Lecture 2 MATH 223 (Lt Col Tahmina Sultana)

 

 

 MATH 223

Lecture 02 

Laplace Transformation







   Let F(t)F(t) be a function of tt specified for t>0t>0 , then the laplace transform of F(t)F(t)

denoted by  L{f(t)}L{f(t)} , is defined by 

  L{F(t)}=0est F(t) dt=f(s)L{F(t)}=0est F(t) dt=f(s)  


 


    

     Example: If F(t)=1F(t)=1, then find out  L{F(t)}L{F(t)} .

Solution:  By the definition of Laplace Transformation

L{F(t)}=0estF(t) dtL{1}=0est 1 dt=[ests]0Putting the value of t =[es.ses.0s]=[es1s]=01s    [e=1e=1=0]=1s=f(s);   s>0  





    

    Example: If  F(t)=t, then find out L{F(t)}.

Solution:  By Definition 

L{F(t)}=0est F(t) dtL{t}=0est t dt=[t est dt[ddtt est dt]dt]=[t estsests dt]0=[t estsests2]0=[ eses20.e0s+e0s2]=0001s2 L{t}=1s2  





    Home Work: Find L{F(t)}, where F(t)=t2.

Solution:  by the definition of laplace Transformation,L{F(t)}=0est F(t) dtL{sin at}=0est sinat dt   Using the property of Integration,           eatsinβt dt=eαtα2+β2(αsinβtβ cosβt) Comparing them,   α=s and β=aso, we'll get,  [ests2+a2(ssinatacosat)]0= [ests2+a2(ssinacos)][e0s2+a2(ssin0acos0)]= 01s2+a2(0a)= as2+a2  



   

    Example: Find L{F(t)}, where F(t)=eat

Solution: By the definition of Laplace Transformation, 

  L{F(t)}=0est F(t) dtL{eat}=0est.eat dt          [ea×eb=ea+b]=0e(sa)t dt=[e(sa)t(sa)]0=[e(sa)(sa)e(sa).0(sa)]=e(sa)1(sa)=0+1saL{eat}=1sa  


   




    Example: If F(t)=sinat, then find L{F(t)}.

Solution: By the definition of Laplace Transformation,

  L{F(t)}=0est F(t) dtL{sin at}=0est sinat dt   Property of Integration           eatsinβt dt=eαtα2+β2(αsinβtβ cosβt) Comparing them,   α=s and β=aso, we'll get,  [ests2+a2(ssinatacosat)]0= [es2+a2(ssinacos)]          [e0s2+a2(ssin0acos0)]= 01s2+a2(0a)= as2+a2  


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