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Find the particular integral of given differential equation (D3 + 3D + 2)y = 2 cos (2x + 3) + 2ex + x2. |
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Answer» P.I. = \(\frac1{D^3+3D+2}[2cos(2x+3)+2e^x+x^2]\) = \(\frac{2cos(2x+3)}{-4D+3D+2}+\frac{2e^x}{1+3+2}\) + \(\frac12(1+\frac{3D+D^3}2)^{-1}x^2\) = \(\frac{2(2+D)}{4-D^2}\)cos(2x + 3) + \(\frac{e^x}3\) + \(\frac12(1-(\frac{3D+D^3}2)+(\frac{3D+D^3}2)^2)x^2\) = \(\frac{2(2+D)}{4-(-4)}\)cos(2x + 3) + \(\frac{e^x}3\) + \(\frac12(x^2-\frac{6x}2+\frac94\times2)\) = \(\frac14\) (2cos(2x + 3) - 2sin(2x + 3)) + \(\frac{e^x}3\) + \(\frac{x^2}2-\frac{3x}2+\frac94\) = \(\frac14\) (cos (2x + 3) - sin (2x + 3)) + \(\frac{e^x}3\) + \(\frac{x^2}2-\frac{3x}2+\frac94\) |
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