How To Calculus
Part II: Return of The Integral (AM2270)
By Tom Machado
2
Table of Contents
1.1
Order and Linearity of DEs ................................................................................................................ 4
1.2
Existence and Uniqueness of a Solution ........................................................................................... 5
2.2
Separable DEs ................................................................................................................................... 6
2.3
First Order Linear DEs ....................................................................................................................... 7
2.4
Exact DEs ........................................................................................................................................... 9
Test of Exactness ..................................................................................................................................... 12
Converting to Exact DE with Integrating Factor...................................................................................... 12
2.5
Solving by Substitution ................................................................................................................... 13
Homogeneous DEs by Substitution......................................................................................................... 15
Bernoulli DEs by Substitution.................................................................................................................. 18
2.7
Applications of 1st Order Linear DEs ............................................................................................... 21
Population Growth.................................................................................................................................. 21
Radioactive Decay ................................................................................................................................... 21
Temperature Change .............................................................................................................................. 21
Solution Mixtures .................................................................................................................................... 22
RL and RC Circuits ................................................................................................................................... 22
Leaking Tank............................................................................................................................................ 22
3.1
Theory of Linear Equations ............................................................................................................. 23
Boundary Value Problems....................................................................................................................... 23
Superposition Principle ........................................................................................................................... 23
Testing for Linear Independence Using the Wronskian.......................................................................... 24
3.3
Homogeneous Linear DEs with Constant Coefficients.................................................................... 26
3.4
Method of Undetermined Coefficients........................................................................................... 28
3.5
Variation of Parameters .................................................................................................................. 31
3.6
Cauchy-Euler DEs ............................................................................................................................ 34
3.8
Applications of 2nd Order Linear DEs............................................................................................... 37
Mass-Spring Systems .............................................................................................................................. 37
LRC Circuits.............................................................................................................................................. 37
4.1
Laplace Transform........................................................................................................................... 38
4.2
Inverse Laplace Transform and Transform of Derivatives .............................................................. 39
Larch’s Theorem...................................................................................................................................... 40
3
4.3
Translation of Laplace Transforms .................................................................................................. 42
Translation Along the s Axis .................................................................................................................... 42
Translation Along the t Axis .................................................................................................................... 42
4.4
Additional Laplace Transform Properties and Operations ............................................................. 44
Derivatives of Transforms ....................................................................................................................... 44
Convolution ............................................................................................................................................. 45
Integral Equations ................................................................................................................................... 46
4.6
Systems of Linear Differential Equations ........................................................................................ 49
12.1
Orthogonal Functions ..................................................................................................................... 51
Norm of Functions .................................................................................................................................. 52
12.2
Fourier Series .................................................................................................................................. 54
Review and Other Useful Things ................................................................................................................. 57
Cover-up Method of Partial Fractions .................................................................................................... 57
Trig Identities .......................................................................................................................................... 58
Integration by Parts ................................................................................................................................ 59
Written by Tom Machado
4
1.1
Order and Linearity of DEs
The order of a differential equation is simply equal to the highest derivative in the differential equation.
For example, the order of
is 2 because
is the highest derivative, whereas the
order of
is 5 because
is the highest derivative. Below is the general form of an nth
order differential equation:
( )
Where ( )
variable.
( )
( )
(
)
( )
( )
( )
( ) are the coefficients of each term and ( ) is a function of the dependant
The linearity of a differential equation is determined by the exponent on the dependant variables,
similar to - but not the same as - how you determine if a polynomial is linear or non-linear. A differential
equation is linear if the exponent on each of its dependant variables is 1. It’s non-linear when the
exponent on any term is greater than 1. For example,
is linear, but
(
)
is not because once simplified it has a ( ) term which has an exponent
greater than 1.
tl;dr:


