Answer:
[tex]y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint [/tex]
Step-by-step explanation:
We are given that linear differential equation
[tex]y''-3y=8e^{3t}+4 sint[/tex]
Auxillary equation
[tex]D^2-3=0[/tex]
[tex] D=\pm \sqrt3[/tex]
C.F=[tex]C_1e^{\sqrt3t}+C_2e^{-\sqrt3}[/tex]
P.I=[tex]\frac{8e^{3t}+4sin t}{D^2-3}[/tex]
P.I=[tex]\frac{8e^{3t}}{9-3}+4\frac{sint }{-1-3}[/tex]
P.I=[tex] e^{ax}{\phi (D+a)}[/tex] and [tex]P.I=\frac{sinax}{(\phi D)}[/tex]where D square is replace by - a square
P.I=[tex]-\frac{4}{3}e^{3t}- sint [/tex]
Hence, the general solution
G.S=C.F+P.I
[tex]y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint [/tex]
