Question

2D²+3D+4=x²-2x

Answer 1

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Answer 2

2D²+3D+4 = x²-2x

It's auxiliary equation is 2m²+3m+4 = 0 (By putting min place of D in L.H.S. of equation)

⇒ m \(=\frac{-3\pm\sqrt{9-32}}{4}\)
\(=\frac{-3\pm\sqrt{23}i}{4}\) (∵ √-1 = i)

∴ Complementary function of the differential equation is C.F. \(=e^{-\frac{3}{4}x}\)(c₁ cos\(\frac{\sqrt{23}}{4}\)x + c₂ sin\(\frac{\sqrt{23}}{4}\)x)

And P.I. \(=\frac{1}{2D^2+3D+4}\) (x²-2x)

= 1/4 (1+\(\frac{2D^2+3D}{4})\)^-1 (x² - 2x)

= 1/4 (1 - \(\frac{2D^2+3D}{4}\) \(+(\frac{2D^2+3D}{4})^2\)....) (x²-2x) (∵ (1+x)^-1 = 1-x+x²-x³+....)

= 1/4 (1 - \(\frac{3D}{4}-\frac{2D^2}{4}+\frac{9D^2}{16})\) (x² - 2x) (∵ Dⁿ (x² - 2x) = 0 if n > 2)

= 1/4 (x²-2x-3/4 D(x²-2x)+1/16 D²(x²-2x))

= 1/4 (x² - 2x - 3/2 x + 3/2 + 2/16) (∵ D = d/dx and D² \(=\frac{d^2}{dx^2})\)

= 1/4 (x² - 7/2 x+13/8)

= x²/4 - 7x/8 + 13/32

∴ Complete solution of given differential equation is y = e.F.+P.I. \(=e^{-\frac{3}{4}x}\)(c₁ cos\(\frac{\sqrt{23}}{4}\)x + c₂ sin\(\frac{\sqrt{23}}{4}\)x) + x²/4 - 7x/8 + 13/32.