Question

Solve the particular integral:

(D² - D + 1)y = 3x² - 1

Answer 1

P. I. = \(\frac1{D^2-D+1}(3x^2-1)\)

= (1 - (D² - D) + (D² - D)² - (D² - D)³+...)(3x² - 1)

(\(\because\) \(\frac1{1+x}\) = 1 - x + x² - x³ + ....)

 = (3x² - 1) - (D² - D)(3x² - 1) + (D⁴ - 2D³ + D²)(3x² - 1)

 = 3x² - 1 - D.D(3x²) - D3x² + D.D(3x²)

 = 3x² - 1 - D(6x) - 6x + D.6x (\(\because\) D(3x²) = \(\frac{d}{dx}3x^2=6x\))

 = 3x² - 1 - 6 - 6x + 6

 = 3x² - 6x - 1