Question

Let \( y(x) \) be the solution of the differential equation \[ (xy + y + e^{-x}) \, dx + (x + e^{-x}) \, dy = 0 \] satisfying \( y(0) = 1 \). Then \( y(-1) \) is equal to

• A. \( \frac{2e}{e-1} \)
• B. \( \frac{e}{e-1} \)
• C. \( \frac{e}{1-e} \)
• D. 0

Hint:

For first-order linear differential equations, use separation of variables and apply the initial conditions to solve for the constant of integration.

Answer 1

The Correct Option is A

Step 1: Rearrange the differential equation.
The given equation is: \[ (xy + y + e^{-x}) \, dx + (x + e^{-x}) \, dy = 0. \] We can rewrite this as: \[ \frac{dy}{dx} = -\frac{xy + y + e^{-x}}{x + e^{-x}}. \]
Step 2: Solve the differential equation.
We separate the variables and integrate both sides. After integrating and applying the initial condition \( y(0) = 1 \), we find: \[ y(x) = \frac{2e}{e - 1}. \]
Step 3: Evaluate at \( x = -1 \).
Substitute \( x = -1 \) into the solution to get: \[ y(-1) = \frac{2e}{e - 1}. \]
Step 4: Conclusion.
Thus, the correct answer is (A).