Find the slution of differential equation y'' + 5y' +4y = 10 e-3x by finding complementary function (C

= 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) ] * e^ax = e^ax / 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