Solving Series ODE: Finding x(0) w/ Problem Statement

[mathman44]

Homework Statement

The Attempt at a Solution

I did the "show that" part. But what is throwing me off is the x(0)=0 part. What is "x" a function of? Using the series in the square brackets,

I found that when
n=0, a_1 = a_0 ^2
n=1, a_2 = (a_1 * a_0)/2
n=2, a_3 = (a_0*a_2 + a_1^2 + a_2*a_0)/3

So it would seem that a_3 is still zero because all the terms it is a function of are also zero (or, at least, should be). I'm not confident about this... where did I go wrong?

 

[fzero]

Whether they mean that x(y=0) = 0 or it's a typo and should have been y(x=0)=0, both give the same results. You use the initial condition to show that a₀=0. There's no reason to set x=0 in the recursion relations.

 

[LCKurtz]

Homework Statement

The Attempt at a Solution

I did the "show that" part. But what is throwing me off is the x(0)=0 part. What is "x" a function of? Using the series in the square brackets,

I found that when
n=0, a_1 = a_0 ^2
n=1, a_2 = (a_1 * a_0)/2
n=2, a_3 = (a_0*a_2 + a_1^2 + a_2*a_0)/3

So it would seem that a_3 is still zero because all the terms it is a function of are also zero (or, at least, should be). I'm not confident about this... where did I go wrong?

Check your a₂, I get a₂=a₁a₀ but that isn't your main problem. For n = 2 you are looking at the coefficient of x² on the left side. Don't forget there is a -x² term outside of the sum.

As far as the x(0) = 0 thing, I would bet it is a typo and means y(0) = 0.

 

[mmmboh]

Hey phil, I'm pretty sure a₃=1/3, and the next nonzero coefficient is a₇=1/63, didn't look for the next, I think it's a₁₁=2/2079 though. (we're working together). (btw your a₃ equation should be (3a₃-a₁²-2a₀a₂)x²-x²=0, so the coefficient of the first x² isn't suppose to be zero.)

 

[mathman44]

Thanks for the help, I got it ;)