I need help to use the method of variation of parameters to find the solution y(x) of the non- homogeneous differential equation y'' + 4y' - 5y = 8e^3x, with initial conditions y(0) = 5 and y'(0) = 0 Please post as much steps as possible if you can't get the final answer.

 
According to the  variations of parameters method, the formula for the particular solution of the equation of the form
 y''+q(x)y' + r(x) y = g(x)
is
 Y_P = -y_1 int y_2(x)g(x)/(W(y_1, y_2)) dx + y_2 int y_1(x)g(x)/(W(y_1, y_2)) dx
Here, 
y_1 and y_2
are the solutions of the corresponding homogeneous equation and W(y_1, y_2)
is the Wronskian.
(Please see the reference link for the derivation and the discussion of this formula.)
The given equation is
y'' + 4y' - 5y = 8e^(3x)
To use the method of variations, we need to first find the solutions of the corresponding homogeneous equation
y" + 4y' - 5y = 0
This equation has the characteristic equation lambda^2 + 4lambda - 5 = 0 , with the roots lambda_1 = -5  and lambda_2 = 1
This means its solution is
y(x) = c_1e^(-5x) +c_2e^x , so
y_ 1 = e^(-5x)  and y_2 = e^x .
The Wronskian of these two functions equals
y_1y_2' - y_1' y_2 = e^(-5x)e^x - (-5e^(-5x))e^x = e^(-4x) + 5e^(-4x) = 6e^(-4x)
 
Now we can proceed to find the particular solution using the formula for the method for the variation of parameters. To make writing easier, I will calculate the two integrals first and then put them back in the formula.
The first integral is
int y_2(x)g(x)/(W(y_1, y_2))dx = int (e^x*8*e^(3x))/(6e^(-4x))dx = 4/3int e^(8x)dx = 1/6e^(8x)
The second integral is 
int y_1(x) g(x)/(W(y_1, y_2)) dx = int (e^(-5x)*8*e^(3x))/(6e^(-4x)) dx = 4/3int e^(2x)dx = 2/3e^(2x)
So, putting these results into the formula for the particular solution, we obtain
Y_p = -e^(-5x)*1/6*e^(8x) + e^x * 2/3e^(2x) = -1/6e^(3x) + 2/3e^(3x) = =1/2e^(3x)
 We have found the particular solution. The general solution of the non-homogeneous equation is the general solution of the homogeneous equation, which we already have, plus the particular solution of the non-homogeneous equation.
Thus, the general solution is
y(x) = c_1e^(-5x) + c_2e^x + 1/2e^(3x)
For the given initial conditions, we can find the constants:
y(0) = 5, so y(0) = c_1 + c_2 + 1/2 = 5
y'(0) = 0, so 
y'(0) = -5c_1 + c_2 + 3/2 = 0
Solving this system of equations results in the values of the constants equal 1 and 3.5.
So, the solution of the given equation satisfying the given initial conditions is
y(x) = e^(-5x) + 7/2e^x + 1/2e^(3x)
 
 
 
 
 
y'(0) = -5c_1+c_2 + 3/2 = 0
romeF




http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx