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Homework answers / question archive / Find the slution of differential equation y'' + 5y' +4y = 10 e-3x by finding complementary function (C
Find the slution of differential equation y'' + 5y' +4y = 10 e-3x by finding complementary function (C.F.) and particular integral ( P.I.).
= C.F + P.I
for calculation of C.F we need auxiliary equation i.e.
(D2 + 5D + 4 )y = 10e-3x
now doing the auxiliary equation = 0;
D2 + 5D + 4 = 0;
D2 + 4D + D + 4 = 0;
D(D+4) +1(D+4) = 0;
(D+4)(D+1) =0;
D = -4, -1
since both roots are real and different so C.F will be ,
C.F. = Ae-4x + Be-x ( where A and B are constants.)
now calculation of P.I.;
P.I. = [(1 / (D2 + 5D + 4) )] * 10e-3x
we use the standard form to calculate P.I. = [1 / F(D) ] * eax = eax / F(a) when F(a) ≠ 0;
here, F(-3) = (-3)2 + 5(-3) + 4
=9-15+4
=-2
since F(-3) ≠ 0;
then P.I. = 10 [{1/-2}]e-3x
= -5 e-3x.
solution of this differential equation will be C.F + P.I
y = Ae-4x + Be-x - 5 e-3x