Question

The general solution of the differential equation $\sqrt{1-x^2{y}^2}\cdot dx=y\cdot dx+x\cdot dy$ is :

• A. $sin\left(xy\right)=x+c$
• B. $si{n}^{-1}\left(xy\right)+x=c$
• C. $sin(x+c)=xy$
• D. $sin\left(xy\right)+x=c$

Answer 1

Answer: (C)

$sin(x+c)=xy$

We have, $\sqrt{1-(xy{)}^2}=y+x\frac{dy}{dx}$

Substitute $xy=u\Rightarrow y+x\frac{dy}{dx}=\frac{du}{dx}$

So above differential equation becomes,

$\sqrt{1-u^2}=\frac{du}{dx}$

$dx=\frac{du}{\sqrt{1-u^2}}$

$x+c={\sin }^{-1}u$

${\sin }^{-1}\left(xy\right)=x+c$

$\sin (x+c)=xy$