Question

Find a particular solution to y′′+25y=−40sin(5t). y″+25y=−40sin⁡(5t).

Answer 1

Answer:

Step-by-step explanation:

Non homogeneous differential

$y''+25y = -40sin(5t)---(1)$

To find homogeneous solution consider auxilary equation:

$D^{2}+25=0\\D=\pm5i$

Homogeneous solution is:

$y_{h}=Acos(5t)+Bsin(5t)$

Particular soution for -40 sin(5t)

$y_{p}= Csin(5t)+Dcos(5t)\\y'_{p}=5Ccos(5t)-5Dsin(5t)\\y''_{p}=-25Csin(5t)-25Dcos(5t)$

substituting value of $y_{p}, y''_{p}$  in (1)

$-25Csin(5t)-25Dcos(5t)+25Csin(5t)+25Dcos(5t)=-40sin(5t)\\\implies\\C=0\\D=0\\\implies \\y_{p}=0$

General solution is

$y=Acos(5t)+Bsin(5t)$