Laplace Transforms
Tuesday, July 19, 2022
2:00 PM
The Definition of the Laplace Transform
The Laplace transform is a method that lets you turn a differential equation into an algebraic one,
among other uses. It is an operator that transforms one function into a different type of function. It also
automatically takes care of initial conditions. It is especially useful with a step function, which is a
discontinuous function where the value instantaneously changes.
Given an appropriate function f(t), its Laplace transform is defined as
Where s is a parameter greater than 0. So when we find a Laplace transform, it is always an expression
in terms of s. Sometimes the limit that comes from this integral can't be evaluated unless we assume s is
greater than some other value, like 3, and this is a perfectly acceptable when finding Laplace transforms
so long as the assumption is stated in the final answer.
Once we solve the "algebraic version" of the differential equation obtained by finding its transform, we
convert it back by taking the inverse Laplace transform, denoted as
.
Properties of the Laplace Transform
The Laplace transform of a function exists so long as that function is:
1. Piecewise continuous
2. Of exponential order, which means that the absolute value of the function is bounded by some
other function of the form
, where M and c are constants and e is Euler's number.
The Laplace transform can be factored out or distributed, the terms inside can be split up, and constants
can be pulled out, just like with integration:
Laplace Transforms With a Table
In practice, Laplace transforms are not typically calculated using the indefinite integral definition.
Instead, we employ the use of tables, in the same way that integrals are often evaluated using
integration tables. Examples using a table will come later. That means that unless a problem requires
you to find the Laplace transform using the definition (that is, the improper integral), Laplace
transforms should not be calculated this way. Use a table instead.
Example 1: Laplace Transform of a Piecewise Function Via the Definition
Find the Laplace transform of the piecewise function
using the definition.
1. We need one Laplace transform for each piece of the piecewise function, using the formula
. We can simply find the transform for each piece, and then sum them together for
the transform for the whole function:
. Notice that the
limits of integration reflect the domain for each piece of the function. The first integral, for f(t) = t,
has bounds from 0 to 1, since the function is defined as t only on that interval. From 1 to infinity,
the function is simply 1, and the second integral's bounds reflect that.
, either by parts of using tabular integration. We get
2. Integrate
so we now have the Laplace transform for the first part of the function.
3. Integrate
rewrite it as
. We always evaluate improper integrals with limits, so we first need to
. Evaluating by u-substitution, we have
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.
rewrite it as
. Evaluating by u-substitution, we have
.
4. The key to evaluating the limit is to remember that s must be greater than 0. That means
must have a negative exponent (b is also positive since it approaches infinity) , so we can rewrite
as . Now as b approaches infinity, the whole exponent approaches infinity, and therefore
the whole denominator approaches infinity. That means at infinity the expression
approaching
limit,
, or
. Since
is
is 0, the whole term approaches 0. The other term in the
, doesn't have a b in it, so it approaches itself.
5. Add the transform for each piece of the piecewise function:
. Simplifying, we get
. This holds so long as s > 0.
Example 2: Laplace Transform Via the Definition
Find the Laplace transform of
using the definition.
1. Start by plugging the function into the definition for the Laplace transform:
. Simplify by adding the exponents:
. We'll also factor the exponent:
.
2. Evaluate the integral by parts or tabular integration. We are left with the limit:
3. To evaluate this limit, we note that if
had a negative exponent, we would be able to
rewrite it as
. Since b must be positive as it approaches infinity, we can ensure a negative
exponent so long as s > 4. So, we'll assume s > 4. Then the whole denominator of the expression
goes to , which is 0. So then every term in our limit involving
approaches 0,
leaving us with
approaches
. There are no more b's in this limit, so it simply
. That's the answer:
, assuming s > 4.
Laplace Transforms With Tables
The following table has the most common Laplace transforms and will be given on the test:
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Although using a table can make it possible to find a Laplace transform in 10 seconds rather than 10+
minutes, it can take a little practice to be able to use the table correctly.
