Question

The general solution of the differential equation
\(\sqrt{1+x^2+y^2+x^2y^2}+xy\dfrac{dy}{dx}=0\) is :
(where C is a constant of integration)

• A. \(\sqrt{1+y^2}+\sqrt{1+x^2}=\frac12\log_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+C\)
• B. \(\sqrt{1+y^2}-\sqrt{1+x^2}=\frac12\log_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+C\)
• C. \(\sqrt{1+y^2}+\sqrt{1+x^2}=\frac12\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C\)
• D. \(\sqrt{1+y^2}-\sqrt{1+x^2}=\frac12\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C\)

Hint:

Note \(1+x^2+y^2+x^2y^2=(1+x^2)(1+y^2)\); separate variables to get \(\frac{y\,dy}{\sqrt{1+y^2}}=-\frac{\sqrt{1+x^2}}{x}dx\) and substitute \(t=\sqrt{1+x^2}\).

Answer 1

The Correct Option is C

\(1+x^2+y^2+x^2y^2=(1+x^2)(1+y^2)\), so \(\sqrt{(1+x^2)(1+y^2)}+xy\dfrac{dy}{dx}=0\).
Separate variables: \(\dfrac{y\,dy}{\sqrt{1+y^2}}=-\dfrac{\sqrt{1+x^2}}{x}\,dx\).
LHS integrates to \(\sqrt{1+y^2}\).
RHS: put \(t=\sqrt{1+x^2}\), so \(x\,dx=t\,dt\) and \(dx/x=\dfrac{t\,dt}{t^2-1}\). Then \(\int\dfrac{\sqrt{1+x^2}}{x}dx=\int\dfrac{t^2}{t^2-1}dt=\int\left[1+\dfrac12\left(\dfrac1{t-1}-\dfrac1{t+1}\right)\right]dt=t+\dfrac12\log_e\dfrac{t-1}{t+1}\).
So \(\sqrt{1+y^2}=-\sqrt{1+x^2}-\dfrac12\log_e\dfrac{t-1}{t+1}+C\).
Hence \(\sqrt{1+y^2}+\sqrt{1+x^2}=\dfrac12\log_e\left(\dfrac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C\).
Answer: (3) \(\sqrt{1+y^2}+\sqrt{1+x^2}=\frac12\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C\)