SATHEE: Differential Equations 3 Question 13

Differential Equations 3 Question 13

13.

A curve passing through the point $(1, 1)$ has the property that the perpendicular distance of the origin from the normal at any point $P$ of the curve is equal to the distance of $P$ from the $X$ -axis. Determine the equation of the curve.

(1999, 10M)

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Answer:

Correct Answer: 13. ( $x^{2} + y^{2} = 2 x$ )

Solution:

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

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

Distance of perpendicular from the origin to Eq. (i)

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

Also, distance between $P$ and $X$ -axis is $| y |$ .

$\begin{aligned} ∴ \frac{| y + \frac{d x}{d y} \cdot x |}{\sqrt{1 + (\frac{d x}{d y})^{2}}} = | y | \\ \Rightarrow y^{2} + \frac{d x}{d y} \cdot x^{2} + 2 x y \frac{d x}{d y} = y^{2} [1 + (\frac{d x}{d y})^{2}] \\ \Rightarrow (\frac{d x^{2}}{d y}) (x^{2} - y^{2}) + 2 x y \frac{d x}{d y} = 0 \\ \Rightarrow \frac{d x}{d y} [\frac{d x}{d y} (x^{2} - y^{2}) + 2 x y = 0] \\ \Rightarrow \frac{d x}{d y} = 0 or \frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y} \\ But \frac{d x}{d y} = 0 \end{aligned}$

$\Rightarrow x = c$ , where $c$ is a constant.

Since, curve passes through $(1, 1)$ , we get the equation of the curve as $x = 1$ .

The equation $\frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y}$ is a homogeneous equation.

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

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

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

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

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

$\Rightarrow \log | x | (v^{2} + 1) = c_{1} \Rightarrow | x | (\frac{y^{2}}{x^{2}} + 1) = e^{c_{1}}$

$\Rightarrow x^{2} + y^{2} = \pm e^{c_{1}} x$ or $x^{2} + y^{2} = \pm e^{c} x$ is passing through $(1, 1)$ .

$\begin{aligned} ∴ & 1 + 1 & = \pm e^{c} \cdot 1 \\ \Rightarrow & \pm e^{c} & = 2 \end{aligned}$

Hence, required curve is $x^{2} + y^{2} = 2 x$ .

Differential Equations 3 Question 13

13.

Answer:

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Differential Equations 3 Question 13

13.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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