Name: Exact Differential Equations
Uploaded: 2019-07-19
Duration: 14 min 3 s
Description: An exact differential equation is a certain kind of ordinary differential equation that is widely used in mathematics. Watch this video and learn about the e...

Question

The light rays emanating from a point source situated at origin when reflected from the mirror of a search light are reflected as beam parallel to the

x

-axis. Show that the surface is parabolic, by first forming the differential equation and then solving it.

Accepted Answer

Let the curve be $y = f (x)$ and point on curve be $P \equiv (x, y)$ .

We know that from law of reflection that angle of incidence is equal to angle of reflection.

Equation of normal at point $P (x, y)$ is

$Y - y = - \frac{d x}{d y} (X - x)$

$\Rightarrow X (\frac{d x}{d y}) + Y - y - x \frac{d x}{d y} = 0 . . . (i)$

It is meeting $x - axis$ at $N$ ,

$X (\frac{d x}{d y}) + 0 - y - x \frac{d x}{d y} = 0$

$\Rightarrow X = \frac{y + x (\frac{d x}{d y})}{(\frac{d x}{d y})}$

Hence, $N \equiv (\frac{y + x (\frac{d x}{d y})}{(\frac{d x}{d y})}, 0)$ .

In $O P N$ ,

$O P = O N$

$\Rightarrow \sqrt{x^{2} + y^{2}} = \frac{y + x (\frac{d x}{d y})}{(\frac{d x}{d y})}$

$\Rightarrow (\frac{d x}{d y}) \sqrt{x^{2} + y^{2}} = y + x (\frac{d x}{d y})$

$\Rightarrow (\frac{d x}{d y}) (\sqrt{x^{2} + y^{2}} - x) = y$

$\Rightarrow (\frac{d x}{d y}) = \frac{y}{\sqrt{x^{2} + y^{2}} - x}$

$\Rightarrow (\frac{d y}{d x}) = \frac{\sqrt{x^{2} + y^{2}} - x}{y} . . . . (i)$

Put $y = v x \Rightarrow \frac{d y}{d x} = v \times 1 + x \frac{d v}{d x}$ , hence by $(i)$

$v + x \frac{d v}{d x} = \frac{\sqrt{x^{2} + v^{2} x^{2}} - x}{v x}$

$\Rightarrow x \frac{d v}{d x} = \frac{\sqrt{1 + v^{2}} - 1}{v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{\sqrt{1 + v^{2}} - (1 + v^{2})}{v}$

$\Rightarrow \frac{v d v}{\sqrt{1 + v^{2}} - (1 + v^{2})} = \frac{d x}{x}$

$\Rightarrow \int \frac{v d v}{\sqrt{1 + v^{2}} (1 - \sqrt{1 + v^{2}})} = \int \frac{d x}{x}$

$\Rightarrow I_{1} = \int \frac{d x}{x}$

$\Rightarrow I_{1} = \log x + \log C . . . (ii)$

Now,

$I_{1} = \int \{\frac{v}{\sqrt{1 + v^{2}} (1 - \sqrt{1 + v^{2}})}\} d v$

Put $1 + v^{2} = t^{2} \Rightarrow v d v = t d t$

$I_{1} = \int \{\frac{t}{t (1 - t)}\} d t$

$\Rightarrow I_{1} = \int \{\frac{1}{(1 - t)}\} d t$

$\Rightarrow I_{1} = - \log |1 - t|$

$\Rightarrow I_{1} = - \log |1 - \sqrt{1 + v^{2}}|$

$\Rightarrow I_{1} = - \log |1 - \sqrt{1 + \frac{y^{2}}{x^{2}}}|$

$\Rightarrow I_{1} = - \log |\frac{x - \sqrt{x^{2} + y^{2}}}{x}|$

$\Rightarrow I_{1} = - \log |x - \sqrt{x^{2} + y^{2}}| + \log x$

Putting in $(ii)$

$\Rightarrow - \log |x - \sqrt{x^{2} + y^{2}}| + \log x = \log x + \log C$

$\Rightarrow - \log |x - \sqrt{x^{2} + y^{2}}| = \log C$

$\Rightarrow \log |\frac{1}{x - \sqrt{x^{2} + y^{2}}}| = \log C$

$\Rightarrow \frac{1}{x - \sqrt{x^{2} + y^{2}}} = C$

$\Rightarrow x - \sqrt{x^{2} + y^{2}} = K$

$\Rightarrow x^{2} + y^{2} = {(x - K)}^{2}$

$\Rightarrow y^{2} = K^{2} - 2 K x$

This is the equation of parabola.

The light rays emanating from a point source situated at origin when reflected from the mirror of a search light are reflected as beam parallel to the $x$ -axis. Show that the surface is parabolic, by first forming the differential equation and then solving it.

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

The light rays emanating from a point source situated at origin when reflected from the mirror of a search light are reflected as beam parallel to the x-axis. Show that the surface is parabolic, by first forming the differential equation and then solving it.

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

The light rays emanating from a point source situated at origin when reflected from the mirror of a search light are reflected as beam parallel to the $x$ -axis. Show that the surface is parabolic, by first forming the differential equation and then solving it.