Answer:
The integral form of the given differential equation is:
yp(x) = x * ∫[x0 to x] G(x, t) * f(t) dt
where G(x,t) is the Green's function for the differential equation, which depends on the specific form of the equation and the boundary conditions.
To find the Green's function, we first solve the homogeneous equation y'' + 2y' + y = 0, which has the characteristic equation r^2 + 2r + 1 = 0. This has a double root of -1, so the general solution to the homogeneous equation is yh(x) = c1e^(-x) + c2x*e^(-x).
Next, we need to find a particular solution to the non-homogeneous equation y'' + 2y' + y = f(x). We can use the method of undetermined coefficients to guess a particular solution of the form yp(x) = A*x, where A is a constant. Taking the first and second derivatives of yp(x), we get yp'(x) = A and yp''(x) = 0. Substituting these into the differential equation, we get:
0 + 2A + Ax = f(x)
Solving for A, we get A = (1/x) * ∫[x0 to x] f(t) dt.
Therefore, the general solution to the non-homogeneous equation is y(x) = yh(x) + yp(x) = c1e^(-x) + c2xe^(-x) + (1/x) * ∫[x0 to x] G(x, t) * f(t) dt, where G(x,t) = e^(-(x-t)) - e^(-(x-t))(x-t).
Using this expression for yp(x), we can now substitute it into the integral form of the differential equation to get:
yp(x) = x * ∫[x0 to x] (e^(-(x-t)) - e^(-(x-t))*(x-t)) * f(t) dt
Therefore, the solution to f(x) = g(x) is x = 1.
Explanation: