Question

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

Answer 1

`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`