Question

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

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

Hint:

For first-order linear differential equations, calculate the integrating factor and solve the equation step by step.

Answer 1

The Correct Option is A

Rearrange the given equation as follows: \[ \sqrt{1 - y^2} \, dx = (\sin^{-1} y - x) \, dy. \]
Next, divide both sides by \( \sqrt{1 - y^2} \): \[ dx = \frac{\sin^{-1} y - x}{\sqrt{1 - y^2}} \, dy. \]
Let \( P(y) = \frac{1}{\sqrt{1 - y^2}} \) and \( Q(y) = \frac{\sin^{-1} y}{\sqrt{1 - y^2}} \). Step 1: Compute the integrating factor (IF)
The integrating factor is given by: \[ \text{IF} = e^{\int P(y) \, dy} = e^{\sin^{-1} y}. \] Step 2: Solve the differential equation
Multiply through by the integrating factor \( \text{IF} \): \[ e^{\sin^{-1} y} \, dx + e^{\sin^{-1} y} \frac{x}{\sqrt{1 - y^2}} \, dy = e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}} \, dy. \]
This simplifies to: \[ \frac{d}{dy} \left( x e^{\sin^{-1} y} \right) = e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}}. \]
Now, integrate both sides: \[ x e^{\sin^{-1} y} = \int e^{\sin^{-1} y} \frac{\sin^{-1} y}{\sqrt{1 - y^2}} \, dy + c. \]
Simplifying further, we obtain: \[ x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}. \] Final Answer: \[ \boxed{x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}}. \]