Differential Equation:
The given equation is a first-order linear homogeneous differential equation:

dy/dx + exp(x)y = 1

Homogeneous Solution:
To find the homogeneous solution, we set the right-hand side equal to zero:

dy/dx + exp(x)y = 0

This is a separable differential equation. Rearranging the terms, we have:

dy/y = -exp(x)dx

Integrating both sides, we get:

ln|y| = -exp(x) + C

where C is the constant of integration. Taking the exponential of both sides, we have:

|y| = e^(-exp(x) + C)

Since the absolute value can be positive or negative, we rewrite the equation as:

y = ±e^(-exp(x) + C)

Particular Solution:
To find the particular solution, we consider the right-hand side of the original equation:

dy/dx + exp(x)y = 1

This is a non-homogeneous equation, so we need to find a particular solution. We can use the method of undetermined coefficients. Assume a particular solution of the form:

y_p = A

where A is a constant. Substituting this into the equation, we have:

0 + exp(x)A = 1

Solving for A, we get:

A = 1/exp(x)

Thus, the particular solution is:

y_p = 1/exp(x)

General Solution:
The general solution of the differential equation is the sum of the homogeneous and particular solutions:

y = y_h + y_p

Substituting the expressions for y_h and y_p, we have:

y = ±e^(-exp(x) + C) + 1/exp(x)

where C is the constant of integration.

Initial Condition:
The initial condition given is y(0) = 1. Substituting this into the general solution, we have:

1 = ±e^(-exp(0) + C) + 1/exp(0)

Simplifying, we get:

1 = ±e^(-1 + C) + 1

1 - 1 = ±e^(-1 + C)

0 = ±e^(-1 + C)

Since e^(-1 + C) is always positive, the equation 0 = ±e^(-1 + C) has no solutions. Therefore, the given initial condition is not compatible with the general solution of the differential equation.

Conclusion:
In conclusion, the differential equation dy/dx + exp(x)y = 1 has a general solution of y = ±e^(-exp(x) + C) + 1/exp(x), but the given initial condition y(0) = 1 does not have a solution.