The Solution of d y d x = y+x^2-y^2 2 is...... - Vidyadip

MCQ

The Solution of $\frac{d y}{d x}=\frac{y+\sqrt{x^2-y^2}}{2}$ is......

A
$\sin ^{-1}\left(\frac{y}{x}\right)=2 \log |x|+c$
✓
$\sin ^{-1}\left(\frac{y}{x}\right)=\log |x|+c$
C
$\sin \left(\frac{x}{y}\right)=\log |x|+c$
D
$\sin \left(\frac{y}{x}\right)=\log |y|+c$

✓

Answer

Correct option: B.

$\sin ^{-1}\left(\frac{y}{x}\right)=\log |x|+c$

$\sin ^{-1}\left(\frac{y}{x}\right)=\log |x|+c$
Hint :
$\frac{d y}{d x}=\frac{y+\sqrt{x^2-y^2}}{x}$
Put $y=v x \quad \therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^2-v^2 x^2}}{x}=v+\sqrt{1-v^2}$
$\therefore x \frac{d v}{d x}=\sqrt{1-v^2}$
$\therefore \int \frac{1}{\sqrt{1-v^2}} d v=\int \frac{1}{x} d x$
$\therefore \sin ^{-1} v=\log |x|+c$
$\therefore \sin ^{-1}\left(\frac{y}{x}\right)=\log |x|+c$

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