Question

The general solution of the differential equation \[ (1 + y^2) \, dx = ( \tan^{-1} y - x ) \, dy \] is:

• A. \( x = \tan^{-1} y - 1 + c e^{\tan^{-1} y} \)
• B. \( x = \tan^{-1} y - 1 + c e^{-\tan^{-1} y} \)
• C. \( x = \tan^{-1} y + c e^{\tan^{-1} y} \)
• D. \( x = c \tan^{-1} y + e^{-\tan^{-1} y} \)

Hint:

When solving first-order differential equations using the method of integrating factors, always check if the equation can be rewritten as an exact differential. This allows for easy integration and obtaining the general solution.

Answer 1

The Correct Option is B

We are given the differential equation: \[ (1 + y^2) \, dx = ( \tan^{-1} y - x ) \, dy \] Step 1: Rearrange the terms. Rearrange the equation to separate the variables \( x \) and \( y \): \[ (1 + y^2) \, dx + x \, dy = \tan^{-1} y \, dy \] Now we will solve this using the method of integrating factors. Step 2: Use the method of integrating factors. To solve the equation, we look for an integrating factor. We multiply through by \( \frac{1}{1 + y^2} \): \[ dx + \frac{x}{1 + y^2} \, dy = \frac{\tan^{-1} y}{1 + y^2} \, dy \] Now, the left-hand side is an exact differential, and we can integrate both sides with respect to \( y \). Step 3: Integrate both sides. The integral of \( dx \) is simply \( x \), and the integral of \( \frac{x}{1 + y^2} \, dy \) is \( \tan^{-1} y \). On the right side, the integral of \( \frac{\tan^{-1} y}{1 + y^2} \) is: \[ \int \frac{\tan^{-1} y}{1 + y^2} \, dy = e^{-\tan^{-1} y} \] Step 4: General solution. We now combine the results to get the general solution: \[ x = \tan^{-1} y - 1 + c e^{-\tan^{-1} y} \] Thus, the general solution is: \[ x = \tan^{-1} y - 1 + c e^{-\tan^{-1} y} \] Therefore, the correct answer is option (B)