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

Find the equation of a curve such that the projection of its ordinate upon the normal is equal to its. abscissa.

Accepted Answer

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

Ordinate $= P M$

Projection of ordinate on normal $= P N$

$∴ P N = P M \cos θ = x$

$\Rightarrow \frac{y}{\sec θ} = x$

$\Rightarrow \frac{y}{\sqrt{1 + \tan^{2} θ}} = x$

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

$\Rightarrow \frac{y^{2}}{x^{2}} = 1 + {(\frac{d y}{d x})}^{2}$

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

Put $y = v x \Rightarrow v + x \frac{d v}{d x}$

Then, by $(i)$ , we have

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

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

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

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

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

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

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

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

Find the equation of a curve such that the projection of its ordinate upon the normal is equal to its. abscissa.

Important Questions on Differential Equations

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