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&#039;-2xy &amp; = \sum_{n=0}^{\infty} (na_nx^{n-1} - 2xa_nx^n )\\<br /> &amp;= \sum_{n=0}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} 2xa_nx^n\\<br /> &amp;= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}\\<br /> &amp;= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}<br /> \end{align*}<br />

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&#039;-2xy &amp; = \sum_{n=0}^{\infty} (na_nx^{n-1} - 2xa_nx^n )\\<br /> &amp;= \sum_{n=0}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} 2xa_nx^n\\<br /> &amp;= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}\\<br /> &amp;= \sum_{n=1}^{\infty} na_nx^{n-1} - 2\sum_{n=0}^{\infty} a_nx^{n+1}\\<br /> &amp;= \sum_{n=0}^{\infty} (n+1)a_{n+1}x^n - 2\sum_{n=0}^{\infty} a_nx^{n+1}<br /> \end{align*}<br />

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, &amp; \;n\;\text{odd}\\ <br /> \displaystyle \pm\frac{1}{ \left( \frac{n}{2} \right) !}a_0, &amp; \;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&#039;-y &amp; = \sum_{n=0}^{\infty} (na_nx^{n} - a_nx^n )\\<br /> &amp;= \sum_{n=0}^{\infty} (n-1)a_nx^{n}<br /> \end{align*}<br />

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.