1.

Solve the particular integral:(D2 - D + 1)y = 3x2 - 1

Answer»

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



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