Answer:
Step-by-step explanation:
Non homogeneous differential
[tex]y''+25y = -40sin(5t)---(1)[/tex]
To find homogeneous solution consider auxilary equation:
[tex]D^{2}+25=0\\D=\pm5i[/tex]
Homogeneous solution is:
[tex]y_{h}=Acos(5t)+Bsin(5t)\\[/tex]
Particular soution for -40 sin(5t)
[tex]y_{p}= Csin(5t)+Dcos(5t)\\y'_{p}=5Ccos(5t)-5Dsin(5t)\\y''_{p}=-25Csin(5t)-25Dcos(5t)\\[/tex]
substituting value of [tex]y_{p}, y''_{p}[/tex] in (1)
[tex]-25Csin(5t)-25Dcos(5t)+25Csin(5t)+25Dcos(5t)=-40sin(5t)\\\implies\\C=0\\D=0\\\implies \\y_{p}=0[/tex]
General solution is
[tex]y=Acos(5t)+Bsin(5t)[/tex]
