Answered: "Is there any known method to solve such..."

Minerva Kline

Answered question

2021-11-20

Is there any known method to solve such second order non-linear differential equation?

y_{n} - n x \frac{1}{\sqrt{y_{n}}} = 0

Answer & Explanation

Jeffrey Jordon

Expert2021-11-25Added 2605 answers

Let ${\begin{cases} u = \frac{x^{3}}{y_{n}^{\frac{3}{2}}} \\ v = \frac{x}{y_{n}} \frac{d y_{n}}{d x} \end{cases}$
Then
$\frac{d v}{d u} = \frac{\frac{d v}{d x}}{\frac{d u}{d x}}$
$= \frac{\frac{x}{y_{n}} \frac{d^{2} y_{n}}{d x^{2}} + \frac{1}{y_{n}} \frac{d y_{n}}{d x} - \frac{x}{y_{n}^{2}} (\frac{d y_{n}}{d x})^{2}}{\frac{3 x^{2}}{y_{n}^{\frac{3}{2}}} - \frac{3 x^{3}}{2 y_{n}^{\frac{5}{2}}} \frac{d y_{n}}{d x}}$
$= \frac{\frac{x}{y_{n}} \frac{d^{2} y_{n}}{d x^{2}} + \frac{v}{x} - \frac{v^{2}}{x}}{\frac{3 u}{x} - \frac{3 u v}{2 x}}$
$= \frac{\frac{x^{2}}{y_{n}} \frac{d^{2} y_{n}}{d x^{2}} + v - v^{2}}{3 u (1 - \frac{v}{2})}$
$3 u (1 - \frac{v}{2}) \frac{d v}{d u} = \frac{x^{2}}{y_{n}} \frac{d^{2} y_{n}}{d x^{2}} + v - v^{2}$
$\frac{x^{2}}{y_{n}} \frac{d^{2} y_{n}}{d x^{2}} = 3 u (1 - \frac{v}{2}) \frac{d u}{d v} + v^{2} - v$
$\frac{d^{2} y_{n}}{d x^{2}} = \frac{y_{n}}{x^{2}} (3 u (1 - \frac{v}{2}) \frac{d v}{d u} + v^{2} - v)$
$\frac{y_{n}}{x^{2}} (3 u (1 - \frac{v}{2}) \frac{d v}{d u} + v^{2} - v) - \frac{n x}{\sqrt{y_{n}}} = 0$
$\frac{y_{n}}{x^{2}} (3 u (1 - \frac{v}{2}) \frac{d v}{d u} + v^{2} - v) = \frac{n x}{\sqrt{y_{n}}}$
$3 u (1 - \frac{v}{2}) \frac{d v}{d u} + v^{2} - v = \frac{n x^{3}}{y_{n}^{\frac{3}{2}}}$
$3 u (1 - \frac{v}{2}) \frac{d v}{d u} + v^{2} - v = n u$
$3 u (\frac{v}{2} - 1) \frac{d v}{d u} = v^{2} - v - n u$
Let $w = \frac{w}{2} - 1$
Let $w = \frac{v}{2} - 1$
Then $v = 2 w + 2$
$\frac{d v}{d u} = 2 \frac{d w}{d u}$
$6 u w \frac{d w}{d u} = 4 w^{2} + 6 w + 2 - n u$

user_27qwe

Skilled2021-11-29Added 375 answers

This is indeed a Emden-Fowler equation as found at EqWorld.
Polyanin provides a parametric solution to $y^{″} = n x y^{- 1 / 2}$ as (2.3.1.11) in his book 'Exact Solutions for Ordinary Differential Equations' :
$x (t) = a C_{1} e^{- t} (e^{3 t} + C_{2} \sin (\sqrt{3} t))$
$y (t) = b (C_{1})^{2} e^{- 2 t} (2 e^{3 t} - C_{2} \sin (\sqrt{3} t) + \sqrt{3} C_{2} \cos (\sqrt{3} t))^{2}$
with $n = 16 a^{- 3} b^{\frac{3}{2}}$
In fact there are more constants than necessary and one of the constants $a, b, C_{1}$ may be set to 1 without restricting the solutions. To verify this set $b := d^{2}$ getting $x (t) = (a C_{1})^{2} . . ., y (t) = (d C_{1})^{2}$ with constraint $n = 16 (\frac{d}{a})^{3}$ so that d and a are defined up to a multiplicative constant even without $C_{1}$
To verify that the solution provided 'works' we may evaluate :
$\frac{d y}{d x} = \frac{y^{'} (t)}{x^{'} (t)} = \frac{4 C_{1} b}{a} e^{- t} (2 e^{3 t} - C_{2} \sin (\sqrt{3} t) - \sqrt{3} C_{2} \cos (\sqrt{3} t))$
$\frac{d \frac{d y}{d x}}{d x} = \frac{16 b}{a^{2}} \frac{e^{3 t} + C_{2} \sin (\sqrt{3} t)}{2 e^{3 t} - C_{2} \sin (\sqrt{3} t) + \sqrt{3} C_{2} \cos (\sqrt{3} t)}$
that we may compare with $\frac{x (t)}{\sqrt{y (t)}}$
Of course a detailed proof would be better.

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