Question

please help to solve and i will fan and give a medal
Y" + x y' + y= 0 when y(0)=3 and y'(0)=1

Answer 1

what have you tried so far ?

Answer 2

i have tried to solve it but i have not found a good book for it

Answer 3

you may find similar example problems at below site useful
http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx

Answer 4

its too complex for me and its an exam question

Answer 5

please help me

Answer 6

suppose the solution can be represented as power series :
\[y = \sum\limits_{n=0}^{\infty} a_nx^n\]

Answer 7

please continue sir

Answer 8

find \(y'\) and \(y''\)
then plug them into the given differential equation

Answer 9

ok but can you please help with solution.. its easier if it is solved and i follow thesame step please

Answer 10

\(y = \sum\limits_{n=0}^{\infty} a_nx^n\)

\(y' = \sum\limits_{n=1}^{\infty} na_nx^{n-1}\)

\(y'' = \sum\limits_{n=2}^{\infty} n(n-1)a_nx^{n-2}\)


the given differential equation :
\[y" + x y' + y= 0\]
becomes :
\[ \sum\limits_{n=2}^{\infty} n(n-1)a_nx^{n-2} + x \sum\limits_{n=1}^{\infty} na_nx^{n-1}+ \sum\limits_{n=0}^{\infty} a_nx^n= 0\]

Answer 11

ok that is the y" ,y' and y in the equation above. thanks sir please continue

Answer 12

Next the trick is to make the exponent of "x" equal in all terms :

\[\sum\limits_{n=2}^{\infty} n(n-1)a_nx^{n-2} + \color{red}{x} \sum\limits_{n=1}^{\infty} na_n\color{Red}{x^{n-1}}+ \sum\limits_{n=0}^{\infty} a_nx^n= 0\]

Multiplying them gives

\[\sum\limits_{n=2}^{\infty} n(n-1)a_nx^{n-2} + \sum\limits_{n=1}^{\infty} na_n\color{Red}{x^{n}}+ \sum\limits_{n=0}^{\infty} a_nx^n= 0\]

Answer 13

Notice the exponent in first term is \(x^{n-2}\),
lets see how to make it \(n\) ?

Answer 14

ok prof

Answer 15

please ride on

Answer 16

substitute \(n-2 = k\)
then \(n = k+2\)

\[\sum\limits_{k=0}^{\infty} (k+2)(k+1)a_{k+2}x^{k} + \sum\limits_{n=1}^{\infty} na_n\color{Red}{x^{n}}+ \sum\limits_{n=0}^{\infty} a_nx^n= 0\]

which is same as

\[\sum\limits_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n} + \sum\limits_{n=1}^{\infty} na_n\color{Red}{x^{n}}+ \sum\limits_{n=0}^{\infty} a_nx^n= 0\]

yes ?

Answer 17

sir please note that the intial conditions given are y(0)=1 and y'(0)=0 please forgive me for my mistake . continue please

Answer 18

we will use initial conditions shortly, did u get how we managed to get all the terms to have same exponent ?

Answer 19

yes very well sir

Answer 20

\[\sum\limits_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n} + \sum\limits_{n=1}^{\infty} na_n\color{Red}{x^{n}}+ \sum\limits_{n=0}^{\infty} a_nx^n= 0\]

which is same as

\[\sum\limits_{n=0}^{\infty} \left[ (n+2)(n+1)a_{n+2}x^{n} + na_n\color{Red}{x^{n}}+ a_nx^n\right]= 0\]

Answer 21

yes.

Answer 22

what next sir

Answer 23

sir please continue and complete the solution . my system battery is dying . i will be viwing it with my phone but might not type any thing .... please hope you will complete it for me. GOD bless you....this is the best site

Answer 24

\[\sum\limits_{n=0}^{\infty} \left[ (n+2)(n+1)a_{n+2}x^{n} + na_n\color{Red}{x^{n}}+ a_nx^n\right]= 0\]
\[\sum\limits_{n=0}^{\infty} \left[ (n+2)(n+1)a_{n+2} + (n+1)a_n\right] x^n = 0\]
set the coefficient to 0 and solve the recurrence relation

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

Answer 25

ok. i get this step sir. what next

Answer 26

please complete it for me sir. i will look at how you solved every thing and learn. GOD bless

Answer 27

isolating \(a_{n+2}\) gives :

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

use the initial conditions and see if you can sovle above

Answer 28

Non-power series approach: Notice that you have two derivatives in the ODE:
\[\frac{d}{dx}[y']=y''\quad\text{and}\quad\frac{d}{dx}[xy]=xy'+y\]
So
\[\begin{align*}
y''+xy'+y&=0\\\\
\frac{d}{dx}[y'+xy]&=0\\\\
y'+xy&=C_1
\end{align*}\]
which is linear in \(y\).

Answer 29

@ rational, i understood all you did here but i could not finish it. please continue and finish it for me. i have less time before my exam. please help me on this . thanks and God bless

Answer 30

please do not forget i made a modification on the initial conditions given