Question

19. Solve (D2 + 7D + 12)y = 14e-3x​

Answer 1

The solution of the equation is $y=Ae^{-4x} +Be^{-3x}+\frac{e^{-3x} }{3}$

Given that $(D^{2} +7D+12)y=14e^{-3x}$

To solve this equation,

$(D^{2} +7D+12)y=14e^{-3x}$

First find the characteristic equation,

$m^{2} +7m+12=0$

Which impies, $(m+4)(m+3)=0$

That is, $m=-4,m=-3$

Therefore the characteristic function (C.F) is $Ae^{-4x} +Be^{-3x}$

To find the particular integral P.I,

$P.I=\frac{1}{(D^{2} +7D+12)} 14e^{-3x}\\=\frac{14e^{-3x}}{(D+4)(D+3)}$       [put D=a=3]

$=\frac{14e^{-3x} }{(3+4)(3+3)} \\=\frac{14e^{-3x} }{7*6}\\ =\frac{e^{-3x}}{3}$

The general solution= C.F+P.I

                                 $=Ae^{-4x} +Be^{-3x}+\frac{e^{-3x} }{3}$

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