Question

Solution of the differential equation $\left(\sqrt{1-x^2 y^2}-y\right) d x=x d y$ is
Where ' $c$ ' is an arbitrary constant.

• A. $y=\frac{\sin (x+c)}{x}$
• B. $x =\frac{\sin ( y + c )}{ y }$
• C. $y=\frac{\cos (y+c)}{x}$
• D. $x=\frac{\tan (x+c)}{y}$

Answer 1

Answer: (A)

$\sqrt{1-x^2 y^2}=x d y+y d x ; \frac{d(x y)}{\sqrt{1-x^2 y^2}}=1 $
$\sin ^{-1}(x y)=x+c$
$\therefore y=\frac{\sin (x+c)}{x}$