Question

Find the general solution by the method of undetermined coefficients.

(D² − 3D + 2)y = x²sin(2x).

Answer 1

Write the equation as 

(∗) P(D)y = x²sin(2x), 

where P(D) = D² − 3D + 2. Note that x²sin(2x) is the imaginary part of x²e^2ix. Thus, to solve (∗), we want to solve the complex equation 

(♥) P(D)y = x²e^2ix, and then take the imaginary part of the solution. To solve equation (♥), we move the exponential to the left side are write the equation as 

e^−2ixP(D)y = x². 

By the shifting rule, this is equivalent to P(D + 2i)[e^−2ixy] = x².

We set z = e^2ixy and calculate 

P(D + 2i) = (D + 2i)² − 3(D + 2i) + 2 

= D² + 4iD − 4 − 3D − 6i + 2 

= D² + (−3 + 4i)D + (−2 − 6i). 

Thus, the equation for z is 

(D² + (−3 + 4i)D + (−2 − 6i))z = x² 

Since the right-hand side is a polynomial of degree 2, we try a polynomial of degree 2, z = Ax² + Bx + C. Plugging into the equations and collecting coefficients yields 

(− 2 − 6i)Ax² + ((− 6 + 8i)A + (− 2 − 6i)B)x + (2A + (− 3 + 4i)B +(− 2 − 6i)C) = x², 

so we get the equations 

(−2 − 6i)A = 1 

(−6 + 8i)A + (−2 − 6i)B = 0 

2A + (−3 + 4i)B + (−2 − 6i)C = 0 

(use the calculator!) The solutions are