# ODE and series

1. Nov 2, 2005

### Benny

Ok I'm giving these another go. I found the following DE from a reduction of order problem and figured that it would be an alright question if I turned it into one requiring a series solution. However I'm stuck. I think it's just a matter of index shifts to get an appropriate recurrence relation between the coefficients. Here is the equation and what I've done.

$$2x^2 y'' + 3xy' - y = 0$$

$$y = \sum\limits_{n = 0}^\infty {c_n x^n } \Rightarrow y' = \sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } \Rightarrow y'' = \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} }$$

The reason why I haven't just straight away shifted the starting indices for y' and y'' is that the coefficients of those derivatives in the DE, are not constant. So I think that initially leaving the derivatives with different starting values makes it easier to decide on index shifts later on. Is that the right way to start?

$$2x^2 y'' = 2x^2 \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right)x^{n - 2} } = \sum\limits_{n = 2}^\infty {2c_n n\left( {n - 1} \right)x^n }$$

$$3xy' = 3x\sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } = \sum\limits_{n = 1}^\infty {3c_n nx^n }$$

I know that I need to get a 'common power' for x in each of the summations eg. x^n, x^(n+1), x^(n+2) etc. I can't think of a way to do this. Does anyone have any suggestions?

2. Nov 2, 2005

### HallsofIvy

Staff Emeritus
You can't think of a way to do that? Isn't it obvious that you already have "a common power"- n? If it's the fact that each sum starts with a different value, you could do the cases n= 0, n= 1, n> 1 separately.
Actually, this is an "equipotential" equation. also known as an "Euler-type equation", (the coefficient of each derivative is a power of x- the power is the same as the order of the derivative): you don't need to use a power series method.
Either: Let y= xr, substitute into the equation to get a "characteristic equation" just as you do with erx for equations with constant coefficients or
Make the substitution u= ln x to convert to an equation in u with constant coefficients.

Last edited: Nov 2, 2005
3. Nov 2, 2005

### saltydog

Benny, your equation has a singular point at x=0 and therefore if you wished to solve it via power series, you would need to use:

$$y(x)=\sum_{n=0}^\infty a_n x^{n+c}$$

where c is a root of the indicial equation. Depending if the difference of the roots is nonintegral, 0 or integer, logarithms may enter into the solutions. Check out "solutions near regular singular points" and the indicial equation for these types of DEs.

4. Nov 2, 2005

### Benny

I don't know what I was thinking. Yeah, I already had a common power. The starting index being different for the summations shouldn't be a problem. I could always just do something starting each summation at a common value of n by taking out the first few terms from the relevant summations. This will allow me to set some coefficients equal to zero and from the recurrence this should mean that a whole lot of other coefficients are also zero. Something like that.

Saltydog - From the reduction of order question, a given solution is y = 1/x. So that is probably part of the problem with my series 'solution' attempt.

Now instead of just trying to make up a question (there's not much point in taking the reduction of order questions I have because the equations are all Cauchy - Euler equations), I'll take an actual series question. The examples associated with this question only use the standard trial series for y so the following shouldn't require any modifications to be made to y. Here it is.

$$x^2 y'' + x^2 y' + xy = 0,y\left( 0 \right) = 1,y'\left( 0 \right) = 1$$

$$y = \sum\limits_{n = 0}^\infty {c_n x^n } \Rightarrow y' = \sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } \Rightarrow y'' = \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} }$$

Substitute into the equation.

$$x^2 \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} } + x\sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } + x^2 \sum\limits_{n = 0}^\infty {c_n x^n } = 0$$

$$\sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)} x^n + \sum\limits_{n = 1}^\infty {c_n nx^n } + \sum\limits_{n = 0}^\infty {c_n x^{n + 2} } = 0$$

$$\sum\limits_{n = 0}^\infty {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right)x^{n + 2} } + \sum\limits_{n = 0}^\infty {c_n x^{n + 2} } + \sum\limits_{n = 1}^\infty {c_n nx^n } = 0$$

$$\sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right) + c_n } \right]} x^{n + 2} + \sum\limits_{n = - 1}^\infty {c_{n + 2} \left( {n + 2} \right)x^{n + 2} } = 0$$

Extract the n = -1 term to get:

$$\sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right) + \left( {n + 3} \right)c_n } \right]} x^{n + 1} + c_1 x = 0$$

Edit 2: I wrote this problem down on paper and basically got down to the following.

$$\sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)^2 + c_n } \right]} x^{n + 2} + c_1 x = 0$$

This implies c_1 = 0 which is consistent with the ICs. This leaves me with $$c_{n + 2} \left( {n + 2} \right)^2 + c_n = 0$$.

$$c_{n + 2} = - \frac{{c_n }}{{\left( {n + 2} \right)^2 }}$$

$$n = 1:c_3 = - c_1 \times stuff = 0$$ since c_1 = 0.

$$\Rightarrow c_{2n + 1} = 0,\forall n \in N$$

Concentrating on the even coefficients:

$$c_{2n} = - \frac{{c_{2n - 2} }}{{\left( {2n} \right)^2 }}$$

Iterating a few times I find that $$c_{2n} = \frac{{\left( { - 1} \right)^n c_0 }}{{\left( {2^n n!} \right)^2 }}$$.

Using c_0 = 1 I get:

$$y\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n }}{{\left( {2^n n!} \right)^2 }}x^{2n} }$$

That's my best attempt. I'm really being bogged down by this final topic.

Last edited: Nov 3, 2005
5. Nov 3, 2005

### HallsofIvy

Staff Emeritus
Benny, this is again an equation that is singular at x= 0 (the coefficient of y" is 0 at x= 0) so a simple power series may not be enough. As salty dog said, you should use "Frobenius' method": try a series of the form
$$y(x)=\sum_{n=0}^\infty a_n x^{n+c}$$
Because that "c" can be chosen in many different ways, assume that a0 is not 0. Put that into your differential equation, collect like terms and set the coefficient of the lowest power term equal to 0 (setting n=0). That will typically give you a formula in c times a0. Since a0 is not 0 you can solve for c (and, typically, since this is a second order equation, that "indicial equation" will be quadratic and will give two different solutions for c).

6. Nov 3, 2005

### Benny

Ok I'll try that method out. It's just that from the examples that were given in the section containing this problem, only a standard power series was required. But I'll try saltydog's suggestion anyway. Thanks for the help.

7. Nov 3, 2005

### saltydog

Benny, you sure your initial conditions are correct? I suspect y(0) can't be 0. This is why:

When I use the trial solution:

$$\sum_0^{\infty} a_nx^{n+c}$$

and substitute into the DE and shifted indexes so that all the sums have the same power of x, I obtain:

**** edited to show c's ****

$$\sum_0 a_n(n+c)(n+c-1)x^{n+c}+\sum_1 a_{n-1}(n-1+c)x^{n+c}+\sum_1 a_{n-1}x^{n+c}$$

The indicial equation comes from the n=0 term when this method is used so therefore it's:

$$c(c-1)=0$$

The roots are therefore 0 and 1. The difference is integral so there is a likelyhood a logarithm will enter into the solution.

Letting c=0:

$$\sum_0^\infty a_n n(n-1)x^n+\sum_1^\infty a_{n-1}(n+1-1)x^n$$

Sure enough, when I solve for the coefficients, I obtain:

$$a_0=0$$

$$a_1=\text{arbitrary}$$

$$a_n=-\frac{a_{n-1}}{n-1};\quad n\ge 2$$

That's only 1 arbitrary constant so at most I can obtain only one solution unless I use another method (such as the logarithmic method).

Anyway, when I form the product for n ge 2:

$$a_2a_3a_4\cdots a_n=\frac{(-1)a_1}{1}\frac{(-1)a_2}{2}\frac{-1)a_3}{3}\cdots\frac{(-1)a_{n-1}}{n-1}$$

and cancel, I'm left with:

$$a_n=\frac{(-1)^na_1}{(n-1)!}$$

Thus, one solution is:

$$y(x)=\sum_1^\infty \frac{(-1)^na_n}{(n-1)!} x^n$$

Using the magic of Mathematica:

Code (Text):

$$y(x)=\sum_1^\infty \frac{(-1)^na_n}{(n-1)! }x^n$$

yd1[x_]=D[y[x],x];

yd2[x_]=D[y[x],{x,2};

Simplify[x^2 yd2[x]+x^2 yd1[x]+x y[x]

Mathematica returns 0 so I have some confidence this series solution is correct.

Note that y(0) can only be 0 and the logarithm solution will not admit x=0.

Last edited: Nov 3, 2005