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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section C.5 Laplace Transforms

s=0,e0t=1
= 15limb→∞∫0b1 dt=15limb→∞t|0b=15limb→∞b=∞.
s=0,does not exists.
s=0,
= limb→∞∫0bt dt=limb→∞t22|0b=12limb→∞b2=∞.
s≠0
s<0,b→∞,e−sb→∞
limb→∞b↑∞ e−sb↓∞=∞.
s<0.s>0.
b,s−1s
b→∞,(7−s)⏞negativeb→−∞e(7−s)⏞negativeb→0
L{t2}= ∫0∞e−st⋅t2dt=(1)t2⋅−1se−st|0∞−∫0∞−1se−st⋅2tdt=(2)−1s[t2e−st|0∞−2∫0∞e−st⋅tdt]=(3)−1s[t2e−st|0∞−2L{t}]=(4)−1s[limb→∞t2e−st|0b−2L{t}]= −1s[limb→∞(b2e−sb−02⋅e−s⋅0)−2⋅1s2]= −1s[limb→∞b2e−sb−0−2s2]= −1s[limb→∞b2esb−2s2]=(5)−1s[limb→∞2bsesb−2s2]= −1s[2slimb→∞besb−2s2]= −1s[2slimb→∞1sesb−2s2]= −1s[2s⋅0−2s2]= 2s3
L{t3}= ∫0∞e−st⋅t3dt=(1)t3⋅−1se−st|0∞−∫0∞−1se−st⋅3t2dt=(2)−1s[t3e−st|0∞−3∫0∞e−st⋅t2dt]=(3)−1s[limb→∞t3e−st|0b−3L{t2}]=(4)−1s[limb→∞(b3e−sb−03⋅e−s⋅0)−3⋅2s3]= −1s[limb→∞b3e−sb−0−6s3]= −1s[limb→∞b3esb−6s3]=(5)−1s[limb→∞3b2sesb−6s3]= −1s[3slimb→∞b2esb−6s3]= −1s[3slimb→∞2bsesb−6s3]= −1s[6s2limb→∞besb−6s3]= −1s[6s2limb→∞1sesb−6s3]= −1s[6s2⋅0−6s3]= 6s4
L{t4}= ∫0∞e−st⋅t4dt=(1)t4⋅−1se−st|0∞−∫0∞−1se−st⋅4t3dt=(2)−1s[t4e−st|0∞−4∫0∞e−st⋅t3dt]=(3)−1s[limb→∞t4e−st|0b−4L{t3}]=(4)−1s[limb→∞(b4e−sb−04⋅e−s⋅0)−4⋅6s4]= −1s[limb→∞b4e−sb−0−24s4]= −1s[limb→∞4b3sesb−24s4]= −1s[4slimb→∞b3esb−24s4]=(5)−1s[4slimb→∞3b2sesb−24s4]= −1s[12s2limb→∞b2esb−24s4]= −1s[12s2limb→∞2bsesb−24s4]= −1s[24s3limb→∞besb−24s4]= −1s[24s3limb→∞1sesb−24s4]= −1s[24s3⋅0−24s4]= 24s5
Key Steps
=[1]
Integration by parts with u=f(t) and dv=e−stdt
=[2]
s is a constant in this integral, so we can bring it out.
=[3]
e−stf(t)=f(t)est must go to zero.
∫L1.
L{1}=1s,s>0
∫L2.
L{eat}=1s−a,s>a
∫L3.
L{tn}=n!sn+1,s>0
∫L4.
L{sin⁡(bt)}=bs2+b2,s>0
∫L5.
L{cos⁡(bt)}=ss2+b2,s>0
∫L6.
L{tneat}=n!(s−a)n+1,s>a
∫L7.
L{eatsin⁡(bt)}=b(s−a)2+b2,s>a
∫L8.
L{eatcos⁡(bt)}=s−a(s−a)2+b2,s>a
∫L9.
L{af(t)+bg(t)}=aL{f(t)}+bL{g(t)},a,b constant
∫L10.
L{f′(t)}=sF(s)−f(0)
∫L11.
L{f″(t)}=s2F(s)−sf(0)−f′(0)
∫L12.
L{f(n)(t)}=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0)
∫L13.
L{tnf(t)}=(−1)ndndsn(L{tnf(t)})
∫L14.
L{U(t−a)}=e−ass
∫L15.
L{f(t)U(t−a)}=e−asL{f(t+a)}
∫L16.
L{f(t−a)U(t−a)}=e−asF(s)

graph of e−1.2b vs. e+1.2b.

