y%3Dx%5E3.jpeg "(d^4-2d^3+d^2)y=x^3")
Solving the Differential Equation: (d^4 - 2d^3 + d^2)y = x^3
In this article, we will explore the solution to the differential equation (d^4 - 2d^3 + d^2)y = x^3. This is a fourth-order linear homogeneous differential equation with variable coefficients.
The General Form of the Solution
The general form of the solution to this differential equation is:
y = y_c + y_p
where y_c is the complementary function and y_p is the particular solution.
Finding the Complementary Function (y_c)
To find the complementary function, we need to solve the homogeneous differential equation:
(d^4 - 2d^3 + d^2)y = 0
The characteristic equation for this differential equation is:
r^4 - 2r^3 + r^2 = 0
Factoring the characteristic equation, we get:
(r - 1)^2 (r^2 - r) = 0
This gives us three distinct roots: r = 0, r = 1, and r = 1. Therefore, the complementary function is:
y_c = c1 + c2e^x + c3xe^x + c4x^2e^x
where c1, c2, c3, and c4 are arbitrary constants.
Finding the Particular Solution (y_p)
To find the particular solution, we will use the method of undetermined coefficients. Assume that the particular solution is of the form:
y_p = Ax^3 + Bx^2 + Cx + D
Substituting this into the original differential equation, we get:
(d^4 - 2d^3 + d^2)(Ax^3 + Bx^2 + Cx + D) = x^3
Simplifying and collecting terms, we get:
A(x^3) + (-6A + B)x^2 + (11A - 2B + C)x + (6A - 2B + C - D) = x^3
Equating coefficients, we get:
A = 1, B = -6, C = 25, and D = -25
Therefore, the particular solution is:
y_p = x^3 - 6x^2 + 25x - 25
The General Solution
The general solution to the differential equation (d^4 - 2d^3 + d^2)y = x^3 is:
y = c1 + c2e^x + c3xe^x + c4x^2e^x + x^3 - 6x^2 + 25x - 25
where c1, c2, c3, and c4 are arbitrary constants.
