y%3Dxsinx.jpeg "(d^2-1)y=xsinx")
Solving the Differential Equation (d^2-1)y=xsinx
In this article, we will explore the solution to the differential equation (d^2-1)y=xsinx. This equation is a second-order linear ordinary differential equation with a non-homogeneous term. We will use the method of undetermined coefficients to find the general solution of this equation.
The Homogeneous Equation
First, let's consider the homogeneous equation associated with the given equation:
(d^2-1)y = 0
This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:
r^2 - 1 = 0
Solving for r, we get:
r = ±1
Therefore, the general solution of the homogeneous equation is:
y_h = c1e^x + c2e^(-x)
where c1 and c2 are arbitrary constants.
The Particular Solution
Now, let's find the particular solution of the non-homogeneous equation. We will use the method of undetermined coefficients. Assume that the particular solution has the form:
y_p = Axcosx + Bxsinx
where A and B are constants to be determined.
Substituting this into the original equation, we get:
(d^2-1)(Axcosx + Bxsinx) = xsinx
Expanding and simplifying, we get:
(-A + B)xsinx + (A + B)xcosx = xsinx
Equating the coefficients of xsinx and xcosx, we get:
-A + B = 1 A + B = 0
Solving this system of equations, we get:
A = -1/2 B = 1/2
Therefore, the particular solution is:
y_p = (-1/2)xcosx + (1/2)xsinx
The General Solution
The general solution of the differential equation is the sum of the homogeneous solution and the particular solution:
y = y_h + y_p = c1e^x + c2e^(-x) - (1/2)xcosx + (1/2)xsinx
where c1 and c2 are arbitrary constants.
This is the general solution of the differential equation (d^2-1)y=xsinx.
