(b) Solve (d^(2)-2d+1)y=x^(2)e^(3x)
In this problem, we are asked to solve a second-order linear differential equation with an exponential function on the right-hand side. The differential equation is given as:
(d^(2)-2d+1)y=x^(2)e^(3x)
Step 1: Identify the type of differential equation
The given differential equation is a second-order linear differential equation with constant coefficients. The general form of such an equation is:
a(d^(2)y) + b(dy) + cy = f(x)
where a, b, and c are constants, and f(x) is a function of x.
Step 2: Find the homogeneous solution
The homogeneous solution is the general solution of the associated homogeneous equation:
(d^(2)-2d+1)y=0
Let's find the roots of the characteristic equation:
r^(2) - 2r + 1 = 0
(r - 1)^2 = 0
r = 1 (repeated root)
The homogeneous solution is:
y_h = (c1 + c2x)e^(x)
where c1 and c2 are arbitrary constants.
Step 3: Find the particular solution
The particular solution is a specific solution of the non-homogeneous equation:
(d^(2)-2d+1)y=x^(2)e^(3x)
Let's use the method of undetermined coefficients to find the particular solution.
Assume the particular solution has the form:
y_p = (Ax^2 + Bx + C)e^(3x)
where A, B, and C are constants to be determined.
Step 4: Substitute into the differential equation
Substitute the particular solution into the differential equation:
(d^(2)-2d+1)(Ax^(2) + Bx + C)e^(3x) = x^(2)e^(3x)
Simplify and equate coefficients
After simplifying, we get:
9Ax^(2) + (2B - 6A)x + (C - 2B + 1) = x^(2)
Equating coefficients, we get:
9A = 1, 2B - 6A = 0, and C - 2B + 1 = 0
Solving these equations, we get:
A = 1/9, B = 1/3, and C = 2/9
Step 5: Write the general solution
The general solution is the sum of the homogeneous solution and the particular solution:
y = y_h + y_p
(c1 + c2x)e^(x) + (1/9x^(2) + 1/3x + 2/9)e^(3x)
Therefore, the general solution of the differential equation (d^(2)-2d+1)y=x^(2)e^(3x) is:
(c1 + c2x)e^(x) + (1/9x^(2) + 1/3x + 2/9)e^(3x)
where c1 and c2 are arbitrary constants.