y%3Dxe%5E3x%2Bsin2x.jpeg "(d^2-3d+2)y=xe^3x+sin2x")
Solving the Differential Equation (d^2-3d+2)y=xe^3x+sin2x
In this article, we will explore the solution to the differential equation (d^2-3d+2)y=xe^3x+sin2x. This equation is a second-order linear differential equation with non-constant coefficients, which makes it a bit more challenging to solve.
Step 1: Homogeneous Solution
To start, we need to find the homogeneous solution to the equation, which is the solution to (d^2-3d+2)y=0. This can be done by using the characteristic equation:
r^2 - 3r + 2 = 0
Factoring the quadratic, we get:
(r-1)(r-2)=0
This gives us two distinct roots: r=1 and r=2. Therefore, the homogeneous solution is:
y_h = c1e^x + c2e^(2x)
where c1 and c2 are arbitrary constants.
Step 2: Particular Solution
Next, we need to find a particular solution to the non-homogeneous equation (d^2-3d+2)y=xe^3x+sin2x. To do this, we can use the method of undetermined coefficients.
Assuming a particular solution of the form:
y_p = Ae^3x + Bxsin2x + Cxcos2x
where A, B, and C are constants to be determined.
Substituting this into the original equation, we get:
(d^2-3d+2)(Ae^3x + Bxsin2x + Cxcos2x) = xe^3x + sin2x
Equating coefficients, we can solve for A, B, and C:
A = 1/9
B = -1/4
C = 1/4
Therefore, the particular solution is:
y_p = (1/9)e^3x - (1/4)xsin2x + (1/4)xcos2x
Step 3: General Solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:
y = y_h + y_p
y = c1e^x + c2e^(2x) + (1/9)e^3x - (1/4)xsin2x + (1/4)xcos2x
This is the final solution to the differential equation (d^2-3d+2)y=xe^3x+sin2x.
Conclusion
In this article, we have successfully solved the differential equation (d^2-3d+2)y=xe^3x+sin2x by finding the homogeneous and particular solutions and combining them to obtain the general solution. This solution involves a combination of exponential and trigonometric functions, demonstrating the complexity and richness of second-order linear differential equations.
