Ricatti differential equation $y'=-13(x-y^{2})$

Question

$$y'=-13(x-y^{2})$$

I'm a beginner. I know, that this is a kind of Riccati equation, but is it possible to solve it with only simple methods? Thank you

p.s. I know nothing about Bessel functions. Please, be explicit

Answer 1

The derivation in the first section of Ricatti uses only elementary but judiciously chosen substitutions (with malice aforethought).

$$ y' = -13 x + 0 y + 13 y^2\\ q_0 = -13 x\\ q_1 = 0\\ q_2 = 13\\ S = q_2 q_0 = -169 x\\ R = q_1 + \frac{q_2'}{q_2} = 0 + \frac{0}{13} = 0\\ v = 13 y\\ v' = v^2 + 0 v + (-169 x)\\ v = \frac{-u'}{u}\\ u'' - 169 x u = 0 $$

You solve for $u$ and then plug in to get $v$ and from there $y$.

How do you solve a differential equation like

$$ y'' - x*y = 0 $$

It is not an elementary function. It is the Airy function.

So getting

$$ u'' - 169 x u = 0\\ t = A x\\ \tilde{u}(t) = u(\frac{t}{A})\\ \frac{d^2}{dt^2}\tilde{u} (t) = \frac{1}{A^2} u'' (\frac{t}{A})\\ \frac{d^2}{dt^2}\tilde{u} (t) = \frac{1}{A^2} 169 \frac{t}{A} \tilde{u} (t)\\ $$

Let $A^3=169$, $A=13^{2/3}$ to give

$$ \frac{d^2}{dt^2}\tilde{u} (t) - t \tilde{u} (t) = 0\\ \tilde{u}(t) = Airy(t)\\ u(x) = \tilde{u}(Ax) = Airy(Ax) $$

Use that to get back to $v$ and $y$.