Help with frobenius series

In summary, the question is asking for help on finding two linearly independent solutions for 4xy+2y=0. The user has tried almost everything but has not been successful. They need to assume a solution and find the derivatives. Once they have the derivatives, they can substitute them into the equation and solve for y' and y".f
  • #1
10
0
the question is
" find two linearly independent of frebenius series solutions for 4xy''+2y'+y=0"

I try almost everything to slove this, but could't figure it
any help is appercaited

thank you
 
  • #2
Well, what have you done on it? You say you've tried "almost everything". Okay, what have you tried?
 
  • #3
Well, what have you done on it? You say you've tried "almost everything". Okay, what have you tried?


well, i divided the whole thing by 4x, trying to get it to the general format, but it didnt work out.
 
  • #4
In other words, you really haven't done anything!

Start by writing
[tex]y= \sum_{n=0}^\infty a_n x^{n+c}[/tex]
Find y' and y" from that and put them into the equation. What do you get?
 
  • #5
In other words, you really haven't done anything!

Start by writing
[tex]y= \sum_{n=0}^\infty a_n x^{n+c}[/tex]
Find y' and y" from that and put them into the equation. What do you get?

i don't know how to find y' and y''
im slef learning def.Q
thanks for the reply tho.
 
  • #6
In other words, you really haven't done anything!

Start by writing
[tex]y= \sum_{n=0}^\infty a_n x^{n+c}[/tex]
Find y' and y" from that and put them into the equation. What do you get?

by the way, i don't think y' and y'' is needed here
i divided the whole thing by 4x, then slove for the indicial,
which are r=0,-.5
but i don't know what to do from this point on,
i think i need to find the equation for cn, but don't know how.
can you help
 
  • #7
You have to assume a solution as you have been advised
You don't need to divide by 4x.
When you assume the solution you will find d second and first derivatives of y, you will then subtitute into the equation.
I solved it and that's the way it goes.
 
  • #8
Though not the question it is interesting to know that this DE has a solution in closed form. One can find this by substituting [itex]x=t^2[/itex] in the equation. A very simple DE will appear and can be solved directly. Doing the inverse substituting on this solution gives the result of the original DE. This solution can then be compared to the series solution.
 
  • #9
i don't know how to find y' and y''
im slef learning def.Q
thanks for the reply tho.
It's probably not a good idea to try to learn differential equations, by your self or not, if you do not know Calculus!

Are you seriously saying that you do not know how to find the derivatives of xn?

And, to even attempt a problem like this you should have had enough Calculus to know that a power series is "term by term" differentiable inside its radius of convergence.
 

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