Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}+4 y^{\prime}+5 y=26 e^{3 x}\)

To solve a second-order linear non-homogeneous differential equation, we start with the corresponding homogeneous equation.
We assume that the solution to this homogeneous equation has the form of an exponential function: \( y = e^{rx} \).
Substituting this assumption into the homogeneous differential equation converts it to a polynomial equation in terms of \(r\), known as the *characteristic equation*.
In this exercise, the homogeneous equation is:
y'' + 4y' + 5y = 0. Replacing \(y = e^{rx}\), the characteristic equation becomes:
\[ r^2 + 4r + 5 = 0 \]
This quadratic equation allows us to solve for the roots (\(r\)) using the quadratic formula. The solutions to this polynomial will help us construct the complementary function.

The complementary function (C.F.) is the general solution to the homogeneous part of the differential equation.
The roots of the characteristic equation we found were complex:
\[ r = -2 \text{±} i \]
For complex roots of the form \(r = \text{α} \text{±}βi\), the complementary function is:
\[ y_c = e^{ \text{α} x} (C_1 \text{cos} (\text{β}x) + C_2 \text{sin}(\text{β}x)) \].
Substituting \( \text{α} = -2 \) and \( \text{β} = 1 \) into this form gives:

\[ y_c = e^{ -2x} (C_1 \text{cos}(x) + C_2 \text{sin}(x)) \]
Here, \(C_1\) and \(C_2\) are arbitrary constants that we can determine if initial conditions are provided.

The particular solution (P.S.) is a specific solution that satisfies the non-homogeneous differential equation.
For this equation, we assume a particular solution based on the form of the non-homogeneous term \(26e^{3x}\).
Assume a solution of the form:
\( y_p = Ae^{3x} \).
To find \(A\), we differentiate:
\( y_p' = 3Ae^{3x} \) \( y_p'' = 9Ae^{3x} \)
Substitute these into the original non-homogeneous equation:
\( 9Ae^{3x} + 12Ae^{3x} + 5Ae^{3x} = 26e^{3x} \)
Combining coefficients, we get:
\( 26Ae^{3x} = 26e^{3x} \).
Thus, \(A = 1\), giving the particular solution:
\( y_p = e^{3x} \)

A homogeneous differential equation is one where all terms depend only on the function and its derivatives.
There is no term that is a standalone function of the independent variable. The equation is set to zero.
For our given problem, the homogeneous form is:
\( y'' + 4y' + 5y = 0 \).
This is derived by removing the non-homogeneous term \(26e^{3x}\). By solving this, we find the complementary function, which combined with the particular solution, forms the general solution.
The general solution of the given second-order linear non-homogeneous differential equation
is then given by:
\( y = y_c + y_p \)
where \( y_c = e^{-2x}(C_1 \text{cos}x + C_2 \text{sin}x) \) and \( y_p = e^{3x} \).