Trig Identities
There are three trig identities that especially come up for these types of problems:
•
•
•
These identities are so useful because we have table entries for expressions like
and
,
but not for other trig expressions. So these identities let us rewrite certain expressions to make them fit
the table.
It's also helpful to know the exponential definitions of the hyperbolic sine and cosine:
•
•
Functions Multiplied by a Power of t
With the exception of
, we don't have any useful table entries for factors multiplied by a power of
t, such as
. How is the Laplace transform calculated here? We can take the negative of the
derivative of the other factor's transform (that is the derivative of
in the example). In
general:
What this means:
1. Identify the two factors in the function, which are 1) and 2) everything in the function that's not
, which we call
.
, as if the term weren't there.
2. Find the Laplace transform
, and add a negative sign out front
3. Take n derivatives of the result from the previous step,
if n is odd. Or you could simply add a negative sign after taking every derivative, and it'll cancel out
when n is even anyway.
Example 1: Simple Laplace Transforms Using a Table
Find the Laplace transforms
.
For the first transform, apply table entry #1,
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. We can pull constants out of the Laplace
1. For the first transform, apply table entry #1,
transform operator, so
. We can pull constants out of the Laplace
.
2. For the next one, apply #3,
. Plugging in n = 2, we get
. Simplify:
.
3. For
, apply #2, which is
. Plugging in a = 15, we get
4. For the last one, apply #6, which is
.
. Plugging in a = 2, we get
.
Example 2: Trig
Find the Laplace transform
.
1. This expression doesn't look like anything in our table. Can we rewrite it in a way that is more
familiar? If we apply the identity
. Divide both sides by 2:
. We'll also substitute in 4t in place of x:
2. Nothing in the table matches
1/2, we get
.
, but what about
? Pulling out the constant
, which matches entry #5,
. Plug in a = 8 and
multiplying by 1/2, we get the Laplace transform
. Simplify:
.
Example 3: Hyperbolic Trig
Find the Laplace transform
.
1. Once again we have an expression that doesn't resemble any entries in the table. If we rewrite the
sinh in terms of its exponential definition, it might simplify or cancel with the to give us
something recognizable. We get
. Simplify:
2. Do we have an entry for
, which is
.
? If we break this up into two transforms, yes:
. Now we can apply entry #2 to the first transform, and if we factor out the 1/2 from
get
, we can apply #1. We get
to
.
Example 4: Trig
Find the Laplace transform
.
1. Since this doesn't match any entry in the table, we'll apply the identity
Substituting 4t in place of x, we get
2. So now we want to find
.
.
. This nearly matches entries #6 and #1. We can factor out the
2 from the denominator:
. And we can split this into two transforms:
. Now it matches the table. Applying #1 and #6, we get
and
. Adding these together, the final answer is
.
Example 5: Differentiation
Find the Laplace transform
1. Start by finding
.
. Using table entry #5, we get
2. Find the negative derivative of the result:
. Since the t factor is raised to the 1st
power, we only take a first derivative. We can rewrite it as
quotient rule. We get
, which we can write as
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.
. No need for the
. That is the final answer.
quotient rule. We get
, which we can write as
. That is the final answer.
Example 6: Translation
Find the Laplace transform
.
. It also fits #10, but we'll use #14 for
1. This fits table entry #14, which is
practice. This rule tells us to find the transform of the factor f(t), i.e. the factor that doesn't involve
e, which in this case is cos(2t). We find
, applying entry #6, is
.
2. Next we replace every s with
where c is the coefficient on the exponent of the e factor,
which is 3. So replace every s with s - 3:
.
Example 7: Polynomial
Find the Laplace transform
.
1. First, let's expand the polynomial:
Laplace transform
.
2. Break the terms up into 4 Laplace transforms:
entries #1 and #3 to get
. Now we are looking for the
. We can apply
. Simplify:
.
Inverse Laplace Transforms
After we're done working with the Laplace transform of a function, we'll need to convert it back to its
original form using an inverse Laplace transform. While a function that we want to find the transform of
may be denoted as
, a transform that we want to find the inverse of is often denoted
.
