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$ y_n'' -nx\frac{1}{\sqrt {y_n}}=0$

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

What I tried to solve:

$ 2y_n'y_n''= 2nx\frac{y_n'}{\sqrt {y_n}}$

$ y_n'^2= 4nx\sqrt {y_n}-4n\int\sqrt {y_n} dx$

After that I could not see any way how to proceed.

Please advice what to do to solve the differential equation.

Many thanks

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  • $\begingroup$ from Maple I got: $$ \int^{y(x)} {1 \over \sqrt{4nx \sqrt a - C_1}}da- x - C_2$$ $\endgroup$ – Santosh Linkha Aug 29 '12 at 8:44
  • $\begingroup$ @experimentX : thanks a lot. is the result equal to zero? any idea how to get the result manually? $\endgroup$ – Mathlover Aug 29 '12 at 8:51
  • $\begingroup$ yes it was ... and where were two!! $\endgroup$ – Santosh Linkha Aug 29 '12 at 8:54
  • $\begingroup$ the other was $$\int^{y(x)} -{1 \over \sqrt{4nx \sqrt a - C_1}}da- x - C_2$$ $\endgroup$ – Santosh Linkha Aug 29 '12 at 8:55
  • $\begingroup$ let's try something like this $$ {d^2x \over dy^2} = {\sqrt{y_n} \over nx}$$ $\endgroup$ – Santosh Linkha Aug 29 '12 at 9:00
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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' : $$ \begin{align} x(t)&=a\;C_1\,e^{-t}\left(e^{3t}+C_2 \sin(\sqrt{3}t)\right)\\ y(t)&=b\;(C_1)^2\,e^{-2t}\left(2e^{3t}-C_2 \sin(\sqrt{3}t)+\sqrt{3}C_2 \cos(\sqrt{3}t)\right)^2\\ &\quad\text{with}\ \ n=16\,a^{-3}\,b^{\frac 32} \end{align} $$

(you may find this too at Amazon searching $\ $Axy$\ $ and selecting page 309)

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 $\ x(t)=(a\,C_1)^2\cdots$, $y(t)=(d\,C_1)^2\cdots\ $ with the constraint $n=16\bigl(\frac da\bigr)^3$ 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 : $$\frac {dy}{dx}=\frac {y'(t)}{x'(t)}=\frac {4 C_1 b}a e^{-t}\left(2e^{3t}-C_2 \sin(\sqrt{3}t)-\sqrt{3}C_2 \cos(\sqrt{3}t)\right)$$ $$\frac {d \frac {dy}{dx}}{dx}=\frac {16 b}{a^2} \frac{e^{3t}+C_2\sin(\sqrt{3}t)}{2e^{3t}-C_2 \sin(\sqrt{3}t)+\sqrt{3}C_2 \cos(\sqrt{3}t} $$ that we may compare with $\dfrac {x(t)}{\sqrt{y(t)}}$ (the ratio should be $\frac {16\,b^{3/2}}{a^3}$)

Of course a detailed proof would be better.

Hoping this helped anyway,

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In fact it belongs to an Emden-Fowler equation.

First, according to http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf or http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=6 , all Emden-Fowler equations can be transformed into Abel equation of the second kind.

Let $\begin{cases}u=\dfrac{x^3}{y_n^\frac{3}{2}}\\v=\dfrac{x}{y_n}\dfrac{dy_n}{dx}\end{cases}$ ,

Then $\dfrac{dv}{du}=\dfrac{\dfrac{dv}{dx}}{\dfrac{du}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{1}{y_n}\dfrac{dy_n}{dx}-\dfrac{x}{y_n^2}\left(\dfrac{dy_n}{dx}\right)^2}{\dfrac{3x^2}{y_n^\frac{3}{2}}-\dfrac{3x^3}{2y_n^\frac{5}{2}}\dfrac{dy_n}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{v}{x}-\dfrac{v^2}{x}}{\dfrac{3u}{x}-\dfrac{3uv}{2x}}=\dfrac{\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2}{3u\left(1-\dfrac{v}{2}\right)}$

$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}=\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2$

$\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}=3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v$

$\dfrac{d^2y_n}{dx^2}=\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)$

$\therefore\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)-\dfrac{nx}{\sqrt{y_n}}=0$

$\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)=\dfrac{nx}{\sqrt{y_n}}$

$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=\dfrac{nx^3}{y_n^\frac{3}{2}}$

$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=nu$

$3u\left(\dfrac{v}{2}-1\right)\dfrac{dv}{du}=v^2-v-nu$

Let $w=\dfrac{v}{2}-1$ ,

Then $v=2w+2$

$\dfrac{dv}{du}=2\dfrac{dw}{du}$

$\therefore6uw\dfrac{dw}{du}=(2w+2)^2-(2w+2)-nu$

$6uw\dfrac{dw}{du}=4w^2+6w+2-nu$

$w\dfrac{dw}{du}=\dfrac{2w^2}{3u}+\dfrac{w}{u}+\dfrac{2-nu}{6u}$

In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $w=\dfrac{1}{z}$ ,

Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$

$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2}{3uz^2}+\dfrac{1}{uz}+\dfrac{2-nu}{6u}$

$\dfrac{dz}{du}=\dfrac{(nu-2)z^3}{6u}-\dfrac{z^2}{u}-\dfrac{2z}{3u}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2.

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