Question

The general solution of $x\sqrt{1+y^{2}}\,dx + y\sqrt{1+x^{2}}\,dy = 0$ is

• A. $\sin^{-1}x + \sin^{-1}y = c$
• B. $x^{2} + y^{2} = \sqrt{1+x^{2}} + \sqrt{1+y^{2}} + c$
• C. $\sqrt{1+x^{2}} + \sqrt{1+y^{2}} = c$
• D. $\tan^{-1}x - \tan^{-1}y = c$

Hint:

Standard integral: $\displaystyle\int \frac{x}{\sqrt{1+x^2}}\,dx = \sqrt{1+x^2} + C$. Recognise this pattern in separation-of-variables problems.

Answer 1

The Correct Option is C

Step 1: Conceptual Understanding:
Separate the variables and integrate both sides.
Step 2: Explanation in Detail:
Divide by $\sqrt{(1+x^2)(1+y^2)}$: $\dfrac{x}{\sqrt{1+x^2}}dx + \dfrac{y}{\sqrt{1+y^2}}dy = 0$.
Integrate: $\sqrt{1+x^2} + \sqrt{1+y^2} = c$ \quad (using $\int \dfrac{x}{\sqrt{1+x^2}}dx = \sqrt{1+x^2}$).
Step 3: Therefore, Stating the Final Answer
General solution: $\sqrt{1+x^2} + \sqrt{1+y^2} = c$.