Question

A particular solution of \( 3e^x \tan y \, dx + (1 - e^x)\sec^2 y \, dy = 0 \) with \( y(1) = \frac{\pi}{4} \) is: 

• A. \( \tan y = \left( \frac{1 - e^x}{1 + e^x} \right)^3 \)
• B. \( \tan y = \left( \frac{1 - e^x}{e^x} \right)^3 \)
• C. \( \tan y = \left( \frac{1 - e^x}{1 - e^x} \right)^3 \)
• D. \( \tan y = \left( \frac{1 - e^x}{1 + e^x} \right)^3 \)

Hint:

When solving separable differential equations, always aim to rearrange the equation such that terms involving \( x \) are on one side and terms involving \( y \) are on the other.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to solve a first-order variable-separable differential equation and find the constant using the given initial condition.
Step 2: Key Formula or Approach:
Rearrange terms to separate \( x \) and \( y \):
\[ \frac{\sec^2 y}{\tan y} \text{ d}y = \frac{-3\text{e}^x}{1 - \text{e}^x} \text{ d}x \] Step 3: Detailed Explanation:
1. Integrate both sides:
\[ \int \frac{\sec^2 y}{\tan y} \text{ d}y = \int \frac{3\text{e}^x}{\text{e}^x - 1} \text{ d}x \] 2. Using substitution \( u = \tan y \Rightarrow \text{d}u = \sec^2 y \text{ d}y \) and \( v = \text{e}^x - 1 \Rightarrow \text{d}v = \text{e}^x \text{ d}x \):
\[ \ln|\tan y| = 3 \ln|\text{e}^x - 1| + \ln \text{C} \] \[ \tan y = \text{C}(\text{e}^x - 1)^3 \] Actually, note that \( \frac{-3\text{e}^x}{1 - \text{e}^x} = \frac{3\text{e}^x}{\text{e}^x - 1} \), which integrates to \( 3\ln|\text{e}^x - 1| \).
3. Apply initial condition \( y(1) = \pi/4 \):
\[ \tan(\pi/4) = \text{C}(\text{e}^1 - 1)^3 \Rightarrow 1 = \text{C}(\text{e} - 1)^3 \] \[ \text{C} = \frac{1}{(\text{e} - 1)^3} \] 4. Substitute \( \text{C} \) back:
\[ \tan y = \frac{(\text{e}^x - 1)^3}{(\text{e} - 1)^3} = \left( \frac{\text{e}^x - 1}{\text{e} - 1} \right)^3 \] Since \( (\text{e}^x - 1) / (\text{e} - 1) = (1 - \text{e}^x) / (1 - \text{e}) \), this matches option (D).
Step 4: Final Answer:
The particular solution is \( \tan y = \left( \frac{1-\text{e}^x}{1-\text{e}} \right)^3 \).