Lecture 4 MATH 223 (Lt Col Tahmina Sultana)

 

 MATH 223

Lecture 04 

Laplace Transformation









    ♦ Some Important Results:

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 1sa1sa
sinatsinat as2+a2as2+a2
cosatcosat ss2+a2ss2+a2
sinhatsinhat as2a2as2a2
coshatcoshat ss2a2ss2a2






    ♦ Linear Properties of Laplace Transformation:
  L{F1(t)}=f1(s)=0est.F1(t) dtL{F2(t)}=f2(s)=0est.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)}=0est.{C1F1(t)+C2F2(t)} dt=0est C1F1(t) dt+0estC2F2(t) dt=C10est F1(t) dt+C20F2(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+6t23sin4t+2cos2t}  
Solution:  L{4e5t+6t23sin4t+2cos2t}=L{4e5t}+L{6t3}L{3sin4t}+L{2cos2t}=4L{e5t}+6L{t3}3L{sin4t}+L{2cos2t}=41s5+63!s3+1343+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(sa)=f(s3)=L{e3t t2}f(s)=2s3f(s3)=2(s3)3L{e3t t2}=2(s3)3  



   
    • Example: Find  L{e2t.sin4t}  
Solution:  eat=e2t a=2   L{e2t.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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