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A friend of mine set the question $\sin(xy) =\dfrac {dy}{dx}$ but after quite a long time I have made no headway (so far I have tried $v=xy$ and the solving it as a $1$st order differential equations and also differentiating and trying to solve as a second order ($3$rd order makes things even messier). If anyone is able to tell me whether this is doable that would be fantastic!

Edit: he said he had a closed form solution but if there was an infinite sum solution that would also be fine

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  • $\begingroup$ i think a numerical method is a good way $\endgroup$ – Dr. Sonnhard Graubner Mar 10 '18 at 20:29
  • $\begingroup$ Interesting equation. Do you expect a closed form solution? $\endgroup$ – Yuriy S Mar 10 '18 at 20:30
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Let us try the substitution the OP mentioned:

$$v=y x$$

Then:

$$y'=\frac{v'}{x}-\frac{v}{x^2}$$

So the equation becomes (assuming $x\neq 0$):

$$\frac{v'}{x}-\frac{v}{x^2}=\sin v$$

Or:

$$v'-\frac{v}{x}=x \sin v$$

It's a nonlinear 1st order equation and it's unlikely to have a closed form solution, however you could search for something similar here http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc1.htm.


I find it useful to consider some asymptotic cases.

1) For $|x| \ll 1$ we have an approximation:

$$v'-\frac{v}{x}=0$$

With a solution:

$$v(x)=c_1 x$$

2) For $|x| \gg 1$ we have an approximation:

$$v'=x \sin v$$

$$\frac{dv}{\sin v}=x dx$$

Integrating we have:

$$\ln \left( \tan \frac{v}{2} \right)=\frac{x^2}{2}+C$$

$$\tan \frac{v}{2}=c_2~ e^{x^2/2}$$

$$v(x)=2 \tan^{-1} \left( c_2 ~e^{x^2/2} \right)$$


Depending on the initial condition we have, we can use one or another approximation.


If an approximation won't do, we'll need to use numerical methods.


Though we might also notice that the asymptotic solutions we have are bounded (since $v_1$ is small for small $x$ and $v_2$ approaches a constant for large $x$).

Which means we can search for another approximation by expanding $\sin v$ as a series.

3) For $|v| \ll 1$ we have a linear equation:

$$v'-\frac{v}{x}=x v$$

With the solution:

$$v(x)=c_3 x e^{x^2/2}$$

4) The next approximation will be nonlinear:

$$v'-\frac{v}{x}=x v-x \frac{v^3}{6}$$

This equation also has an exact solution (I've got it with Wolfram Alpha):

$$v(x)= \pm \frac{ x e^{x^2/2}}{\sqrt{c_4+\frac{1}{6}(x^2-1)e^{x^2}}}$$

We can try to get more approximations for different conditions. Still, I wouldn't expect a general analytic solution, much less a closed form.

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  • $\begingroup$ Complete answer +1 $\endgroup$ – LostInSpace Mar 11 '18 at 0:57
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    $\begingroup$ @Isham, thank you, though I wouldn't call it complete $\endgroup$ – Yuriy S Mar 11 '18 at 1:38
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Hint:

Let $u=xy$ ,

Then $y=\dfrac{u}{x}$

$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$

$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\sin u$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{x^2\sin u+u}{x^2}$

$(x^2\sin u+u)\dfrac{dx}{du}=x$

Let $v=x^2$ ,

$\dfrac{dv}{du}=2x\dfrac{dx}{du}$

$\therefore\dfrac{(x^2\sin u+u)}{2x}\dfrac{dv}{du}=x$

$(x^2\sin u+u)\dfrac{dv}{du}=2x^2$

$(v\sin u+u)\dfrac{dv}{du}=2v$

Let $w=v+u\csc u$ ,

Then $v=w-u\csc u$

$\dfrac{dv}{du}=\dfrac{dw}{du}+(u\cot u-1)\csc u$

$\therefore(\sin u)w\left(\dfrac{dw}{du}+(u\cot u-1)\csc u\right)=2(w-u\csc u)$

$(\sin u)w\dfrac{dw}{du}+(u\cot u-1)w=2w-2u\csc u$

$(\sin u)w\dfrac{dw}{du}=(3-u\cot u)w-2u\csc u$

$w\dfrac{dw}{du}=(3\csc u-u\csc u\cot u)w-2u\csc^2u$

This belongs to an Abel equation of the second kind.

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

$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{3\csc u-u\csc u\cot u}{z}-2u\csc^2u$

$\dfrac{dz}{du}=2z^3u\csc^2u+(u\csc u\cot u-3)z^2$

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

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  • $\begingroup$ But is that final equation solvable? $\endgroup$ – Ethan Horsfall Mar 12 '18 at 13:11
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Firstly, let's assume that you wanted to know if a slightly less intimidating ODE did indeed have a solution, you would firstly need a set of initial conditions say, for $dx/dt = f(x)$, $x(t_0)=x_0$. most times solutions will only exist for specific sets of initial conditions. However, your'e asking if this ODE has an explicit general form using the normal analytical methods. No. It won't be doable. most differential equations won't have an analytical solution, you would most likely have to numerically approximate your answer to this one.

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