Order of DE = highest derivative in equation
Non-linear if a dependant variable has an exponent >1, linear otherwise
5
1.2
Existence and Uniqueness of a Solution
If you have some differential equation
define a rectangle
that contains (
solution on , but maybe more. If
(
). If (
) that passes through some point ( )
you can
) is continuous on , then there is at least one
is continuous on , then there is a single unique solution on .
Most often you will be asked to find where a unique solution for a differential equation is defined, so
here’s an example:
Let’s say we want to find where there is a unique solution for
√ . For this example, our
)
differential equation is (
(these will always be given
√ and our initial condition is ( )
to you in an actual question). First, we determine where ( ) is continuous:
. Therefore, (
By inspection, we should be able to see that √ is no longer defined when
has at least one solution when
.
Second, we find
)
and determine where it is continuous:
√
Once again, inspection tells us that this equation is only defined when
(
) has a unique solution when those conditions are met.
and
. This means that
Now we take into account one final condition: the area that the unique solution is defined on must
contain the initial condition ( )
. This means that even though (
we can’t include anything past the
contain the initial condition.
TM wrote this!
) and
are continuous when
asymptote because the resultant rectangle won’t
6
2.2
Separable DEs
A separable differential equation is an equation of the form: Tom wrote this!
( ) ( )
To solve a separable DE, rearrange the equation to separate the x and y terms and then integrate.
( )
∫
( )
( )
∫ ( )
Practical example:
∫
∫
Separable DEs can also come in other forms, such as the following:
( )
( )
( )
( )
However all of these forms can be rearranged into the general form above.
tl;dr:


Separable DEs are those where the two variables can be separated from each other
To solve: separate the variables; integrate; isolate for function; ???; profit!
7
2.3
First Order Linear DEs
st
A 1 order linear differential equation has the form:
( )
( )
To solve, you must create an integrating factor of the form:
∫ ( )
( )
And multiply the entire equation by this integrating factor.
∫ ( )
∫ ( )
( )
( )
∫ ( )
The left half of the equation is equivalent to the derivative of
substitute that in.
(
∫ ( )
( )
)
∫ ( )
(via product rule), so we can
∫ ( )
Then we integrate both sides of the equation and isolate for y.
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
Practical example:
∫
( )
(
)
∫
Continued on next page
8
tl;dr:

1st order linear DEs come in the form

To solve:
( )
( )
∫ ( )
o
o
Define an integrating factor ( )
Multiply the whole equation by ( )
o
Substitute
o
Integrate both sides of the equation and then solve for y
( ( )) for
( )
( ) ( )
9
2.4
Exact DEs
Exact differential equations come in one of the two following forms:
(
(
)
)
OR
(
)
(
)
A differential equation is exact only if it passes the test of exactness (see below).
If the DE doesn’t pass the test of exactness, it can be made exact using an integrating factor (see below).
To solve, start with either
(
) or (
) and integrate with respect to
(
(
(
)
)
∫
∫
(
respectively.
)
(
)
)
( )
Where ( ) is an unknown function that is constant with respect to
*If you chose to use (
or
(aka it contains no x terms).
) and integrate with respect to , then there will be ( ) instead of ( ).
Next, take the derivative of (
) with respect to
(
∫
if you chose (
(or
)
)).
( )
This equation is equal to whichever one of (
) or ( ) that you didn’t integrate earlier, so set
them equal to each other and solve for ( ) ( ( ) if you chose to use (
)).
∫
( )
(
)
(
( )
)
Next, integrate to find ( ) (or ( ) if you used (
( )
∫( (
)
∫
(
(
)
)
) at the start).
∫
(
)
)
Continued on next page
10
Finally, substitute the function you just found into the equation for (
(
)
(
∫
)
∫( (
)
) to finish.
(
∫
)
)
I know that was a rather hideous equation, so here’s a concrete example with explanations:
(
(
)
(
)
)
(
)
First use test of exactness to make sure it’s exact. For a full explanation of the test of exactness, see
below.
and are both continuous, so the DE is exact. Next chose whether to use (
) or ( ) and
integrate it with respect to or respectively. It doesn’t matter which one you choose, you’ll get the
same answer in the end. I’m going to use ( ).
(
)
(
∫
( )
)
( )
Now take the partial derivative with respect to the other variable. If you chose ( ) take the partial
derivative with respect to , but if you chose (
) then take the partial derivative with respect to .
( )
Now equate this to whichever function you didn’t chose at the start and isolate for
example the other function is ( ).
( )
( ). For this
( )
( )
Integrate to solve for ( )
( ).
( )
Continued on next page
11
Substitute ( )
( ) into the equation for (
(
) and rearrange for the constant to finish.
)
tl;dr:
(
(
)
)