Review: Partial Fraction Decomposition
Inverse Laplace transforms tend to require the use of partial fraction decomposition. Remember these
steps:
1. Factor the denominator into linear, irreducible quadratic, linear repeated, and/or irreducible
quadratic repeated factors.
a. A linear factor looks like this:
. The variable isn't raised to a power, nor is the whole
expression.
. The variable is squared, but the whole
b. An irreducible quadratic factor looks like
expression isn't.
. The variable isn't raised to a power, but the
c. A linear repeated factor looks like
whole expression is, which makes it repeated.
. The variable and the whole
d. An irreducible quadratic repeated factor looks like
expression are both squared.
2. Once the denominator is factored, split the fraction up, giving each factor its own fraction. But
repeated fractions are split up into the same number of fractions as their degree. In the
numerator of the fraction:
a. If the denominator has a linear factor, give the fraction a constant numerator, e.g. .
b. If the denominator has a quadratic factor, give the fraction a linear numerator, e.g.
.
c. Linear/quadratic repeated factors get as many fractions as their degree. For example,
would become
. Three fractions, because the degree is 3.
The same applies for a quadratic. The factor
would be split up as
. All the numerators are linear terms of the form
because
the x is squared, not because the term is repeated.
3. Multiply all the new fractions by the denominator of the original fraction (after it was factored).
This will make the denominators all cancel. For example,
. Multiplying the denominator across, we have
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splits into
.
4. Find all the variables A, B, C, ….
a. The simplest method is to plug in values for your variable that make all but one term cancel.
For example, plugging x = 7 into
will make the term 0 and eliminate the unknown
B, possibly leaving you free to then solve for A.
b. If this doesn't work, you can equate coefficients. For example, the equation
makes it clear that A = 7, B = 3 because the coefficients are aligned to make it
obvious.
5. Plug the values of the unknowns into each new fraction, completing the decomposition.
Additionally, if the degree of the numerator is greater than the degree of the denominator, you must
use long division to force the numerator to have a lesser or equal degree to the denominator. But none
of the problems in this section will need this treatment.
Example 1
Find
.
1. Which table entry does this expression resemble? It looks like #5, which is
But what is the value of a? Is it 6 or the square root of 24? Let's make
the numerator so that it also is
, by multiplying by it. We'll divide by
and then adjust
outside the L-
operator to preserve equality, and pull out the 6 in the numerator:
2. Now the expression fits entry #5, so plugging in
.
.
, the solution is
.
Example 2
Find
.
1. We'll start by expanding the polynomial in the numerator:
.
2. Break up the polynomial into three inverse transforms:
and pull out the constants:
. Simplify
.
3. All of these transforms (mostly) fit the table. Applying entry #1 to the first one, we get
. We can use entry #3 on
if we rewrite it into the form
. We must have n = 1:
. We can use #3 again on
if we again make it fit
. The
denominator of the inverse transform has s raised to the 3rd power, so subtracting 1, we get n = 2.
That gets us
equality:
, but our numerator is 1, not 2!, so we must divide by 2 to preserve
.
4. Add all the transforms together for the final answer:
.
Example 3
Find
, but our expression has a
1. This expression resembles entry #6, which is
coefficient of 4 in the denominator and the numerator. We need to remove both of them. First
pull out the 4 in the numerator to get
.
Factoring out 4 from the denominator, we get
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. Then we pull the 4 out:
.
2. Factoring out 4 from the denominator, we get
. Then we pull the 4 out:
3. Now that the expression matches the form
and the final answer is
, we see
so
.
. Plug a = 1/2 into
.
Example 4: Partial Fraction Decomposition
Find
.
1. Factor the denominator:
. Now we can apply partial fraction decomposition to find
the inverse transform. The linear term s will have a constant numerator A, and the quadratic term
will need a linear numerator
. We get
.
2. Multiply both sides by the denominator:
.
3. If we set s = 0, we get
, so A = 1/2. So now we have
. Simplify:
.
