Q. 71 - Calculus

Question

Airy’s equation $y^{\text{'}\text{'}}=xy$ was developed to describe the patterns of light that pass through a circular aperture and that are caused by diffraction at the edges of the aperture.

Show that the function 

$y_0\left(x\right)=1+\sum _{k=1}^{\infty }\frac{x^{3k}}{(2·3)(5·6)···\left(\right(3k-1)·3k)}$

is a solution of Airy’s equation. 

Step-by-Step Solution

Answer

These are the same, so $y_0\left(x\right)$ is the solution of the Airy's equation $y\text{'}\text{'}=xy$

To show that the function y 0 ( x ) is the solution of y ' ' = x y , let us first find second derivative of the function.

Therefore,

$$\begin{gathered} y_0\text{'}\left(x\right)=\frac{d}{dx}\left[1+\sum _{k=1}^{\infty }\frac{x^{3k}}{(2·3)(5·6)···\left(\right(3k-1)·3k)}\right] \\ {y}_0\text{'}\left(x\right)=\frac{d}{dx}\left[1\right]+\sum _{k=1}^{\infty }\frac{1}{(2·3)(5·6)···\left(\right(3k-1)·3k)}\frac{d}{dx}\left[x^{3k}\right] \\ {y}_0\text{'}\left(x\right)=0+\sum _{k=1}^{\infty }\frac{1}{(2·3)(5·6)···\left(\right(3k-1)·3k)}3k{x}^{3k-1} \end{gathered}$$

Hence,

$y_0\text{'}\left(x\right)=\sum _{k=1}^{\infty }\frac{3k}{(2·3)(5·6)···\left(\right(3k-1)·3k)}{x}^{3k-1}$

Again differentiate y 0 ' ( x ) with respect to x

So,

$$\begin{gathered} y_0\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[\sum _{k=1}^{\infty }\frac{3k}{(2·3)(5·6)···\left(\right(3k-1)·3k)}{x}^{3k-1}\right] \\ {y}_0\text{'}\text{'}\left(x\right)=\sum _{k=1}^{\infty }\frac{3k(3k-1)}{(2·3)(5·6)···\left(\right(3k-1)·3k)}{x}^{3k-2} \end{gathered}$$

Or,

$y_0\text{'}\text{'}\left(x\right)=x+\sum _{k=2}^{\infty }\frac{x^{3k-2}}{(2·3)(5·6)···\left(\right(3k-4)·(3k-3\left)\right)}$

Finally, calculate $x{y}_0\left(x\right)$

So,

$$\begin{gathered} x{y}_0\left(x\right)=x\left[1+\sum _{k=1}^{\infty }\frac{x^{3k}}{(2·3)(5·6)···\left(\right(3k-1)·3k)}\right] \\ x{y}_0\left(x\right)=x+\sum _{k=1}^{\infty }\frac{x^{3k+1}}{(2·3)(5·6)···\left(\right(3k-1)·3k)} \end{gathered}$$

Now, change the index

Therefore,

$$\begin{gathered} x{y}_0\left(x\right)=x+\sum _{k=2}^{\infty }\frac{x^{3(k-1)+1}}{(2·3)(5·6)···\left(3\right(k-1)-1)·3(k-1))} \\ x{y}_0\left(x\right)=x+\sum _{k=2}^{\infty }\frac{x^{3(k-1)+1}}{(2·3)(5·6)···(3k-4)·(3k-3))} \end{gathered}$$