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Find the general solution of {y}''-3y=8e^{3t}+4sin(t) .

1 Answer

4 votes

Answer:


y(x)=C_1e^(\sqrt3t)+C_2e^(-\sqrt3t)-(4)/(3)e^(3t)-sint

Explanation:

We are given that linear differential equation


y''-3y=8e^(3t)+4 sint

Auxillary equation


D^2-3=0


D=\pm \sqrt3

C.F=
C_1e^(\sqrt3t)+C_2e^(-\sqrt3)

P.I=
(8e^(3t)+4sin t)/(D^2-3)

P.I=
(8e^(3t))/(9-3)+4(sint )/(-1-3)

P.I=
e^(ax){\phi (D+a)} and
P.I=(sinax)/((\phi D))where D square is replace by - a square

P.I=
-(4)/(3)e^(3t)- sint

Hence, the general solution

G.S=C.F+P.I


y(x)=C_1e^(\sqrt3t)+C_2e^(-\sqrt3t)-(4)/(3)e^(3t)-sint

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