How to Solve $(x^2 + 1)y'' - 6xy' + 10y = 0$ Using Series Methods?

[shamieh]

Use series methods to solve $(x^2 + 1) y" - 6xy' + 10y = 0$
a) Give the Recursion formula
b) give the first two non zero terms of the solution corresponding to $a_0 = 1$ and $a_1 =0$
c) give the first three non zero terms of solution corresponding to $a_0 = 0$ and $a_1 = 1$

So I posted this also on math exchange and someone was helping me out but i think they've made some mistakes. I'm really confused on this problem.

Where I got stuck with my series :(

 

[I like Serena]

Use series methods to solve $(x^2 + 1) y" - 6xy' + 10y = 0$
a) Give the Recursion formula
b) give the first two non zero terms of the solution corresponding to $a_0 = 1$ and $a_1 =0$
c) give the first three non zero terms of solution corresponding to $a_0 = 0$ and $a_1 = 1$

So I posted this also on math exchange and someone was helping me out but i think they've made some mistakes. I'm really confused on this problem.

Where I got stuck with my series :(

$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + \sum ^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - 6\sum ^{\infty}_{n=0} na_nX^n + 10 \sum ^{\infty}_{n=0}a_nX^n$$

Hi shamieh,

Consider that:
$$
\sum_{n=0}^\infty b_n = b_0 + b_1 + b_2 + ... = b_0 + \sum_{n=1}^\infty b_n
$$

And also that:
$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n = \sum ^{\infty}_{n=1} n(n-1)a_nX^n$$
since the term for $n=1$ is zero anyway. (Thinking)

 

[shamieh]

so would that give me $0 + 2a_2\sum^{\infty}_{n=0} n(n-1)a_nX^n$ ?

 

[I like Serena]

so would that give me $0 + 2a_2\sum^{\infty}_{n=0} n(n-1)a_nX^n$ ?

Let's see...

$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + \sum ^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - 6\sum ^{\infty}_{n=0} na_nX^n + 10 \sum ^{\infty}_{n=0}a_nX^n \\
= \sum ^{\infty}_{n=1} n(n-1)a_nX^n + \left(2a_2 + \sum ^{\infty}_{n=1} (n+2)(n+1)a_{n+2}X^n\right)
- 6\left(0 + \sum ^{\infty}_{n=1} na_nX^n\right) + 10 \left(a_0 + \sum ^{\infty}_{n=1}a_nX^n\right) \\
= 2a_2 + 10a_0 + \sum ^{\infty}_{n=1} \Big(n(n-1)a_n + (n+2)(n+1)a_{n+2} - 6na_n + 10 a_n\Big)X^n \\
= 0
$$

 

[shamieh]

Ooh, I think I see what you're saying. Gonna try it out now.

 

[shamieh]

In the last summation where is the $n$? that goes with the series being multiplied by $10$?

I got:

$y = \sum^{\infty}_{n=0} na_nX^n$

$y' = \sum^{\infty}_{n=1} na_nX^{n-1}$

$y'' = \sum^{\infty}_{n=2} n(n-1)a_nX^{n-2}$

so we have: $(x^2+1)[\sum^{\infty}_{n=2} n(n-1)a_nX^{n-2}] - (6x) \sum^{\infty}_{n=1} na_nX^{n-1} + 10\sum^{\infty}_{n=0} na_nX^n $

After doing some series manipulation I got:

$= \sum^{\infty}_{n=2} n(n-1)a_nX^n + \sum^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - \sum^{\infty}_{n=0} 6na_nX^n + \sum^{\infty}_{n=0} 10na_nX^n$

 

[I like Serena]

In the last summation where is the $n$? that goes with the series being multiplied by $10$?

I got:

$y = \sum^{\infty}_{n=0} na_nX^n$

That should be $$y = \sum^{\infty}_{n=0} a_nX^n$$.

Is that your missing $n$? (Wondering)

 

[shamieh]

Oh yep. My brain is working extra good today.

 

[shamieh]

So finally for the Recursion Formula I got: $a_{n+2} = -a_n\frac{(n-5)(n-2)}{(n+2)(n+1)}$

 

[I like Serena]

So finally for the Recursion Formula I got: $a_{n+2} = -a_n\frac{(n-5)(n-2)}{(n+2)(n+1)}$

Looks correct to me. (Smile)

Note that it also satisfies $2a_2+10a_0=0$, which is necessary.

 

[shamieh]

Also for part b) I got:

$a_0 = 1$

$a_1 = 0$

$a_2 = -5$

$a_3 = 0$

$a_4 = 0$

so: $y_1 = 1 + 0 + (-5x^2) + .. $

and for part c i got:

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

$a_3 = \frac{-4}{6}$

$a_4 = 0$

so $y_2 = x + 0 + (\frac{-4}{6}x^3) + 0 .. +..$

 

[I like Serena]

Part b) looks fine. (Smile)

For c), shouldn't you get at least 3 non-zero terms? Indeed, I get a different $a_5$. (Worried)

 

[shamieh]

How does this look?
$a_0 = 0$

$a_1 = 1$

$a_2 = -5$

$a_3 = \frac{-4}{6}$
so $y_2 = x + -5x^2 + (\frac{-4}{6}x^3) + 0 .. +..$

 

[I like Serena]

How does this look?
$a_0 = 0$

$a_1 = 1$

$a_2 = -5$

$a_3 = \frac{-4}{6}$

$a_4 = 0$

so $y_2 = x + -5x^2 + (\frac{-4}{6}x^3) + 0 .. +..$

Let's see... (Thinking)

We had $2a_2 + 10 a_0 = 0$.
Substituting, we get $2 \cdot -5 + 10 \cdot 0 = -10 \ne 0$.

Nope. It looks more wrong. (Sweating)

 

[shamieh]

Let's see... (Thinking)

We had $2a_2 + 10 a_0 = 0$.
Substituting, we get $2 \cdot -5 + 10 \cdot 0 = -10 \ne 0$.

Nope. It looks more wrong. (Sweating)

Let me try again lol

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

$a_3 = \frac{-4}{6}$

$a_4 = 0$

$a_5 = -1/15$

so $y_2 = x + \frac{-2}{3} x^3 + \frac{1}{15} x^5..$

 

[I like Serena]

That looks much better! (Whew)

 

[shamieh]

You know when you finish doing differential equations by series and then you forget how to add and subtract and do algebra? Lol that is my life right now. Thanks for all your help I like Serena(Malthe)(Time)

 

[shamieh]

UPDATE!

The correct solutions for part b) and part c) are:

b)
$a_0 = 1$
$a_1 = 0$
$a_2 = -2/3$
$a_3 = 0$
$a_4 = 0$

$\therefore$ $y_1 = 1 + (-\frac{2}{3}x^2) + ... + ...$

c)
$a_0 = 0$
$a_1 = 1$
$a_2 = 0$
$a_3 = -2/3$
$a_4 = 0$
$a_5 = 1/15$

$\therefore$ $y_2 = x + (-\frac{2}{3}x^2) + \frac{1}{15}x^5 + ...$