MATH 223
Lecture 02
Laplace Transformation
Let F(t) be a function of t specified for t>0 , then the laplace transform of F(t),
denoted by L{f(t)} , is defined by
L{F(t)}=∫∞0e−st F(t) dt=f(s)
Solution: By the definition of Laplace Transformation
L{F(t)}=∫∞0e−stF(t) dtL{1}=∫∞0e−st 1 dt=[e−st−s]∞0Putting the value of t =[e−s.∞−s−e−s.0−s]=[e−∞−s−1−s]=0−1−s [e−∞=1e∞=1∞=0]=1s=f(s); s>0
Solution: By Definition
L{F(t)}=∫∞0e−st F(t) dtL{t}=∫∞0e−st t dt=[t ∫e−st dt−∫[ddtt ∫e−st dt]dt]=[t e−st−s−∫e−st−s dt]∞0=[t e−st−s−e−sts2]∞0=[∞ e−∞−s−e−∞s2−0.e0−s+e0s2]=0−0−0−1s2∴ L{t}=1s2
Home Work: Find L{F(t)}, where F(t)=t2.
Solution: by the definition of laplace Transformation,L{F(t)}=∫∞0e−st F(t) dtL{sin at}=∫∞0e−st sinat dt Using the property of Integration, ∫eatsinβt dt=eαtα2+β2(αsinβt−β cosβt) Comparing them, α=−s and β=aso, we'll get, [e−sts2+a2(−ssinat−acosat)]∞0= [e−sts2+a2(−ssin∞−acos∞)]−[e−0s2+a2(−ssin0−acos0)]= 0−1s2+a2(0−a)= as2+a2
Solution: By the definition of Laplace Transformation,
L{F(t)}=∫∞0e−st F(t) dtL{eat}=∫∞0e−st.eat dt [ea×eb=ea+b]=∫∞0e−(s−a)t dt=[e−(s−a)t−(s−a)]∞0=[e−(s−a)∞−(s−a)−e−(s−a).0−(s−a)]=e−∞−(s−a)−1−(s−a)=0+1s−a∴L{eat}=1s−a
Solution: By the definition of Laplace Transformation,
L{F(t)}=∫∞0e−st F(t) dtL{sin at}=∫∞0e−st sinat dt Property of Integration ∫eatsinβt dt=eαtα2+β2(αsinβt−β cosβt) Comparing them, α=−s and β=aso, we'll get, [e−sts2+a2(−ssinat−acosat)]∞0= [e−∞s2+a2(−ssin∞−acos∞)] −[e−0s2+a2(−ssin0−acos0)]= 0−1s2+a2(0−a)= as2+a2
Post a Comment