L14:L{U(t−a)}=e−assL15:L{f(t)U(t−a)}=e−asL{f(t+a)}L16:L{f(t−a)U(t−a)}=e−asF(s)
4s3−13s2+74s+27= (As+B)(s+1)2+C(s+1)(s2−6s+25)+D(s2−6s+25)= (As+B)(s2+2s+1)+(Cs+C)(s2−6s+25)+Ds2−6Ds+25D= As3+2As2+As+Bs2+2Bs+B+Cs3−6Cs2+25Cs+Cs2−6Cs+25C+Ds2−6Ds+25D= (A+C)s3+(2A+B−6C+C+D)s2+(A+2B+25C−6C−6D)s+(B+25C+25D)= (A+C)s3+(2A+B−5C+D)s2+(A+2B+19C−6D)s+(B+25C+25D)
Equating coefficients gives us four equations in four unknowns.
A+C= 42A+B−5C+D= −13A+2B+19C−6D= 74B+25C+25D= 27A= 4−C2(4−C)+B−5C+D= −13(4−C)+2B+19C−6D= 74B−7C+D= −212B+18C−6D= 70B= 7C−D−212(7C−D−21)+18C−6D= 70(7C−D−21)+25C+25D= 2732C−8D= 11232C+24D= 4832C= 8D+112(8D+112)+24D= 4832D= −64D= −232C= 8(−2)+112= 96C= 3B= 7(3)−(−2)−21= 2A= 4−(3)= 1
Partial fraction decomposition has the form,
54(s+1)(s2−4s+13)=As+1+Bs+Cs2−4s+13,
and we find A, B, and C by
54= A(s2−4s+13)+(Bs+C)(s+1)54= As2−4As+13A+Bs2+Bs+Cs+C0s2+0s+54= (A+B)s2+(−4A+B+C)s+(13A+C)
A+B= 0−4A+B+C= 013A+C= 54A= −B−4(−B)+B+C= 013(−B)+C= 545B+C= 0−13B+C= 54C= −5B−13B+(−5B)= 54−18B= 54B= −3A= −(−3)C= −5(−3)= 3= 15
x(t)= 3e−t−3e2tcos⁡(3t)+3e2tsin⁡(3t)= 3e−t+e2t[3sin⁡(3t)−3cos⁡(3t)]x′(t)= −3e−t+e2t[9cos⁡(3t)+9sin⁡(3t)]= −3e−t+e2t[9cos⁡(3t)+9sin⁡(3t)+6sin⁡(3t)−6cos⁡(3t)]= −3e−t+e2t[3cos⁡(3t)+15sin⁡(3t)]x″(t)= 3e−t+e2t[−9sin⁡(3t)+45cos⁡(3t)]= 3e−t+e2t[−9sin⁡(3t)+45cos⁡(3t)+6cos⁡(3t)+30sin⁡(3t)]= 3e−t+e2t[21sin⁡(3t)+51cos⁡(3t)]
LHS= x″−4x′+13x= (3e−t+e2t[21sin⁡(3t)+51cos⁡(3t)])= 3e−t+e2t[21sin⁡(3t)+51cos⁡(3t)]= 3e−t+e2t[21sin⁡(3t)+51cos⁡(3t)]= 54e−t= RHS
We also verify the initial conditions:
x(0)= 3e−0+e2⋅0[3sin⁡(3⋅0)−3cos⁡(3⋅0)]= 3+[3⋅0−3⋅1]= 3−3= 0x′(0)= −3e−0+e2⋅0[3cos⁡(3⋅0)+15sin⁡(3⋅0)]= −3+[3⋅1+15⋅0]= −3+3= 0

L{eat} Details.

Using the definition of the Laplace transform:
L{eat}= ∫0∞e−st⋅eat dt= limb→∞∫0be(−s+a)t dt
For the integral to converge, the exponent −s+a must be negative, leading to the condition s>a. Proceeding with the integration:
L{eat}= limb→∞∫0be(−s+a)t dt= limb→∞1a−se(−s+a)t|0b= 1a−s[limb→∞(e(a−s)b−1)]= 1a−s[0−1]= −1a−s=1s−a.for s>a
Thus, the Laplace transform of eat is:
L{eat}=1s−a,s>a.

L{t} Details.

Using the definition of the Laplace transform:
L{t}=limb→∞∫0be−st⋅t dt⏟I
Integration by parts on I, gives us
L{t}= limb→∞[−bse−sb+1s∫0be−stdt]L{t}= −1slimb→∞[−besb]+1slimb→∞∫0be−st dtL{t}= −1slimb→∞besb⏟L+1s∫0∞e−st dt⏟L{1}
L’Hopital’s Rule shows L=0 and L{1} is known. Therefore,
L{t}=1s2,s>0.
u=t,dv=e−stdt,du=dt,v=−1se−st
∫0be−st⋅t dt= t⋅(−1se−st)|0b−∫0b(−1se−st)dt= −bse−sb+1s∫0be−stdt
bbss>0,
L=limb→∞b↑∞esb↓∞=LHlimb→∞1sesb=1slimb→∞1esb↓∞=0

L{t2} Details.

Using the definition of the Laplace transform:
L{t2}=limb→∞∫0be−st⋅t2 dt⏟I
Integration by parts on I, gives us
L{t}= limb→∞[−b2se−sb+2s∫0be−st⋅t dt]L{t}= −1slimb→∞[−b2esb]+2slimb→∞∫0be−st⋅t dtL{t}= −1slimb→∞b2esb⏟L+2s∫0∞e−st⋅t dt⏟L{t}
L’Hopital’s Rule shows L=0 and L{t}=1/s2. Therefore,
L{t2}=2s3,s>0.
u=t2,dv=e−stdt,du=2t dt,v=−1se−st
∫0be−st⋅t2 dt= t2⋅(−1se−st)|0b−∫0b(−1se−st)2t dt= −b2se−sb+2s∫0be−st⋅t dt
s>0,
L=limb→∞b2↑∞esb↓∞ =LH limb→∞2bsesb =2slimb→∞b↑∞esb↓∞=LH 2slimb→∞1sesb=2s2limb→∞1esb↓∞=0

factorial.

n,n!,n.5!=5⋅4⋅3⋅2⋅1=120.
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