Exact DEs have the form



Use the test for exactness to make sure the DE is exact
If it’s not exact, you can make it exact using an integrating factor
To solve:
o Pick either ( ) or (
) to work with. It doesn’t really matter which
o Integrate your chosen function with respect to (if you chose (
)) or (if you
chose (
))
o Add an unknown constant function on the end of your integration. The function will be
( ) if you chose (
) or ( ) if you chose (
)
o Take the partial derivative of the whole thing with respect to the same variable as your
( ) function. You should end up with a ( ) or ( ) term in your result
o Set this equation equal to whichever of (
) or ( ) you didn’t chose earlier and
isolate for ( ) or ( ) respectively
o Integrate to solve for ( ) ( )
o Substitute ( ) ( ) into the equation for (
) where it first appeared
o Isolate for the constant to finish
This particular type of DE seems really crappy when described on paper, but it’s actually not as
bad as it sounds when you actually do it. I encourage you to try it out for yourself; once you get
the hang of it it’s not so bad

or
(
)
(
)
12
Test of Exactness
)
( )
A differential equation (
continuous, where
is the partial derivative of
derivative of (
) with respect to .
For example, test if (
)
From the equation above we can see that
is exact only if
and they are both
(
) with respect to , and
is the partial
(
)
(
is exact:
)
and (
)
.
Take the partial derivatives and compare to see if they’re equal.
The partial derivatives are equal, so the equation must be an exact DE.
Converting to Exact DE with Integrating Factor
To convert a differential equation that failed the test of exactness into an exact DE, you must multiply it
by an integrating factor . The integrating factor will be a function of or .
( )
∫
( )
∫
Multiplying the original DE by the integrating factor will convert it into an exact DE.
For example, (
)
(
)
is not an exact DE.
∫
( )
( )
∫
( )
( )
Multiplying the equation by ( ) yields:
(
Which is an exact DE.
)
(
)
13
2.5
Solving by Substitution
Substitution can be used to solve differential equations of the following form:
(
Where ( ) is some function, such as ( ) , or
)
( ).
To solve, start by defining some substitutions:
( )
Next, take the derivative of
with respect to
to find
.
Substitute in ( ).
( )
You now have a separable DE. Separate and then integrate.
( )
∫
( )
Finally, rearrange the result to solve for . Remember to substitute to remove
from the equation!
And now for the concrete example:
(
)
Define substitution and its derivative:
Continued on next page
14
Substitute in ( ):
Separate and then integrate:
∫
(
Substitute to remove
)
from the equation:
(
)
Solve for :
(
)
tl;dr:

Use substitution with DEs of the form

To solve:
(
)
( )
o
Define substitutions
o
Take derivative of
o
Substitute in ( ) for
o
o
Separate so that all terms are on one side and all terms are on the other
Integrate and then isolate for to finish
and
with respect to
15
Homogeneous DEs by Substitution
A function (
) is homogeneous of degree
(
if and only if for any :
)
(
)
A homogeneous differential equation will come in the form:
(
(
Where both
(
) and (
)
)
) are homogeneous functions of the same degree.
To solve using substitution, start by defining the following:
Which allows us to do this:
(
(
)
)
(
(
)
)
.
/
.
/
. /
Next we define the following substitutions:
Using these substitutions, we can convert the original DE into a separable DE:
(
(
)
)
. /
( )
( )
( )
∫
( )
| |
Continued on next page
16
Now that the homogeneous DE has been converted into a separable DE, you can solve it using those old
techniques. Just remember to substitute back in!
Here’s an example to give you the idea:
Substitute in
and
:
Rearrange and solve as separable DE:
(
(
Substitute in
)
)
( )
and isolate for :
(
)
√
tl;dr:

Homogeneous DEs come in the form:

Both
(
(
)
(
(
)
)
) and ( ) are homogeneous functions, which means for any ,
(
)
Continued on next page
17
tl;dr cont’d:

To solve a homogeneous DE:
(
(
)
)
. /
o
Start by defining
o
Define the substitutions
o
o
Make the substitutions you just defined to convert the homogeneous DE into a
separable DE in terms of and
Solve as a separable DE
o
Use the substitution
o
Isolate for
,
to remove
, and
from your solution
18
Bernoulli DEs by Substitution
A Bernoulli DE is a differential equation of the form:
( )
Where
( )
.
To solve a Bernoulli DE, you must convert it into a 1st order linear DE and solve it using those methods.
First, divide the entire equation by
.
( )
( )
Next define the following substitutions:
(
)
→
Once you’ve defined the substitutions, use them to replace all
( )
Multiply both sides by (
terms in the DE.
( )
).
(
) ( )
(
) ( )
You now have a 1st order linear DE, which you can solve using those methods. Just remember to
substitute back into the equation when you’re done!
Here’s an example:
Rearrange it into Bernoulli form (not always necessary):
Continued on next page
19
Divide the entire equation by
.
Now that it’s been rearranged into Bernoulli form, we can see that
substitutions we’ll need.
Make the substitutions to replace all
Multiply by (
. Now we define the
terms.
).
Now solve as a 1st order linear DE.
∫
( )
( )
(
)
∫
Continued on next page
20
Now that the 1st order linear DE has been solved, substitute
back into the equation.
√
tl;dr:

Bernoulli DE is of the form

To solve:
o Divide the entire equation by
( )
( )
o
Define the substitutions
o
o
o
o
Substitute using the substitutions you just defined
)
Multiply the whole equation by (
You now have a 1st order linear DE that you can solve using those methods
Once you’ve solved it, substitute back into the equation using your earlier
substitutions isolate for
and
21
2.7
Applications of 1st Order Linear DEs
Population Growth
Population growth can be modelled using the logistic model, which has the form:
(
Where is the growth rate constant, and
)
is the carrying capacity or maximum population.
The form above can be rearranged into the Bernoulli DE below:
Radioactive Decay
Radioactive decay can be modelled using the following separable DE:
Where is the decay constant.
The solution to the decay equation is:
( )
Where
is the initial amount of radioactive material.
Temperature Change
Temperature change can be modelled using Newton’s cooling law (a separable DE):
(
Where
is the ambient temperature, and
)
is a constant that is always greater than 0.
The solution to the cooling law is:
( )
22
Solution Mixtures
The mixture of two solutions can be modelled with the following DE:
( )
Where
is the amount of solute as a function of , ( )
is the rate of change of the amount of solute,
is the total volume as a function of ,
are the volume of solution going in and out per unit of
time respectively, and
is the concentration (mass per volume) of solute going in per unit of time.
Rearrange into a 1st order linear DE and solve as such:
( )
RL and RC Circuits
RL and RC circuits are modelled using 1st order linear DEs. RL circuits are modelled using:
( )
Where
is the inductance,
is the resistance, and
are the current and rate of change of current
respectively, and ( ) is the impressed voltage as a function of .
RC circuits are modelled using a similar equation:
( )
Where
is the capacitance,
is the resistance,
and
are the charge and rate of change of charge
respectively, and ( ) is the impressed voltage as a function of .
Leaking Tank
Leaking tank problems can be solved using the separable DE below:
√
Where
is the area of the hole in the tank,
may not be a function of ),
respectively, and
solves to:
and
is the cross sectional area of the water (which may or
are the height of the water and the rate of change of the height
is acceleration due to gravity (9.8m/s2 or 32ft/s2). If
( )
4 4
√
5(
)
√ ( )5
and
are constants, the DE
23
3.1
Theory of Linear Equations
Boundary Value Problems
A boundary value problem works almost exactly the same as an initial value problem, but with a few
exceptions:


Boundary values are the function (and/or its derivatives) evaluated at different values, rather
than the function and its derivatives evaluated at the same value
A boundary value problem may have many solutions, one solution, or zero solutions depending
on the values given
For example, the following are all valid sets of boundary values you could be given:
( )
(
( )
and
( )
( )
and
. /
)
and
(
)
Although I seriously doubt you’ll ever see anything remotely like the last pair.
As for boundary value problems having different numbers of solutions, this depends entirely on the
values you’re given. For example, the DE
with the boundary values ( )
and
. /
has an infinite number of solutions. This is because any solution with the form
( )
( ) will satisfy the boundary values regardless of the value of ( can be
determined to be 0). This example was taken from the textbook on page 105 if you want a more
thorough explanation.
Superposition Principle
The superposition principle simply states that if
of the form:
Then any linear combination of
and
and
are solutions of a 2nd order homogeneous DE
is also a solution of the DE. These solutions have the form:
24
Testing for Linear Independence Using the Wronskian
A set of functions is linearly independent of one another if none of them can be expressed as scalar
multiples of any of the others. While you could check this using trial and error, it’s much faster to
calculate the Wronskian of the set of functions.
The Wronskian of a set of functions is defined as follows:
(
)
||
||
(
)
(
)
(
)
Which is the determinant of a matrix formed using the set of functions and their derivatives. In case you
don’t remember how to do determinants from linear algebra, here’s a brief refresher:
In this example we will calculate the following general Wronskian:
(
)
|
|
To take the determinant of a matrix that is larger than 2x2, we must first use expansion along a row or
column. It doesn’t matter which row or column you choose, you’ll still get the same answer, so pick the
one that makes your life easiest (usually the one with the most 0’s). Once you’ve chosen a row or
column, move along each element in the row or column one by one. At each element, cover up the row
and column that intersect at it and make a matrix using the uncovered values.
I’m going to use expansion along the first row in this example.
First row, first column:
|
|
|
|
First row, second column:
|
|
|
|
First row, third column:
|
|
|
|
Continued on next page
25
Once you’ve found the smaller matrices, multiply them by the function that was at the intersection of
the covered row and column (the boxed functions above).
|
|
|
|
|
|
Now take the terms that you have just created and add them together into an equation. Multiply ever
other term (2nd, 4th, 6th, etc) by -1.
(
)
|
|
|
|
|
|
If you have only 2x2 matrices in your equation, you are ready to continue. If you’re matrices are larger
than 2x2 you must do expansion along a row or column for each of them until you have a series of 2x2
matrices.
Now that the matrices have been reduced to 2x2, finding the determinant of each is as simple as
computing the cross products.
(
)
,
-
,
-
,
-
Now that you have the Wronskian, it’s quite easy to determine whether or not the functions are linearly
independent or not. The set of functions is linearly independent if and only if their Wronskian is not
equal to 0.
tl;dr:




