Question

Find the general solution by the method of undetermined coefficients.

(D² − 4D + 5)y = e^2xcos(x).

Answer 1

Write the equation as (∗) 

P(D)y = e^2x cos(x), 

where P(D) = D² − 4D + 5. Note that e^2xcos(x) is the real part of e^(2+i)x, so we can solve (∗) by solving the complex equation 

(♥) P(D)y = e^(2+i)x 

and then taking the real part. To solve equation (♥), move the exponential to the left side and rewrite the equation as 

e^−(2+i)xP(D)y = 1. 

By the shifting rule, this is equivalent to 

P(D + (2 + i))[e^−(2+i)x y] = 1. 

Set z = e⁻⁽²⁺ⁱ⁾x and calculate 

P(D + (2 + i)) = (D + (2 + i))² − 4(D + (2 + i)) + 5 

= D² + 2(2 + i)D + (2 + i)² − 4(D + 2 + i) + 5 

= D² + (4 + 2i)D + 3 + 4i − 4d − 8 − 4i + 5 

= D² + 2iD. 

Thus, the equation for z is (D² + 2iD)z = 1.

Factor out a D and write this as 

(D + 2i)[Dz] = 1. 

Set w = Dz, so the equation for w is (D + 2i)w = 1. 

Since the right-hand side is a polynomial of degree 0, we try a polynomial of degree 0, i.e., w = A, where A is a constant. Plugging into the equation gives 2iA = 1, so 

w = 1/2i = − 1/2i. 

Since Dz = w, integrating gives us 

z = − 1/2ix. 

Since z = e^−(2+i)xy, we get y = − 1/2ixe⁽²⁺ⁱ⁾x . 

This is a particular solution of equation (♥). To find the solution of equation (∗), we need to find the real part of y. We have

as a particular solution of (∗). To find the general solution of the homogeneous equation (D²− 4D + 5)y = 0, use the quadratic formula to find the roots of the characteristic polynomial λ² − 4λ + 5. The roots are 2 ± i, so the general solution of the homogeneous equation is