the method of variation of parameters

Question

y′′ - 2y′ = x + 1

Answer 1

Let e^rx be a solution of the homogeneous DE: y" - 2y' = 0.

 

Then,

 

r² - 2r = 0

 

r(r - 2) = 0

 

So, the complementary solution is y_c = Ae^2x + B with A and B constants.

 

 

Construct the particular solution from the complementary solution as 

 

y_p = C(x)e^2x + D(x).

 

y_p' = C'e^2x + 2Ce^2x + D'

 

Setting C'e^2x + D' = 0, we get y_p' = 2Ce^2x.

 

Then, y_p" = 2C'e^2x + 4Ce^2x.

 

Substituting back into the DE,

 

(2C' + 4C)e^2x -  4Ce^2x = x + 1

 

2C'e^2x = x + 1

 

C' = dC/dx = ½(x + 1)e^-2x

 

Integrate to get C(x) = -(1/8)(2x + 3)e^-2x.

 

 

Solving for D:

 

D' = -C'e^2x = -½(x + 1)

 

D(x) = -¼x² - ½x

 

The general solution is y = y_c + y_p =

 

Ae^2x + B - (1/8)(2x +3) - (1/4)(x² + 2x)

 

Ae^2x + B - ¼x² - ¾x - 3/8

 

 

Absorbing the -3/8 into B,

 

y(x) = Ae^2x + B - ¼x² - ¾x

 

 

Check:

 

y' = 2Ae^2x - ½x - ¾

 

y" = 4Ae^2x - ½

 

y" - 2y' = (4Ae^2x - ½) - 2(2Ae^2x - ½x - ¾) = x + 1