The solution of(x2D2+4xD+2)y = ex is

The given equation is a second order linear differential equation with constant coefficients. The general form of such an equation is:

ay'' + by' + c*y = f(x)

where y'' is the second derivative of y with respect to x, y' is the first derivative of y with respect to x, and f(x) is the forcing function. In your case, a = 1, b = 4, c = 2, and f(x) = e^x.

The solution to this type of equation is the sum of the homogeneous solution (the solution to the equation without the forcing function, i.e., ay'' + by' + c*y = 0) and the particular solution (a specific solution to the full equation).

1. Solve the homogeneous equation:

The characteristic equation of the homogeneous equation is ar^2 + br + c = 0. In your case, this is r^2 + 4*r + 2 = 0. Solving this quadratic equation for r gives the roots r = -2 ± √2i. Therefore, the homogeneous solution is:

y_h(x) = e^(-2x) * (Acos(√2x) + Bsin(√2x))

where A and B are constants to be determined by boundary conditions.

2. Find the particular solution:

A common method for finding the particular solution is the method of undetermined coefficients. However, because the forcing function e^x is a solution to the homogeneous equation, we must multiply it by x to find a solution to the full equation. Therefore, we guess that the particular solution has the form:

y_p(x) = x*(C*e^x)

where C is a constant. Substituting this into the full equation gives:

(x^2D^2 + 4xD + 2)(xCe^x) = e^x

Solving this equation for C gives C = 1/2. Therefore, the particular solution is:

y_p(x) = x*(1/2)*e^x

3. Add the homogeneous and particular solutions:

The general solution to the equation is the sum of the homogeneous and particular solutions:

y(x) = y_h(x) + y_p(x) = e^(-2x) * (Acos(√2x) + Bsin(√2x)) + x*(1/2)*e^x

This is the solution to the given differential equation.