What is the power series solution to this differential equation?

[Ted123]

Homework Statement

[PLAIN]http://img59.imageshack.us/img59/2091/diffeq.png
[PLAIN]http://img684.imageshack.us/img684/6748/diffeqp.png

The Attempt at a Solution

Making the substitutions $$y= \sum_{n=0}^{\infty} a_n x^n$$ and $$y^{\prime} = \sum_{n=0}^{\infty}na_nx^{n-1},$$

$$\begin{align*}<br /> y'-2xy & = \sum_{n=0}^{\infty} (na_nx^{n-1} - 2xa_nx^n )\\<br /> &= \sum_{n=0}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} 2xa_nx^n\\<br /> &= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}\\<br /> &= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}<br /> \end{align*}$$

Now I'm having trouble seeing how to shift the summation index to combine the 2 sums into one and to have one coefficient of $$x^{n+1}$$

 

[tiny-tim]

Hi Ted123!

(on this forum, use "tex", or "itex" for inline latex, rather than "latex" …

and then no need for "begin" etc )

You need to get everything with the same exponent …

so change that ∑…x^n-1 to ∑…xⁿ

 

[Ted123]

Hi Ted123!

(on this forum, use "tex", or "itex" for inline latex, rather than "latex" …

and then no need for "begin" etc )

You need to get everything with the same exponent …

so change that ∑…x^n-1 to ∑…xⁿ

Just think it's easier to read if I align the equations

$$\begin{align*}<br /> y'-2xy & = \sum_{n=0}^{\infty} (na_nx^{n-1} - 2xa_nx^n )\\<br /> &= \sum_{n=0}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} 2xa_nx^n\\<br /> &= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}\\<br /> &= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}\\<br /> &= \sum_{n=0}^{\infty} (n+1)a_{n+1}x^n - 2\sum_{n=0}^{\infty} a_nx^{n+1}<br /> \end{align*}$$

The problem is the RHS sum has $$x^{n+1}$$ not $$x^n$$

EDIT: Just realized I can do this:

$$a_1+\sum\limits_{n=1}^{\infty} [(n+1)a_{n+1}x^n-2a_{n-1}x^n]$$
$$a_1+\sum\limits_{n=1}^{\infty} [(n+1)a_{n+1}-2a_{n-1}]x^n$$

Equating the coefficients of $$x_n$$ at every n, we obtain the recurrence relation

$$a_1 = 0$$
$$(n+1)a_{n+1}-2a_{n-1} = 0,\;\;\;\;\;\;\;\;\;\;n=1,2,3,...$$
or
$$a_{n+1} = -\frac{2a_{n-1}}{n+1}$$

 

[Ted123]

Equating the coefficients of $$x_n$$ at every n we obtain the recurrence relation

$$a_1 = 0$$
$$(n+1)a_{n+1}-2a_{n-1} = 0,\;\;\;\;\;\;\;\;\;\;n=1,2,3,...$$
or
$$a_{n+1} = -\frac{2a_{n-1}}{n+1}$$

Setting consecutively $$n=1,2,3,...$$ we determine $$a_2, a_3, a_4$$ etc.:

$$a_2 = -\frac{2a_0}{2} = -a_0$$

$$a_3 = -\frac{2a_1}{3} = 0$$

$$a_4 = -\frac{2a_2}{4} = \frac{1}{2}a_0$$

$$a_5 = -\frac{2a_3}{5} = 0$$

$$a_6 = -\frac{2a_4}{6} = -\frac{1}{6}a_0$$

What is the general solution $$a_n?$$

It seems to be of the form $$a_n = \left\{ \begin{array}{lr} <br /> 0, & \;n\;\text{odd}\\ <br /> \displaystyle \pm\frac{1}{ \left( \frac{n}{2} \right) !}a_0, & \;n\;\text{even}<br /> \end{array} <br /> \right.$$

 

[tiny-tim]

ok, so a_2n = a₀/n! …

total is ∑ (x²)ⁿ/n! …

remind you of anything? ​

 

[Ted123]

ok, so a_2n = a₀/n! …

total is ∑ (x²)ⁿ/n! …

remind you of anything? ​

Actually, I clearly made an algebraic error and there should be no minus sign in the recurrence relation. Then if you follow it through it gives what you have and this equals $$a_0 e^{x^2}$$

I have another:

[PLAIN]http://img685.imageshack.us/img685/2091/diffeq.png

$$\begin{align*}<br /> xy'-y & = \sum_{n=0}^{\infty} (na_nx^{n} - a_nx^n )\\<br /> &= \sum_{n=0}^{\infty} (n-1)a_nx^{n}<br /> \end{align*}$$

This would imply $$(n-1)a_n = 0 \Rightarrow a_n = 0$$ for all n.

But obviously the solution is $$y = a_0 x$$ so what discrepencies arise?

 

[tiny-tim]

Hi Ted123!

$$(n-1)a_n = 0 \Rightarrow a_n = 0$$

nooo …

(n-1)a_n = 0 has two solutions ​

 

[Ted123]

Hi Ted123!


nooo …

(n-1)a_n = 0 has two solutions ​

$$n=1$$ is also a solution...

 

[tiny-tim]

yup!

so when n = 1, a_n (ie a₁) can be anything ​

 

[Ted123]

yup!

so when n = 1, a_n (ie a₁) can be anything ​

So how do I express the coefficients $$a_n$$ in a power series?

$$a_n = 0$$ unless n=1 in which case $$a_1$$ is arbitrary.

so $$y = a_1 x$$

and this is easily verified by solving the equation directly.

What does it mean when it says 'explain any discrepancies which arise'?

 

[tiny-tim]

So how do I express the coefficients $$a_n$$ in a power series?

Just y = a₁x (or y = Cx).

What does it mean when it says 'explain any discrepancies which arise'?

Sorry … I've no idea.