You solve boundary value problems the same way as initial value problems
Boundary value problems may have many, one, or zero solutions depending on the values given
Superposition states that if and are solutions of a 2nd order homogenous DE, then any
linear combination of the two is also a solution.
A set of functions is linearly independent if and only if their Wronskian is not equal to 0
26
3.3
Homogeneous Linear DEs with Constant Coefficients
A homogeneous linear DE has the general form:
( )
Where
(
)
are constants.
To solve one of these, start by create the characteristic equation (a polynomial) for the DE. The
characteristic equation will have one term for each of the terms in the DE. Each term will be made up of
the constant from the corresponding term in the DE, and an unknown variable (usually ) that is raised
to the same power as the order of the derivative in the corresponding term. For example:
The above homogeneous linear DE has a characteristic equation of:
Once you’ve found the characteristic equation, find the roots of the equation.
√
After you’ve found the roots, what you do next depends on what you got for the roots.
Any roots that are unique, real numbers will give solutions of the form:
Where
is a constant and
is the root.
If any roots are real numbers, but not unique, you simply take the solution above and multiply by
increasing powers of to make each solution linearly independent:
…
Where
are constants and
is the common root.
If you end up with complex roots of the form:
Then the solutions have the form:
(
)
In the event of duplicate complex roots, multiply by
(
))
the same way as with duplicate real roots.
Continued on next page
27
Once you’ve found all of the solutions
linear homogeneous DE:
, simply add them together to form the solution to the
An example of what this would look like with roots
,
,
(
,
:
(
)
)
Here’s a concrete example:
( )
( )
Which gives the characteristic equation:
Which factors to:
(
)(
) (
)
Which yields the roots:
Using these roots, we can construct the solutions
:
(
)
(
And then add all of these solutions together to get the final answer:
(
)
(
)
tl;dr:


( )
(
Linear homogenous DEs come in the form
To solve:
o Create a characteristic equation
o Solve for the roots of this equation
o Unique, real roots produce solutions of the form
o Duplicate real roots produce solutions of the form
o
o
)
for the th root
,
, …,
Complex roots of the form
produce solutions of the form
( ),
( )
Once you’ve found the solutions of the roots, add them together to get the final
solution to the homogeneous DE
)
28
3.4
Method of Undetermined Coefficients
The method of undetermined coefficients is used to solve nonhomogeneous linear DEs of the form:
( )
(
)
( )
Where ( ) is a function with a finite number of linearly independent derivatives.
To solve this we break it into two steps: finding the solution to the associated homogeneous DE ( ),
and finding the particular solution to the other part ( ). The final solution to the problem is:
To find the solution to the homogeneous part, see the previous section.
To find the particular solution, you must create a general solution with unknown coefficients and then
determine the coefficients. The form of your general solution will be determined by the form of ( ).
Form of ( )
1 (or any other constant)
Form of
( )
( )
( )
( )
Where the coefficients on the left side can have any real value, and the coefficients on the right side are
undetermined. A table with more examples can be found one page 129 of the textbook.
*Note: If the form of
you choose is the same as one of the solutions to the homogeneous equation,
the equation will break down. To get around this, multiply the invalid
multiplicity of the solution that was the same as .
Once you’ve determine the form of
original DE.
, find as many derivatives of
you found by
, where
is the
as necessary to substitute into the
( )
Once you’ve found the necessary derivatives, substitute them into the original DE and solve for the
unknown coefficients.
( )
(
)
( )
Continued on next page
29
Once you’ve solved for the unknown coefficients, substitute them back into the original equation for
Now that has been solved, construct the final solution using the following:
Where
.
is the solution to the associated homogeneous equation.
Here’s an example:
First lets solve for
:
(
)(
)
Now we must come up with a general form for . Normally we would see ( )
and immediately
assume
, but notice how
is already part of the solution for . Because of this, we must
multiply our original guess for
our original guess (in this case,
Using
by
, where is the multiplicity of the root(s) that are the same as
). Our new value will now work:
, find as many derivatives as needed so that it can be substituted into the original DE.
(
Now substitute
)
(
)
and its derivatives into the original DE and solve for the undetermined coefficients.
(
)
(
)
(
)
(
)
Continued on next page
30
*Note: It will not always be possible to solve for the coefficients this easily. If it’s not possible to solve
directly, arrange the equation so that the coefficients are on one side and ( ) is on the other. Once
you’ve done that, compare the terms in each side to build smaller equations that you can solve. For
example:
Substitute the coefficient(s) back into
Solve the DE by adding together
and
:
:
And you’re done!
tl;dr:


