Question

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

Answer 1

Answer:

$y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint$

Step-by-step 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=$\frac{8e^{3t}+4sin t}{D^2-3}$

P.I=$\frac{8e^{3t}}{9-3}+4\frac{sint }{-1-3}$

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

P.I=$-\frac{4}{3}e^{3t}- sint$

Hence, the general solution

G.S=C.F+P.I

$y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint$