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) $ $\quad$ …….(i)
Distance of perpendicular from the origin to Eq. (i)
$ =\frac{\left|y+\frac{d x}{d y} \cdot x\right|}{\sqrt{1+(\frac{d x}{d y})^2}} $
Also, distance between $P$ and $X$-axis is $|y|$.
$ \begin{aligned} & \therefore \quad \frac{\left|y+\frac{d x}{d y} \cdot x\right|}{\sqrt{1+(\frac{d x}{d y})^2}}=|y| \\ & \Rightarrow \quad 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 \quad (\frac{d x^{2}}{d y})\left(x^{2}-y^{2}\right)+2 x y \frac{d x}{d y}=0 \\ & \Rightarrow \quad \frac{d x}{d y} \quad [\frac{d x}{d y}\left(x^{2}-y^{2}\right)+2 x y=0] \\ & \Rightarrow \frac{d x}{d y}=0 \quad \text { or } \quad \frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y} \\ & \text { 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 $\quad y=v x \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}$
$ v+x \frac{d v}{d x}=\frac{v^{2} x^{2}-x^{2}}{2 x^{2} v} $
$\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v=\frac{v^{2}-1-2 v^{2}}{2 v}=-\frac{v^{2}+1}{2 v}$
$\Rightarrow \quad \frac{-2 v}{v^{2}+1} d v=\frac{d x}{x}$
$\Rightarrow \quad c _1-\log \left(v^{2}+1\right)=\log |x|$
$\Rightarrow \log |x|\left(v^{2}+1\right)=c _1 \quad \Rightarrow|x| (\frac{y^{2}}{x^{2}}+1)=e^{c _1}$
$\Rightarrow x^{2}+y^{2}= \pm e^{c _1} x$ or $x^{2}+y^{2}= \pm e^{c} x$ is passing through $(1,1)$.
$ \begin{aligned} & \therefore & 1+1 & = \pm e^{c} \cdot 1 \\ \Rightarrow & & \pm e^{c} & =2 \end{aligned} $
Hence, required curve is $x^{2}+y^{2}=2 x$.
