First slide

Introduction to Differential equations

Question

Let y=y(x) be a function of x satisfying $y \sqrt{1 - x^{2}} = k - x \sqrt{1 - y^{2}}$ where k is a constant $y (\frac{1}{2}) = - \frac{1}{4}$ and . Then $\frac{d y}{d x}$ at $x = \frac{1}{2}$ is equal to:

Moderate

A

$\frac{2}{\sqrt{5}}$
B

$- \frac{\sqrt{5}}{4}$
C

$- \frac{\sqrt{5}}{2}$
D

$\frac{\sqrt{5}}{2}$

Solution

At $x = \frac{1}{2}, y = \frac{- 1}{4} \Rightarrow x y = \frac{- 1}{8}$ .
Differentiating w.r.t. x, we get
$y . \frac{1. (- 2 x)}{2 \sqrt{1 - x^{2}}} + y' . \sqrt{1 - x^{2}} = - \{1. \sqrt{1 - y^{2}} + \frac{x . (- 2 y)}{2 \sqrt{1 - y^{2}}} y'\}$
$\Rightarrow - \frac{x y}{\sqrt{1 - x^{2}}} + y' \sqrt{1 - x^{2}} = - \sqrt{1 - y^{2}} + \frac{x y . y'}{\sqrt{1 - y^{2}}}$
$\Rightarrow y' (\sqrt{1 - x^{2}} - \frac{x y}{\sqrt{1 - y^{2}}}) = \frac{x y}{\sqrt{1 - x^{2}}} - \sqrt{1 - y^{2}}$
$\Rightarrow y' (\frac{\sqrt{3}}{2} + \frac{1}{8. \frac{\sqrt{15}}{4}}) = \frac{- 1}{8 \frac{\sqrt{3}}{2}} - \frac{\sqrt{15}}{4}$
$\Rightarrow y' (\frac{\sqrt{45} + 1}{2 \sqrt{15}}) = - \frac{(1 + \sqrt{45})}{4 \sqrt{3}}$
$\Rightarrow y' = {\frac{d y}{d x}}_{x = \frac{1}{2}} = - \frac{\sqrt{5}}{2}$

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