1 = 0
\\\Large y_1(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^n n!}
\\\Large\text {if } a_0= 0 ,a_1 = 1

\\\Large y_2(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}(2n+1)!}{2^n n!}
$$

Answer 30

ok

Answer 31

can you start the other now or do you need some rest sir

Answer 32

the nth coefficient of a series is given by A_n= (1.5.9.13....(4n+1))/2^n express a_n in terms of the gamma funtion

Answer 33

interesting, the denominator for a_2n+1 is (2n+1)!! , double factorial

Answer 34

dont understant that sir

Answer 35

the coefficient 3,15,105 we found above are product of odd integers
3 = 3 * 1
15 = 5 * 3 * 1
105 = 7 * 5 * 3 * 1
...

Answer 36

http://en.wikipedia.org/wiki/Double_factorial

Answer 37

ok but do not understand what next you are ding

Answer 38

is the soluction for Y2(x) not correct?

Answer 39

ok

Answer 40

copy and paste that link

Answer 41

does it mean we are ccorrect sir

Answer 42

yes

Answer 43

sir can you please work on any of the two last questions ?

Answer 44

@SithsandGiggles @Michele_Laino
if you can look over, i would greatly appreciate (for errors)

Answer 45

there is a nice simplification for y_1(x) and y_2(x)

Answer 46

$$


\\\Large y_1(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^n n!} \\
\Large = \sum_{n=0}^{\infty}\frac{(\frac{-x^{2}}{2})^n}{n!}\\
\LARGE =e^{\frac{-x^2}{2}}
\\[0.5 in]
\\\Large y_2(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}(2n+1)!}{2^n n!}
\\ \text{this one doesn't have a nice simplification}
$$

Answer 47

Again to recap I got for the general solution:

$$

\\\Large y(x)=a_0e^{\frac{-x^2}{2}} + a_1\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}(2n+1)!}{2^n n!}



$$

Answer 48

if you can find a closed form or some kind of interesting approach for the latter term, maybe in terms of gamma function.

Answer 49

@perl I think there's just a slight mistake, where \(y''=\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}\), you had substituted \(\sum\limits_{n=2}^\infty n\color{red}{(n-2)}a_nx^{n-2}\).

Answer 50

@SithsAndGiggles
yes i see. I don't understand how these typos happen. and i can't edit it now.
i need a large latex editor so i can preview this.
this was quite a tedious task using Latex.

by the way, how come youre name doesn't show up when i type @ ,
hmm, well my name doesn't appear either

Answer 51

Did you see a simple expression for y_2(x)?

im going to check wolfram and compare their approximation to solving the linear diff.
y ' ' + x y ' + y = 0

The subject of power series is quite vast though, we are barely scratching the surface here, i imagine/

Answer 52

ok sir thanks.

Answer 53

but this is an exam question. where can i stop from the solution given, and i understood very well how you solved it but can you explain why ∑n=2∞n(n−2)⋅anxn−2 became ∑n=0∞(n+2)(n+1)⋅an+2xn

Answer 54

[SIR INTENSIFIES]

Answer 55

hello bendicent can you help me with my open question ?

Answer 56

@perl if you're looking for a simpler expression for the answer, you can determine that by solving the ODE a different way.
\[\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0~~\iff~~\frac{d}{dx}\left[\frac{dy}{dx}+xy\right]=0\]
which becomes linear in \(y\) upon integrating:
\[\frac{dy}{dx}+xy=C_1\]
Finding the integrating factor:
\[\ln\mu(x)=\int x\,dx~~\implies~~\mu(x)=e^{x^2/2}\]
hence
\[e^{x^2/2}\frac{dy}{dx}+xe^{x^2/2}y=C_1e^{x^2/2}~~\iff~~\frac{d}{dx}\left[e^{x^2/2}y\right]=C_1e^{x^2/2}\]
Integrating again and solving for \(y\) yields
\[\newcommand{erf}{\,\text{erf}\,}y=C_1e^{-x^2/2}\int e^{x^2/2}\,dx+C_2e^{-x^2/2}\]
where
\[\int e^{x^2/2}\,dx=\sqrt\frac{\pi}{2}\erf\frac{x}{\sqrt 2}=\sum_{n=0}^\infty \frac{\left(-\frac{1}{2}\right)^nx^{2n+1}}{n!(2n+1)}\]