MATH 223
Lecture 04
Laplace Transformation
F(t)F(t) | L{F(t)}=f(s)L{F(t)}=f(s) |
---|---|
tt | 1s21s2 |
t2t2 | 2s32s3 |
t2t2 | 6s46s4 |
tntn | n!sn+1n!sn+1 |
eateat | 1s−a1s−a |
sinatsinat | as2+a2as2+a2 |
cosatcosat | ss2+a2ss2+a2 |
sinhatsinhat | as2−a2as2−a2 |
coshatcoshat | ss2−a2ss2−a2 |
♦ Linear Properties of Laplace Transformation:
L{F1(t)}=f1(s)=∫∞0e−st.F1(t) dtL{F2(t)}=f2(s)=∫∞0e−st.F2(t) dt
Then if, C1 and C2 are any constants
∴L{C1 F1(t)+C2 F2(t)}=C1 f1(s)+C2 f2(s)
• Proof:
L{C1 F1(t)+C2 F2(t)}=∫∞0e−st.{C1F1(t)+C2F2(t)} dt=∫∞0e−st C1F1(t) dt+∫∞0e−stC2F2(t) dt=C1∫∞0e−st F1(t) dt+C2∫∞0F2(t) dt=C1L{F1(t)}+C2L{F2(t)}=C1f1(s)+C2f2(s)
• Example: Find
L{sin4t}
Solution:
We know that,
L{sinat}=as2+a2
∴L{sin4t}=4s2+42=4s2+16
• Example: Find Laplace Transform of
L{4e5t+6t2−3sin4t+2cos2t}
Solution:
L{4e5t+6t2−3sin4t+2cos2t}=L{4e5t}+L{6t3}−L{3sin4t}+L{2cos2t}=4L{e5t}+6L{t3}−3L{sin4t}+L{2cos2t}=41s−5+63!s3+1−343+16+2ss2+22
• Example: Find Laplace Transform of
L{e3tt2}
Solution:
e3t=eat a=3 F(t)=t2we know, f(s)=L{F(t)} =L{t2} =2s3 here, f(s−a)=f(s−3)=L{e3t t2}f(s)=2s3f(s−3)=2(s−3)3∴L{e3t t2}=2(s−3)3
• Example: Find
L{e−2t.sin4t}
Solution:
eat=e−2t∴ a=−2 L{e−2t.sin4t}=f(s+2)And,F(t)=sin4t f(s)=L{sin4t} =4s2+16 ∴f(s+2)=4(s+2)2+16
• Home Work:
L{e4t cosh5t}
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