Question

Find the solution of the IVP
2x^2y"+xy'-3y=0 y(1)=1, y'(1)=4
Describe how the solution behaves as x goes to zero.

Answer 1

How far did you get on this one?

Answer 2

i assumed that the power series would be centred on 1,
so i let t=x-1
and z(t)=y(x)
then i found the power series expansion, and found that
a2=a0/2
a3=a0/3
a4=-11a0/24
a5=-41a0/60
but i don't think thats right….
I am trying to relate the expressions so I can write a general solution for y(x0=…
but I don't see how the a coefficient expressions relate.

Answer 3

Let y=x^m

Answer 4

y'=mx^(m-1)

Answer 5

y''=m(m-1)x^(m-2)

Answer 6

Substitute these into the DE and solve for m.

Answer 7

Don't go by the power series method. What do you get for m?

Answer 8

Reason why you do not need to go by the series method for this DE is because it is a Cauchy Euler DE. For Cauchy Euler DE, letting y=x^m is a simpler approach.

Answer 9

we have to use power series to solve this because that is the section we are learning in class!

Answer 10

in the question it says we have to use power series

Answer 11

OK. Then I would just let
\[y=\sum_{n=0}^{\infty}a _{n}x ^{n}\]

Answer 12

The final answer to the question as to how it should behave as x goes to 0, is y goes to e.

Answer 13

could you maybe write out the first few steps?
I don't get what you mean when you say y goes to e is the answer

Answer 14

Let me work on this. Will revert.

Answer 15

Here is the solution.

Answer 16

The corret response as x tends to 0, is y tends to - infinity.

Answer 17

I took a look at your solution, is there any way you can center around 1 and solve it that way?

Answer 18

Sorry I dont know why you want to do that and I am not about to d that. Cauchy Euler DE is a very standard one.