y%3Dxe%5E4x.jpeg "(d^2-5d+6)y=xe^4x")
Solving the Differential Equation (d^2-5d+6)y = xe^4x
In this article, we will solve the differential equation (d^2-5d+6)y = xe^4x. This is a second-order linear ordinary differential equation, and we will use the method of undetermined coefficients to find the general solution.
Step 1: Find the Homogeneous Solution
First, we need to find the homogeneous solution of the differential equation, which is the solution to the equation (d^2-5d+6)y = 0. This can be done by finding the roots of the characteristic equation:
r^2 - 5r + 6 = 0
Factoring the quadratic equation, we get:
(r - 2)(r - 3) = 0
This gives us two distinct roots: r = 2 and r = 3. Therefore, the homogeneous solution is:
y_h = c1e^(2x) + c2e^(3x)
where c1 and c2 are arbitrary constants.
Step 2: Find the Particular Solution
Next, we need to find the particular solution of the differential equation, which is the solution to the equation (d^2-5d+6)y = xe^4x. We will use the method of undetermined coefficients, which involves guessing the form of the particular solution.
Let's assume that the particular solution has the form:
y_p = Ae^4x(x^2 + Bx + C)
where A, B, and C are constants to be determined.
Substituting this into the differential equation, we get:
(16Ae^4x(x^2 + Bx + C) - 20Ae^4x(x + B) + 6Ae^4x) = xe^4x
Simplifying and equating coefficients, we get:
A = 1/16, B = -1/4, and C = 1/32
Therefore, the particular solution is:
y_p = (1/16)e^4x(x^2 - x/4 + 1/32)
Step 3: Find the General Solution
The general solution is the sum of the homogeneous solution and the particular solution:
y = y_h + y_p
Substituting the expressions we found earlier, we get:
y = c1e^(2x) + c2e^(3x) + (1/16)e^4x(x^2 - x/4 + 1/32)
This is the general solution to the differential equation (d^2-5d+6)y = xe^4x.
Conclusion
In this article, we have solved the differential equation (d^2-5d+6)y = xe^4x using the method of undetermined coefficients. The general solution involves the sum of two exponential functions and a polynomial function. This solution can be used to model various physical systems, such as electrical circuits or mechanical systems, that exhibit oscillatory behavior.
