# Thread: Tricky problem involving 2nd order ODE

1. ## Tricky problem involving 2nd order ODE

Hi, my prof gave us this problem in my ODE course. I have no idea how to solve it so help would be very much appreciated. The question is as follows:

A curved mirror of equation y=y(x) (with respect to some cartesian coordinate frame) has the property that whenever a ray of light emanates from the origin it reflects parallel to the x-axis. Find the equation of the mirror.

Thanks a lot!

2. Here are some ideas. Pick a point on your curve (x,y(x)). You can find the slope of the line that passes through the origin and your point. You also know that the slope of the horizontal line is zero. At the point you picked on your curve you can find the slope of the tangent line. There is an angle between your first curve and your tangent line (angle of incidence) and an angle between your tangent line and your horizontal line (angle of reflection) and these are equal. By relating your angles to your slopes you'll find a differential eqution for y(x).

3. If You observe the figure...

... on the basis of geometrical considerations You can extablish that, defining $BFC= \alpha$ , is $FCA=FDC= \frac{\alpha}{2}$.Now if $x$ and $y$ are the coordinates of C is...

$\displaystyle \sin \alpha = \frac{x}{\sqrt{x^{2}+ y^{2}}}$

$\displaystyle \cos \alpha = \frac{y}{\sqrt{x^{2}+ y^{2}}}$ (1)

... so that the DE of the 'confocal mirror' is...

$\displaystyle y^{'} = \tan \frac{\alpha}{2} = \frac{1-\cos \alpha}{\sin \alpha} = \sqrt{1+\frac{y^{2}}{x^{2}}} - \frac{y}{x}$ (2)

The first order DE of the type (2) was resolved in the eighteenth century by the Italian mathematician Gabriele Manfredi and is commonly called 'homogeneous'. The standard procedure consists in the change of variables $\frac{y}{x}= t \implies dy= t\ dx + x\ dt$, that permits to write the (2) as...

$\displaystyle \frac{dt}{\sqrt{1+t^{2}}-2\ t}= \frac{dx}{x}$ (3)

The problem now is solving (3) and that will be done in a succesive post...

Kind regards

$\chi$ $\sigma$