. Equating coefficients, notice that there is no s2 term on the
4. Simplify:
LHS of the equation, so then the coefficients of s2 on the RHS must add up to 0:
. Solve
for B: B = -1/2.
5. Equating coefficients again, there is no s term on the LHS, so then the coefficient of s on the RHS
must add up to 0:
.
6. Plugging in the other values, the decomposition is
. Does
fit any entries in the table? Applying entries #1 and #6, we get the answer:
.
Example 5: Partial Fraction Decomposition
Find
.
1. Apply partial fraction decomposition:
. Multiplying the denominator across,
we have
. Simplify:
.
2. Solving for A, B, and C, A + B = 8, C = -4, A = 3, and so B = 5. Plugging these numbers into the
decomposition, we have
.
3. Rewrite the terms to make it fit entries in the table:
. Applying table entries, we get
.
Example 6: Partial Fraction Decomposition
Find
.
1. Start by factoring the denominator:
. We have two linear terms in the
denominator, so the decomposition is
. Multiply the denominator across:
.
2. Plugging in s = 1, we get
. So
Plugging these numbers into the decomposition, we have
3. Rewrite the terms to make it fit entries in the table:
entry #2, which is
.
. Applying table
, we identify
for the second one. Then the answer is
. And plugging in s = -3, we see
for the first inverse transform and
.
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.
for the second one. Then the answer is
.
Solving Differential Equations With Laplace Transforms
The purpose of learning Laplace Transforms here is to apply them to solving differential equations. Using
a table is generally the only reasonable way to do these problems, because using the integral definition
would make these problems take very long to solve.
Laplace Transforms of a Function's Derivatives
We'll typically denote the Laplace transform of y as simply Y(s):
.
The Laplace transform of a function's first derivative is written in terms of the Laplace transform of the
original function and an initial condition.
And for second derivatives:
We'll use these formulas for solving differential equations with Laplace transforms.
How to Solve Differential Equations With Laplace Transforms
The general problem-solving steps are as follows:
1. Take the Laplace transform of both sides of the differential equation
,
, and
to make substitutions into the equation
2. Use the formulas for
3. Plug in the given initial conditions and solve for Y(s)
4. Take the inverse Laplace transform of Y(s)
Example 1
Solve
, y(0) = 6, y'(0) = 0.
1. We start by taking the Laplace transform of both sides:
. We'll pull
out constants and break it up into smaller transforms where possible:
.
2. Using the formulas for
and
, make the substitutions:
. We can also take care of the RHS with table entry #5:
. Making this substitution, we have fully transformed the equation:
.
3. Plug in the initial conditions:
. Solve for Y(s):
.
4. Find
. Starting with
, we use partial fraction decomposition. We get A
= 0, B = -2, C = 0, D = 2. The decomposition is
.
5. Add the decomposition to the other term:
make these terms fit the table.
a. The first inverse transform,
. Now we want to
, resembles #5:
. But since
in our expression, we need to force the numerator to be a, or
Now it fits, and the inverse of this transform is
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.
:
.
.
Now it fits, and the inverse of this transform is
b. Applying #5 to
, notice that
, so we force the numerator to match:
. The inverse of this transform is
c. Applying #6 to
out:
.
, we notice that
. Then the inverse is
so
, and the 6 needs to be pulled
.
6. Adding these three terms, the final answer is
.
Example 2
Solve
, y(0) = 1, y'(0) = 0.
. Note that
. We can
1. Take the Laplace transform of both sides:
split up the transform on the LHS into three:
.
and
, substitute:
2. Using the formulas for
. Simplify:
3. Plug in the initial conditions and solve for Y(s):
4. Find
. We can use partial fraction decomposition here. Factor the denominator:
. We get A = 4/3, B = -1/3. So the decomposition is
.
5. We need to find the inverse Laplace transforms
table entry #2, which is
other one is
. These both resemble
. The first inverse is
, where a = -1. The
, where a = -4.
6. Add the inverse transforms to get the final answer:
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.