y%3De%5E2x-cos3x.jpeg "(d^2-4d+13)y=e^2x Cos3x")
Solving the Differential Equation (d^2-4d+13)y=e^2x cos3x
In this article, we will solve the differential equation (d^2-4d+13)y=e^2x cos3x. This is a second-order linear ordinary differential equation with a non-homogeneous term.
Step 1: Find the Homogeneous Solution
To find the homogeneous solution, we need to solve the equation (d^2-4d+13)y=0. Let's assume the solution has the form y=e^(rx). Substituting this into the equation, we get:
(r^2-4r+13)e^(rx)=0
This gives us the characteristic equation:
r^2-4r+13=0
Solving this quadratic equation, we get:
r = 2 ± √(-9)
r = 2 ± 3i
So, the general solution to the homogeneous equation is:
y_c = c1e^(2x)cos(3x) + c2e^(2x)sin(3x)
where c1 and c2 are arbitrary constants.
Step 2: Find a Particular Solution
To find a particular solution, we need to find a function y_p that satisfies the non-homogeneous equation (d^2-4d+13)y=e^2x cos3x. Let's assume y_p has the form:
y_p = Ae^2x cos3x + Be^2x sin3x
Substituting this into the equation, we get:
(4A+12B)e^2x cos3x + (4B-12A)e^2x sin3x = e^2x cos3x
Comparing coefficients, we get:
4A+12B = 1 and 4B-12A = 0
Solving these two equations, we get:
A = 1/16 and B = 1/48
So, the particular solution is:
y_p = (1/16)e^2x cos3x + (1/48)e^2x sin3x
Step 3: Find the General Solution
The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:
y = y_c + y_p
y = c1e^(2x)cos(3x) + c2e^(2x)sin(3x) + (1/16)e^2x cos3x + (1/48)e^2x sin3x
where c1 and c2 are arbitrary constants.
Therefore, we have successfully solved the differential equation (d^2-4d+13)y=e^2x cos3x. The general solution is a combination of the homogeneous solution and the particular solution.
