Question

The solution of the differential equation \( \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0 \) is

• A. \( \cos^{-1}x + \cos^{-1}y = c \)
• B. \( \sin^{-1}x + \sin^{-1}y = c \)
• C. \( \cosh^{-1}x + \cosh^{-1}y = c \)
• D. \( \sinh^{-1}x + \sinh^{-1}y = c \)

Hint:

Recognize the standard integrals for inverse trigonometric functions to simplify integrals involving square roots like \( \sqrt{1 - x^2} \).

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0 \] Step 1: Rearrange the equation
Bring the second term to the right-hand side: \[ \frac{dy}{dx} = -\sqrt{\frac{1 - y^2}{1 - x^2}} \] Step 2: Separate variables
\[ \frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}} \] Step 3: Integrate both sides
Recall that: \[ \int \frac{1}{\sqrt{1 - z^2}}\,dz = \sin^{-1}z \] So we integrate: \[ \int \frac{1}{\sqrt{1 - y^2}}\,dy = - \int \frac{1}{\sqrt{1 - x^2}}\,dx \] Step 4: Solve the integrals
\[ \sin^{-1}y = -\sin^{-1}x + c \] Step 5: Rearranging terms
\[ \sin^{-1}x + \sin^{-1}y = c \] This is the required general solution.