First, separate the variables:

\(\frac{dy}{dx} = e^{2x+y} \Rightarrow e^{-y} dy = e^{2x} dx\).

Integrate both sides:

\(\int e^{-y} \, dy = \int e^{2x} \, dx\).

This gives:

\(-e^{-y} = \frac{1}{2} e^{2x} + C\).

Use the initial condition \(y = 0\) when \(x = 0\):

\(-e^{0} = \frac{1}{2} e^{0} + C \Rightarrow -1 = \frac{1}{2} + C \Rightarrow C = -\frac{3}{2}\).

Substitute back:

\(-e^{-y} = \frac{1}{2} e^{2x} - \frac{3}{2}\).

Rearrange to solve for \(y\):

\(e^{-y} = \frac{3}{2} - \frac{1}{2} e^{2x}\).

\(-y = \ln\left(\frac{2}{3 - e^{2x}}\right)\).

Thus, \(y = \ln\left(\frac{2}{3 - e^{2x}}\right)\).