Laplace Transforms

The Laplace Transforms Calculator allows you to see all of the Laplace Transform equations in one place! The Laplace Transform equations are for functions of certain forms, as denoted in their name. Due to the many different uses for Laplace Transforms, beyond just Differential Equations, we hope that you find this calculator as helpful as we have found it!

• `mathcal{L}(t^{n})(s)`
• `\mathcal{L} (t^{n}e^{at}) (s)`
• `mathcal{L}(sin(at))(s)`
• `mathcal{L}(e^{at}sin(bt))(s)`
• `\mathcal{L} (e^{at}cos(bt)) (s)`
• `\mathcal{L} (cos(at)) (s)`
• `\mathcal{L} (e^{at}) (s)`

Laplace Transforms and Differential Equations

Laplace Transforms "operate on a function to yield another function" (Poking, Boggess, Arnold, 190).
Given a function `f(t)` from `0<t<infty`, the Laplace Transform is:
     `mathcal(L)(f)(s)=F(s)=int_0^infty f(t)e^(-st) dt " for " s>0`

To find the Laplace Transform of a function, `f(t)`, you have to evaluate an improper integral. We will not discuss how you would evaluate this type of integral, but rather we will discuss why Laplace Transforms are useful for differential equations. Our calculator gives you what the Laplace Transform is based on functions of a certain form. 

Since a Laplace Transform is taking a function and "transforming" it into another function, Laplace Transforms are valuable for finding solutions to differential equations that are made up of linear, continuous functions, and discontinuous functions. 

There are many uses for Laplace Transforms, but we are not going to go into heavy detail about how to use them, rather we are going to state some basic rules regarding Laplace Transforms. We will also provide an example of how to find a solution to an initial value problem using Laplace Transforms. 

The Laplace Transform of `frac(dy)(dt)` is as follows, when given a function y(t) with Laplace Transform `cc"L" [y]` :

     `cc"L"[frac(dy)(dt)]=s cc"L"[y]-y(0).` 

Laplace Transforms have linearity in the following way when applied to functions `f`, `g`, and constant `c`:

     `cc"L" [f+g]=cc"L"[f] +cc"L"[g]`
     `cc"L"[c*f]=c* cc"L"[f]`

Example:

Given the initial-value problem
     `frac(dy)(dt) = y-4e^(-t),  y(0)=1, `

the Laplace Transform of both sides is
     `cc"L" [frac(dy)(dt)]= cc"L" [y-4e^(-t)].`

Using the formula(s) with regards to derivatives and linearity we have
     `cc"L"[frac(dy)(dt)]=s cc"L"[y]-y(0) = cc"L"[y]-4 cc"L" [e^-t]`.

Substituting for the initial condition we have
     `s cc"L"[y] -1 = cc"L"[y]-4 cc"L" [e^-t],`

and we know that the Laplace Transform for `e^(at) =frac(1)(s-a),` as you can discover with our calculator, yielding
     `s cc"L"[y] -1 = cc"L"[y]- frac(4)(s+1).`

Subtracting ` cc"L"[y]` to the left side and factoring we get
     `cc"L" [y] = frac(1)(s-1)-frac(4)((s-1)(s+1)).`

Now we do not have the final answer, the solution to the differential equation, but we are close! Unfortunately, our calculator only calculates Laplace Transforms of certain functions and does not do some of the previous steps nor the following steps of the initial value problem for you. However, following along with this page will help you gain a better understanding of how to find solutions based on Laplace Transforms-some of which is work that cannot be done with most calculators. 

With that said, we need to discuss taking the inverse Laplace transform. The inverse Laplace Transform is moving from Laplace Transform to function, denoted ` cc"L"^(-1).` Linearity holds for inverse Laplace Transforms as well as for Laplace Transforms due to a uniqueness property that applies to both. 

Following that, we know that we need to match up functions' inputs to outputs, such that we recognize how the Laplace Transform we found correlates to functions that you could put into the calculator we have above. 

     `cc"L"^(-1)[cc"L"[y]]=y=cc"L"^(-1)[frac(1)(s-1)]-cc"L"^(-1) [frac(4)((s-1)(s+1))]`

From the formula `cc"L"[e^(at)]` we know that
     `cc"L"^(-1)[frac(1)(s-1)]=e^(t),`

but we do not know what the value of the following is yet:
     `cc"L"^(-1)[frac(4)((s-1)(s+1))]`

To compute the inverse Laplace Transform we need to recognize that we should use partial fractions to find the individual functions formed with respective denominators of either `frac(A)((s-1))` or `frac(B)((s+1))` and not `frac(4)((s-1)(s+1)).` Blanchard, Devaney, and Hall write that, "The idea is to rewrite the term `frac(4)((s-1)(s+1))`as a combination of functions that are known Laplace transforms. In this case (and quite often when using Laplace transforms) we use the technique of partial fractions" (568).

With partial fractions we set the following equal to each other, solving for the constants A and B. 

     `frac(4)((s-1)(s+1))=frac(A)((s-1))+frac(B)((s+1))`

However, we can see that `frac(A)(s-1)` and `frac(B)(s+1)` do not have the same denominator. We find a common denominator for both by multiplying the numerators as follows `A*(s+1)` and `B*(s-1)` and making both denominators `(s+1)(s-1)`,  yielding:

     `frac(4)((s-1)(s+1))=frac(A (s+1))((s+1)(s-1))+frac(B (s-1))((s+1)(s-1)).`

We can see that the denominator on the left and right sides of the equation are the same, which means we can work with just the numerators to solve for A and B because the denominators cancel. Doing this we see that:

     `4 = A(s+1) +B(s-1)=As+Bs+A-B`

We can factor out the `s` so that we have 

     `4=s(A+B)+A-B`

We can separate this into a system of equations and solve, based on the factorization of `s` on both sides. 

     `4 = A - B`
     `0= A + B`

We set the left side equal to zero because there is no `s`-term on the left side.
Solving for A and B, we see that `A=2` and `B=-2`.

We now have
     `frac(4)((s+1)(s-1))= frac(2)(s-1)-frac(2)(s+1)`

Hence we now see that
     `cc"L"^(-1)[frac(4)((s+1)(s-1)) ] = cc"L"^(-1)[ frac(2)(s-1)]-cc"L"^(-1)[ frac(2)(s+1)]`

     `cc"L"^(-1)[frac(4)((s+1)(s-1)) ] = 2 cc"L"^(-1)[ frac(1)(s-1)]-2 cc"L"^(-1)[ frac(1)(s+1)]`

     `cc"L"^(-1)[frac(4)((s+1)(s-1)) ] = 2e^(t)-2e^(-t).`

Putting it all together, we know that
     `y=cc"L"^(-1)[frac(1)(s-1)]-cc"L"^(-1) [frac(4)((s-1)(s+1))]`
     `y= e^(t) -( 2e^(t)-2e^(-t))`
     `y=-e^(t)+2e^(-t)`

Sources

Blanchard, Paul, Robert L. Devaney, and Glen R. Hall. Differential Equations. 3rd ed. Belmont, CA: Thomson Brooks/Cole, 2006. Print. 

Polking, John C., Albert Boggess, and David Arnold. Differential Equations with Boundary Value Problems. Upper Saddle River, NJ: Pearson/Prentice Hall, 2006. N. pag. Print.