Airy's equation bounces back

[logistic_guy]

Find a series solution of Airy's equation \(\displaystyle y'' - xy = 0\) in the form \(\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n\).

 

[khansaheb]

Find a series solution of Airy's equation \(\displaystyle y'' - xy = 0\) in the form \(\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n\).

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

A long time ago I solved the Airy's equation from scratch by using the form \(\displaystyle \sum_{n=0}^{\infty}a_nx^n\). I was the only one on the earth to solve it like that, and I will always be the only one. Why? Because it is very difficult for others to understand the art of differential equations!

Now I am challenging myself to solve it again but with the new form \(\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n\). There's a baker who works in the bakery where I always go to buy bread. He is a nice guy and is very interested in my posts in this forum. He follows my solutions in a daily basis. He was once a professor of mathematics, but got bored of teaching and decided to work in the bakery. He advised me to change the form the Airy's equation to \(\displaystyle y'' - (x - 1)y - y = 0\) before I solve it. He did not tell the reason and I did not ask, but I am guessing that the calculations will be easier.

So let us listen to the baker and beat this Airy again.

We have the differential equation:
\(\displaystyle y'' - (x - 1)y - y = 0\)

And
I will let \(\displaystyle y = \sum_{n=0}^{\infty}a_n(x - 1)^n\)

In the next post, we will take the first and the second derivatives of this function \(\displaystyle y\). Be in touch with this thread if you wanna enjoy the show!

 

[logistic_guy]

\(\displaystyle y = \sum_{n=0}^{\infty}a_n(x - 1)^n\)

\(\displaystyle y' = \sum_{n=0}^{\infty}a_n n(x - 1)^{n-1}\)

\(\displaystyle y'' = \sum_{n=0}^{\infty}a_n n(n-1)(x - 1)^{n-2}\)

 

[khansaheb]

\(\displaystyle y' = \sum_{n=0}^{\infty}a_n n(x - 1)^{n-1}\) INCORRECT

 

[logistic_guy]

\(\displaystyle y' = \sum_{n=0}^{\infty}a_n n(x - 1)^{n-1}\)INCORRECT

You're stealing my brand

\(\displaystyle y'' - (x - 1)y - y = 0\)

Now we substitute what we got in the differential equation above.

\(\displaystyle \sum_{n=2}^{\infty}a_n n(n-1)(x - 1)^{n-2} - (x - 1)\sum_{n=0}^{\infty}a_n(x - 1)^n - \sum_{n=0}^{\infty}a_n(x - 1)^n = 0\)

 

[logistic_guy]

Now we will do trick \(\displaystyle 47\). What's that? It's the hitman style \(\displaystyle \rightarrow\) we change any index \(\displaystyle n\) that is not starting at \(\displaystyle 0\) to \(\displaystyle 0\).

\(\displaystyle \sum_{n=0}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n} - \sum_{n=0}^{\infty}a_n(x - 1)^{n+1} - \sum_{n=0}^{\infty}a_n(x - 1)^n = 0\)

The fun has just started!

 

[logistic_guy]

We have solved the index problem, but a new problem popped up. The first and third terms contain \(\displaystyle (x - 1)^n\) while the second term contains \(\displaystyle (x - 1)^{n+1}\). They don't match.

We can do this for now:

\(\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^{n+1} = \sum_{n=1}^{\infty}a_{n-1}(x - 1)^{n}\)

 

[logistic_guy]

\(\displaystyle \sum_{n=0}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n} = 2a_2 + \sum_{n=1}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n}\)

And

\(\displaystyle \sum_{n=0}^{\infty}a_n(x - 1)^n = a_0 + \sum_{n=1}^{\infty}a_n(x - 1)^n\)

Then, we have:

\(\displaystyle 2a_2 + \sum_{n=1}^{\infty}a_{n+2} (n + 2)(n + 1)(x - 1)^{n} - \sum_{n=1}^{\infty}a_{n-1}(x - 1)^{n} - a_0 - \sum_{n=1}^{\infty}a_n(x - 1)^n = 0\)

 

[logistic_guy]

Let us simplify it further.

\(\displaystyle 2a_2 - a_0 + \sum_{n=1}^{\infty}\left[a_{n+2} (n + 2)(n + 1) - a_{n-1} - a_n\right](x - 1)^n = 0\)

 

[fresh_42]

Why do you think [imath] y' [/imath] is incorrect?

 

[khansaheb]

Why do you think [imath] y' [/imath] is incorrect?

Differentiate term by term and you will see.

 

[fresh_42]

Differentiate term by term and you will see.

I don't see any mistakes in either derivation. The factor [imath] n=0 [/imath] eliminates the negative power.

 

[logistic_guy]

Let us simplify it further.

\(\displaystyle 2a_2 - a_0 + \sum_{n=1}^{\infty}\left[a_{n+2} (n + 2)(n + 1) - a_{n-1} - a_n\right](x - 1)^n = 0\)

This gives us:

\(\displaystyle 2a_2 - a_0 = 0\)

And

\(\displaystyle a_{n+2} (n + 2)(n + 1) - a_{n-1} - a_n = 0\)

Or

\(\displaystyle a_2 = \frac{a_0}{2}\)

And

\(\displaystyle a_{n+2} = \frac{a_{n-1} - a_n}{(n + 2)(n + 1)}\)

It seems that we have a recursive formula in here