Answers 1
Let's find the Laplace transform of tsin3ttsin3t.
Let f(t)=tsin3t.f(t)=tsin3t. Then f(0)=0.f(0)=0.
Deriving we have
f′(t)=sin3t+3tcos3tf′(t)=sin3t+3tcos3t. Then f′(0)=0f′(0)=0.
calculating f′′(t)f″(t), we have
f′′(t)=3cos3t+3tcos3t−9tsin3tf′′(t)=6cos3t−9tsin3t.f″(t)=3cos3t+3tcos3t−9tsin3tf″(t)=6cos3t−9tsin3t.
So taking the transform we have
L{f′′(t)}=6L{cos3t}−9L{tsin3t}.L{f″(t)}=6L{cos3t}−9L{tsin3t}.
But L{f′′(t)}=s2L{f(t)}L{f″(t)}=s2L{f(t)} and L{cos3t}=ss2+9.L{cos3t}=ss2+9.
So we have
s2L{tsin3t}=6ss2+9−9L{tsin3t}.s2L{tsin3t}=6ss2+9−9L{tsin3t}.
Solving we have
s2L{tsin3t}+9L{tsin
-
Author:
freewaynavarro
-
Rate an answer:
0
If you know the answer add it here!
Choose a language and a region
How much to ban the user?
1 hour
1 day
100 years