# ODE Help!

1. Nov 14, 2011

### bballing1210

Hi all,

I've been having trouble answering these two ODE problems. Hopefully someone can help me out.

1. Compute the solution of y"' - xy' = 0 which satisfies y(0) = 1, y'(0) = 0, and y"(0) = 0.

I've tried using the power series expansion for y and substituting it in and getting the recurrence relation, but when I substitute back the initial conditions I'm getting the two series cancelling out which I don't think is right.

2. Solve the initial value problem 3y" - y' + (x+1)y = 1 with y(0) = y'(0) = 0

For this one, I know you have to compute both the power series expansion as well as a particular solution through substitution to be able to apply the initial conditions, but I'm seriously stuck in even thinking about what to substitute, let alone how to tackle this. I tried to start developing out the recurrence relationship, but it got really messy and I don't think I'm doing it right.

Thanks a lot to whoever can help, I really appreciate it!

2. Nov 15, 2011

### HallsofIvy

Yes, that's exactly what has to happen! The first thing I would do is let u= y' so the equation becomes u''= xu, u(0)= 0, u'(0)= 0. That satisfies all the conditions for "existence and uniqueness" of solutions and u(x)= 0 for all x is an obvious solution. That means that y is the constant function: y(x)= 1 for all x satisfies the differential equation and initial conditions.

Doesn't look too bad. Letting $y= \sum_{n=0}^\infty a_nx^n$, $y'= \sum_{n=1}^\infty na_nx^n$ and $y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}$ so the equation becomes
$$3\sum_{n=2}^\infty n(n-1)a_nx^{n-2}- \sum_{n=1}^\infty na_mx^{n-1}+ \sum_{n=0}^\infty a_mx^{n+1}+ \sum_{n=0}^\infty a_nx^n= 1$$
To get the same powers, change the dummy indices: in the first sum, let j= n-2 so that n= j+ 2 and it becomes
$$3\sum_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j$$
In the second sum, let j= n- 1 so it becomes
$$\sum_{j= 0}^\infty (j+1)a_{j+1}x^j$$
In the third sum, let j= n+ 1 so that n= j- 1 and it becomes
$$\sum_{j= 1}^\infty a_{j-1}x^j$$.
Finally, in the fourth sum, let j= n so it becomes
$$\sum_{j= 0}^\infty a_jx^j$$

Note that the third sum does not start until j= 1 (since it starts with $x^1$) so for j= 0 we have
$$6a_2- a_1+ a_0= 1$$
so that $a_2= (a_1- a_0+ 1)/6$
since we are given that $a_0= y(0)= 0$ and $a_1= y'(0)= 0$
that tells us that $a_2= 1/6$.

For j> 0, we have
$$3(j+2)(j+1)a_{j+2}- (j+1)a_{j+1}+ a_{j-1}+ a_j= 0$$
or
$$a_{j+2}= \frac{(j+1)a_{j+1}- a_{j-1}- a_j}{3(j+2)(j+ 1)}$$

Then $a_3= (2a_2- a_0- a_1)/18)= 1/18$, $a_4= (2a_3- a_1- a_2)/36= -1/648$, $a_5= (2a_4- a_2- a_3)/90$ etc. Yes, it wil probably be difficult to find a general formula. Are you required to?

Last edited by a moderator: Nov 16, 2011
3. Nov 15, 2011

### bballing1210

Hi. Thanks a lot for the help!

For number 2, what happened to the coefficient of 3 in front of the y"? You don't factor this into the series solution?

And I think we're allowed to keep the answers in a recurrence relation as long as it's correct, so I think that's what I'll do and I'll list out some terms in the series for clarification. Thanks!

4. Nov 16, 2011

### HallsofIvy

Yes, I accidently dropped the 3. I've gone back and corrected that. Hope its right this time!