Respuesta :
Answer:
the particular solution is
Y_{p}= C +D\sin 5t +E\cos 5t + F\exp 4t + G\exp -2t
the differential operator that annihilate the non homogeneous differential equation is
D(D^2+5)
Step-by-step explanation:
hello,
i believe the non homogeneous differential equation is
[tex]U^{''} - 2U^{'} - 8= \cos 5x + 7[/tex]
the homogeneous differential equation of the above is
[tex]u^{''} -2u^{'} -8 =0[/tex]
the differential form of the above equation is
[tex]D^2-2D-8=0[/tex]
[tex](D-4)(D+2)=0[/tex]
thus the roots are 4 and -2.
thus the solution of the homogenous differential equation is given as
[tex]Y_{h} (t)= A\exp{4t} + B\exp{-2t}[/tex]
the differential operator of the non homogeneous equation is given as
[tex](D-4)(D+2)(u)=\cos 5x +7[/tex]
the differential operator [tex]D^2 +5[/tex] annihilates [tex]\cos 5x[/tex] and the differential operator D annihilates 7
applying [tex]D(D^2+5)[/tex] to both sides of the differential equation we have;
(D-4)(D+2)(u)=\cos 5x +7
[tex]D(D^2+5)(D-4)(D+2)=D(D^2+5)(\cos5x+7)[/tex][tex]D(D^2+5)(D-4)(D+2)=0[/tex]
the roots of the characteristic polynomial of the diffrential equation above are [tex]0, \cmplx 5i, -\cmplx 5i, 4, -2[/tex]
thus the particular solution is
[tex]Y_{p}= C\exp{0}+D\sin 5t +E\cos 5t + F\exp {4t} + G\exp {-2t}[/tex]
this gives us the particular solution
[tex]Y_{p}= C +D\sin 5t +E\cos 5t + F\exp 4t + G\exp -2t[/tex]
