Section 6.1 & 6.2

The Table of Elementary Laplace Transforms

Table #11

Integrate[t^n Exp[-(s-a)t],{t,0,Infinity}]
Integrate[t^n Exp[-(s-a)t],{t,0,Infinity},Assumptions->{Re[n]>-1,Re[a]<Re[s]}]

Table #19

F[s_]=Integrate[Exp[-s t] f[t],{t,0,Infinity}]
Derivative[1][F][s]
Derivative[2][F][s]
Derivative[3][F][s]

Integrate[Exp[-s*t]f[t],{t,0,Infinity}]

Mathematica Basics for Laplace Transforms

The Table of Laplace transforms that is used has elements that (for the most part) can be verified by direct application of the definition of Laplace transform and integration techniques. We will use Mathematica to convince ourselves that some of these are true, but we could do these by hand using technqiues for improper integrals in Calculus II.

Using Definition of Laplace Transform Table #11: ℒ[ (t^n)(e^(a t))]=((n!)/(((s-a))^(n+1))), s>a

Let's use the definition of Laplace transform to figure out the Laplace transform of (t^n)(e^(a t))

Integrate[t^n Exp[-(s-a)t],{t,0,Infinity}]

We got some conditions on the integration, which is not unusual for improper integrals. We can include the conditions as Assumptions if we like.

Integrate[t^n Exp[-(s-a)t],{t,0,Infinity},Assumptions->{Re[n]>-1,Re[a]<Re[s]}]

We can use Mathematica to verify the results in Table 6.2.1, although I have asked you to verify some of the results by hand on Assignments and in the homework.

Using Definition of Laplace Transform Table #19: ℒ[ (((-t))^n)f(t)]=(F^((n)))(s)

F[s_]=Integrate[Exp[-s t]f[t],{t,0,Infinity}]

Table[Derivative[i][F][s],{i,0,4}]//TableForm

We can see the pattern, but to prove it we would need to use induction.

Derivative[n][F][s]

Built-in Commands for Laplace Transforms

You can use the following commands to determine Laplace transforms and inverse Laplace transforms:

LaplaceTransform[Cos[a t]Exp[b t],t,s]
InverseLaplaceTransform[%,s,t]

LaplaceTransform[Cos[ t]Exp[2 t],t,s]
InverseLaplaceTransform[%,s,t]

What happened? Well, there is a result for cosine that will show this result (which is a complex valued function) is the same as Table 6.2.1 entry 10. Both these results follow from Euler's relation:

Exp[i t]=Cos[t]+i Sin[t]

Simplify[Cos[t]==((Exp[I t]+Exp[-I t])/2)]
Simplify[Sin[t]==((Exp[I t]-Exp[-I t])/(2I))]

The intrinsic commands for Laplace transforms can be useful since the will extend what's in Table 6.2.1 for you.

LaplaceTransform[Cos[t] Sin[t]^2,t,s]
InverseLaplaceTransform[%,s,t]

Partial Fractions in Mathematica

The Apart command will do partial fractions for you.

Apart[((s^2-7s+11)/((s-2)(s-1)(s-4)))]

The partial fractions are necessary only if you are using Tables to determine Laplace transforms. The partial fractions will get it in a form that appears on the Table. If you are using Mathematica to do the transforms, the partial fraction step can be skipped.

InverseLaplaceTransform[((s^2-7s+11)/((s-2)(s-1)(s-4))),s,t]

Solve the IVP y''+2y'+y=(e^(-t)), y(0)=1,y'(0)=3.

Doing this using Laplace transforms replaces a solution that would involve 
substituting y=(e^(r t))into the homogeneous equation, 
using undetermined coefficients to determine a particular solution,
using the initial conditions to determine the value of the constants in the general solution.

Here is the bit we can easily do by hand, which is to take the Laplace transform of the differential equation. You will need to use Table 6.2.1, or Mathematica to determine the Laplace transforms. Table 6.2.1 will be provided on the final exam.

L [y'']+2L[y']+L [y]=L [(e^(-t))]

LaplaceTransform[y''[t],t,s]

Using the initial conditions, this reduces to L [y'']=(s^2)Y(s)-s-3.

LaplaceTransform[y'[t],t,s]

Using the initial conditions, this reduces to L [y']=s Y(s)-1.

LaplaceTransform[Exp[-t],t,s]

This leads the the following relationship:

(s^2)Y(s)-s-3+2(s Y(s)-1)+Y(s)=(1/(1+s))
(s^2)Y(s)-s+2s Y(s)-5+Y(s)=(1/(1+s))
Y(s)((s^2)+2s +1)=(1/(1+s))+(s+5)
Y(s)((s^2)+2s +1)=(1/(1+s))+(((s+5)(1+s))/(1+s))
Y(s)((s^2)+2s +1)=(((s^2)+6s+6)/(1+s))
Y(s)=(((s^2)+6s+6)/((1+s)((s^2)+2s +1)))

Now we need to use partial fractions:

Apart[(((s^2)+6s+6)/((1+s)((s^2)+2s +1)))]

And our relationship becomes:

Y(s)=(1/(((1+s))^3))+(4/(((1+s))^2))+(1/(1+s))

Now, take the inverse Laplace transform:

(L^(-1))[Y(s)]=(L^(-1))[(1/(((1+s))^3))]+4(L^(-1))[(1/(((1+s))^2))]+(L^(-1))[(1/(1+s))]

Again, you may use Table 6.2.1 or Mathematica to work out the inverse transformations.

InverseLaplaceTransform[(1/(((1+s))^3)),s,t]

InverseLaplaceTransform[(1/(((1+s))^2)),s,t]

InverseLaplaceTransform[(1/(1+s)),s,t]

Combining all these together, we get the final result for the solution to the initial value problem:

y[t_]=(1/2) (E^(-t)) (t^2)+4(E^(-t)) t+(E^(-t))

Here is a plot of the solution:

plot1=Plot[y[t],{t,0,10},PlotStyle->{Red,Thick}]

Here is Mathematica's DSolve solution, which looks a bit different from ours:

Clear[y]
DSolve[{y''[t]+2y'[t]+y[t]==Exp[-t],y[0]==1,y'[0]==3},y[t],t]