Question

solve xy''+y'+y=0

Answer 1

Do you want series solutions?

Answer 2

I am going to have breakfast and will beback

Answer 3

No I mean not there,
\(\large\color{black}{ xy''+y'=\frac{d}{dx}(xy') }\)

\(\large\color{black}{ -x=\frac{d}{dx}(xy') }\)

Answer 4

integrate both sides,
divide both sides by x,
and then integrate both sides again

Answer 5

I want y1 and y2 to write the general solution

Answer 6

Let me contribute what I know about it :)
We are looking for the form \(y = \sum_{n=0}^\infty a_nx^n\)
hence \(y'=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^{n+1}\)
and \(y"=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\\xy"=\sum_{n=2}^\infty n(n-1) a_{n-1}x^{n-1}\)
Make them have the same limits and add them up

Answer 7

\[xy"+y'+y= \sum_{n=0}^\infty (n+2)(n+1)a_{n+1}x^{n+1} +\sum_{n=0}^\infty (n+1)a_{n+1}x^{n+1}+\sum_{n=0}^\infty a_nx^n=0\]

Answer 8

This is the exact question I thought I should get y1 and y2 to find w(y1,y2)
After your answer I got confused :/

Answer 9

hence, we have
\[- \sum_{n=0}^\infty (n+2)(n+1)a_{n+1}x^{n+1} =\sum_{n=0}^\infty (n+1)a_{n+1}x^{n+1}+\sum_{n=0}^\infty a_nx^n\]

Answer 10

we can rewrite the LHS under the limits of m
That is
\[- \sum_{m=0}^\infty (m+2)(m+1)a_{m+1}x^{m+1} \]

Answer 11

Now solve for the coefficient between the 2 of the sides
Notice, if m +1= n, then
if n=1, m =0
hence the coefficient of the LHS is \(-(0+2)(0+1) a_1= -2a_1\)
the coefficient of the RHS is \((1+1)a_2 +a_1=2a_2 +a_1\)
Hence \(-2a_1=2a_2 +a_1\) or \(a_2= (-1/2) a_1\)

Answer 12

if n=2, do the same argument to have a_3
On that way, we can get the series coefficient of the equation.

Answer 13

ok Thanks alot

Answer 14

It looks like you are looking to find the Wronskian. This might help you: http://www.sosmath.com/diffeq/second/linearind/linearind.html

From that site I believe yo can first rewrite your equation as:\[y''+\frac{1}{x}y'+\frac{1}{x}y=0\]compare this to the standard form which is:\[y''+p(x)y'+q(x)y=0\]and you will deduce that:\[p(x)=\frac{1}{x}\]You are also told that:\[W(y_1,y_2)(2)=\frac{1}{20}\]Then make use of:\[W(y_1,y_2)(x)=W(y_1,y_2)(2)e^{-\int_2^x p(t)dt}=\frac{1}{20}e^{-\int_2^x \frac{1}{t}dt}\]You should be able to solve from here

Answer 15

@asnaseer Thank you sooo much

Answer 16

yw :)