Question

Find a particular solution of the differential equation \( (x + 1) \frac{dy{dx} = 2 e^{-y} - 1 \), given that \( y = 0 \) when \( x = 0 \).}

Hint:

When an integrand has \(e^{-y}\), multiply by \(e^y\) to make it simpler to handle. Always use initial values to evaluate the constant \(C\) immediately after integrating to simplify subsequent steps.

Answer 1

Concept:
• This is a variable-separable differential equation.
• A particular solution is found by first finding the general solution and then using the boundary conditions to find the constant of integration \(C\).

Step 1: Separate the variables
\[ \frac{dy}{2e^{-y} - 1} = \frac{dx}{x + 1} \] Multiply numerator and denominator of LHS by \(e^y\): \[ \frac{e^y \, dy}{2 - e^y} = \frac{dx}{x + 1} \]

Step 2: Integrate both sides
Let \(u = 2 - e^y\), then \(du = -e^y \, dy\). \[ \int \frac{-du}{u} = \int \frac{dx}{x + 1} \] \[ -\log|2 - e^y| = \log|x + 1| + \log C \] \[ \log|x + 1| + \log|2 - e^y| = \log K \] Taking exponents: \[ (x + 1)(2 - e^y) = K \]

Step 3: Apply the initial condition
Given \(y = 0\) when \(x = 0\). \[ (0 + 1)(2 - e^0) = K \] \[ 1 \cdot (2 - 1) = K \implies K = 1 \]

Step 4: Write the final particular solution
\[ (x + 1)(2 - e^y) = 1 \] This can be simplified: \[ 2 - e^y = \frac{1}{x + 1} \implies e^y = 2 - \frac{1}{x + 1} = \frac{2x + 1}{x + 1} \] \[ y = \log \left| \frac{2x + 1}{x + 1} \right| \]