Has anybody got a cleaner solution?

[macilrae]

Attempting to solve

dx/dy = x/y - √{1 + (x/y)^2}

I can substitute x/y = z and get

ln(y/c) = arcsinh(z) c is constant of integration

or putting x/y = tan(θ)

I get

ln(y/c) = ln(sec(θ) + tan(θ))

both of which do give a parabola but the interim logs seem so messy and unnecessary.
Is there a better more direct substitution?

 

[JJacquelin]

Hi !
Let x = y sinh(t)
dx/dy = sinh(t) + y cosh(t) (dt/dy)
which leads to a very simple ODE

 

[macilrae]

Thanks, JJaquelin - that's certainly a fast and ingenious idea - you still bump into the Sinh(ln()) thingy which I had hoped to avoid - probably I'm too picky.
The equation arose from asking what curve, symmetric about the x axis, will always reflect back a right-to-left ray, parallel to x axis, through the same point (parabolic reflector - focal point).

 

[AlephZero]

A trig substitution is the "natural" way to solve this if you formulate the DE using geometry. Putting the fixed point at the origin, the equations for the directon of the reflected ray and the tangent to the curve are

dy/dx = cot (t/2)
tan t = y/x

 

[JJacquelin]

sinh(ln(c*y)) = (c*y-(1/(c*y))/2
So, there is no hyperbolic function, nor trigonometric function, in the final result : x = function of y , which is a very simple parabolic function.

 

[macilrae]

AlephZero - Right, those were the starting equations & they look so simple - I don't begrudge a log popping up, but when I immediately have to antilog it, intuitively I think "There has to be a more elegant way!"

 

[macilrae]

JJaquelin - it's the sinh(ln()) that is bugging me! Having to "do and undo" seems redundant - especially because the final expression is the simplest parabola!

 

[Dickfore]

Notice that:
<br /> \mathrm{arcsinh} (x) = \ln \left( x + \sqrt{1 + x^2} \right)<br />

Your equation
<br /> \frac{dx}{dy} = \frac{x}{y} - \sqrt{1 + \left( \frac{x}{y} \right)^2}<br />
after the substitution
<br /> z = \frac{x}{y}, \ z = z(y)<br />
reduces to:
<br /> y \, z&#039; + z = z - \sqrt{1 + z^2}<br />
<br /> \frac{dz}{\sqrt{1 + z^2}}+ \frac{dy}{y} = 0<br />
which integrates to:
<br /> \mathrm{arcsinh}(z) + \ln(y) = C<br />
I think you have the realtive sign opposite to this in your solution. Then, use the above representation of the inverse hyperbolic sine, and take the antilogratihm:
<br /> y \, \left( z + \sqrt{1 +z^2} \right) = C_1<br />
<br /> x + \sqrt{x^2 + y^2} = C_1<br />
<br /> \sqrt{x^2 + y^2} = C_1 - x<br />
Square it:
<br /> x^2 + y^2 = \left(C_1 - x \right)^2<br />
<br /> x^2 + y^2 = C^{2}_{1} - 2 \, x \, C_1 + x^2<br />
<br /> x = \frac{C_1}{2} - \frac{y^2}{2 C_1}<br />

 

[Dickfore]

Another way is to solve it as an implicit equation. Introduce p = dx/dy, and z = x/y. Then:
<br /> p = z - \sqrt{1 + z^2}<br />
<br /> \sqrt{1 + z^2} = z - p<br />
Squaring it gives:
<br /> 1 + z^2 = z^2 - 2 p z + p^2<br />
<br /> z = \frac{p^2 - 1}{2 p}<br />
or
<br /> x = y \, \frac{p^2 - 1}{2 p}<br />
This is an implicit equation of the Lagrange-Clairot type. Differentiate w.r.t. y to get:
<br /> p \, dy = \frac{p^2 - 1}{2 p} \, dy + y \, \frac{2 p \cdot p - (p^2 - 1) \cdot 1}{2 p^2} \, dp<br />
<br /> \frac{2 p^2 - p^2 + 1}{2 p} \, dy = y \, \frac{2 p^2 - p^2 + 1}{2 p^2} \, dp<br />
<br /> \frac{d y}{y} = \frac{d p}{p}<br />
Integrating this equation gives:
<br /> \ln(p) = \ln(y) + C&#039;_1<br />
and taking the antilogarithm gives:
<br /> p = C_1 \, y<br />
Substitute this in the parametric solution for x:
<br /> x = y \, \frac{C^{2}_{1} y^2 - 1}{2 \, C_{1} \, y} = \frac{C_1 \, y^2}{2} - \frac{1}{2 \, C_1}<br />
which is the same general solution as before if you make the substitution C_1 \rightarrow -1/C_1.

But, there might also be a singular solution to the equation. It is obtained as an envelope of the family of general solutions. Differentiate w.r.t. to the arbitrary constant to get:
<br /> 0 = \frac{y^2}{2} + \frac{1}{2 \, C^{2}_{1}}<br />
This sum of non-negative terms can only be zero, if each term is equal to zero separately. This corresponds to:
<br /> y = 0, \ C_1 \rightarrow \infty<br />
To check whether y = 0 is a solution of the original equation, solve it for y&#039; = dy/dx = (dx/dy)^{-1}:
<br /> y&#039; = \frac{1}{x/y - \sqrt{1 + (x/y)^2}} = \frac{y}{x - \mathrm{sgn}(y) \, \sqrt{y^2 + x^2}}<br />
One may argue that it is because of the y in the numerator, or that the original equation is not defined for y = 0.

 

[macilrae]

Thanks to everyone for engaging my problem - seems the way below is the least painful but certainly not the most creative (thanks Dickfore, I had to look up Lagrange-Clairot!)

The original conditions for the mirror profile were:

dx/dy = -tan(2α) and y/x = tan(α)

which boils down to

dx/dy = x/y ± √{1 + (x/y)^2} ... thanks again for the "+" Dickfore, I'll use that.

using x/y = tan(θ)

dy/y = sec(θ)dθ ... really simple, huh?

but then

ln(y/C) = ln[tan(θ) + sec(θ)]

ending up with

y^2 = 2Cx + C^2 where C/2 is the focal length

for such a trivial result it seemed messy to go in and out of logs but hey

 

[macilrae]

correction

dx/dy = tan(α) and y/x = -tan(2α)

sorry

 

[JJacquelin]

Hi !

if you know an easier way...

 

[macilrae]

JJaquelin: It's just that given the chance I prefer trigonometrics to hyperbolicals - but never call me prejudiced - some of my best fiends are quite hyperbolic ...