y” + (1-x) y= 0 (Part II)

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The first terms of the equation presented in the previous post are:

$\displaystyle a_0''(x) + a_0(x) - x a_0(x)\epsilon$

$\displaystyle + a_1''(x)\epsilon + a_1(x)\epsilon - x a_1(x)\epsilon^2$

$\displaystyle + a_2''(x)\epsilon^2 + a_2(x)\epsilon^2 - x a_2(x)\epsilon^3 + \cdots$

the differential equation can be written as:

$\displaystyle a_0''(x) + a_0(x) + (a_1''(x) + a_1(x) - x a_0(x))\epsilon + (a_2''(x) + a_2(x) - x a_1(x))\epsilon^2 + \cdots = 0$

$\displaystyle a_0''(x) + a_0(x) = 0$

$\displaystyle a_1''(x) + a_1(x) - x a_0(x) = 0$

$\displaystyle a_2''(x) + a_2(x) - x a_1(x) = 0$

This implies

$\displaystyle a_0(x) = \cos(x)$

$\displaystyle a_1(x) = \frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x)$

We can check that $a_1(x)$ is correct:

$\displaystyle a_1(x) = \frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x)$

$\displaystyle a_1'(x) = \frac{1}{4}x^2 \cos(x) + \frac{1}{2} x \sin(x) + \frac{1}{4} \cos(x) - \frac{1}{4} x \sin(x) - \frac{1}{4} \cos(x)$

$\displaystyle a_1'(x) = \frac{1}{4}x^2 \cos(x) + \frac{1}{4} x \sin(x)$

$\displaystyle a_1''(x) = \frac{1}{2}x \cos(x) - \frac{1}{4} x^2 \sin(x) + \frac{1}{4} \sin(x) + \frac{1}{4} x \cos(x)$

We can check that the condition:

$\displaystyle a_1''(x) + a_1(x) - x a_0(x) = 0$

is fulfilled. Therefore:

$\displaystyle y_{\epsilon}(x) = \cos(x) + \left(\frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x)\right) \epsilon + \cdots$

Setting $\epsilon = 1$ gives the first-order perturbative approximation to the corresponding solution of the Airy equation.

$\displaystyle y(x) = \cos(x) + \frac{1}{4}x^2 \sin(x) + \frac{1}{4} x \cos(x) - \frac{1}{4} \sin(x) + \cdots$

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