Question 3 Find a particular solution of the following differential equation \( (1+2 x) \frac{\mathrm{d} y}{\mathrm{~d} x}-e^{-y}=0 \) satisfying initial condition \( y(0)=0 \)

To solve the differential equation \( (1 + 2x) \frac{dy}{dx} - e^{-y} = 0 \) under the initial condition \( y(0) = 0 \), we can rearrange the equation as follows: \[ (1 + 2x) \frac{dy}{dx} = e^{-y} \] Next, separate the variables: \[ e^y dy = (1 + 2x) dx \] Now, integrate both sides: \[ \int e^y dy = \int (1 + 2x) dx \] The left side gives us \( e^y \), and the right side evaluates to: \[ \int (1 + 2x) dx = x + x^2 + C \] Thus we have: \[ e^y = x + x^2 + C \] Now, take the natural logarithm to solve for \( y \): \[ y = \ln(x + x^2 + C) \] Next, apply the initial condition \( y(0) = 0 \): \[ 0 = \ln(0 + 0 + C) \Rightarrow C = 1 \] So we find: \[ y = \ln(x + x^2 + 1) \] Therefore, the particular solution satisfying the initial condition \( y(0) = 0 \) is: \[ \boxed{y = \ln(x + x^2 + 1)} \]