Use when solving nonhomogeneous linear DEs of the form:
( )
(
)
( )
Solve the equation in two halves: the homogeneous part with solution
solution to the other part, .
, and the particular


Solve for using the techniques from 3.3 (homogeneous linear DEs)
To solve for
o Create a general form of based on the form of ( ) (see the table above and/or the
one on page 129 of the textbook)
o If the you found is linearly dependant with the solutions of , multiply by
where is the multiplicity of the conflicting solutions of
o Find as many derivatives of as necessary to substitute into the original DE, then do
the substitution
o Try to solve for the undetermined coefficients.
o If that’s not possible, reduce the equation so that all of the undetermined coefficients
are on one side, and ( ) is on the other. Then compare the two halves to create
equations you can solve.
o Once you’ve solved for the coefficients, substitute them into to solve it

Add
and
together to create the final solution of the DE
31
3.5
Variation of Parameters
Variation of parameters is a more complex - but more robust - tool for solving nonhomogeneous linear
DEs. Unlike the method of undetermined coefficients, variation of parameters works even when the DE
contains non-constant coefficients, or when ( ) has an infinite number of linearly independent
derivatives, and it always gives a solution for (provided that can be solved). Use it when method of
undetermined coefficients won’t work, or when given equations of the form:
( )
( )
( )
(
)
( )
( )
( )
Much like method of undetermined coefficients, you must solve the DE as two parts, and . To solve
for , you must first solve for , so do that like you normally would (using the techniques in 3.3
(homogeneous linear DEs) or 3.6 (Cauchy-Euler DEs)). Once you have , you may continue.
will have the same form as the solution of , but with one exception: the constant coefficients that
had will be replaced with the functions ( ) ( )
( ). You will need to solve for these
functions.
To solve for
Where
( )
( )
( ), you must use the Wronskian of the function as follows:
is the Wronskian of solutions of
:
|
|
(
)
(
)
(
)
And
is with the th column replaced by 0’s, except for the lowest one which is replaced with ( ).
For example,
in a 3x3 matrix:
|
|
( )
( )
( )
Once you’ve used the Wronskian to calculate
( ), integrate them and substitute
them into the equation for to solve it. Once has been solved, simply add it together with to get
the solution to the DE.
Continued on next page
32
Now for an example:
( )
*We can’t solve this using the method of undetermined coefficients because
number of linearly independent derivatives.
Let’s start by solving for
:
( )
( )
( )
Now that we have
( )
, we can determine the general form of
( )
( )
To determine
( ) and
:
( )
( )
( ) we first find their derivatives using the Wronskians
|
( )
( )
|
( )
|
( )
( )
( )
|
( )
( )
( )
( )
|
( )
( )
|
( )
( )
( )
( ) and
( ) to get
( )
∫
( ) and
( )
( )
( )
∫
:
( )
( )
( )
Now that we’ve calculated the Wronskians, we can use them to find
Integrate
( ) has an infinite
( ) and
( ).
( )
( )
( ):
( )
( )
|
( )
( )|
( )
Continued on next page
33
Substitute
( ) and
( ) back into
,
( )
( )
:
|
( )
( )
( )|-
( ) |
( ) |
Add
( )
( )
( )
,
( )|
( )( )
( )
( )
( )|
together to get the final answer:
( )
( )
( ) |
( )
( )|
tl;dr:




Use variation of parameters when you can’t use the method of undetermined coefficients.
Usually it’s because the coefficients of the DE are non-constant or ( ) has an infinite number
of linearly independent derivatives.
Solve for using standard methods from other areas of the course (3.3 or 3.6)
To solve for :
o
will have the same form as , but with the constants replaced with the functions
( ) ( )
( )
o You can solve for the derivatives of these functions using the Wronskians
(to see what these look like, see the previous page)
o To solve for the derivatives, divide the corresponding Wronskian ( ) by
o Integrate the derivatives to solve for ( ) ( )
( )
o Substitute these functions into
To solve the DE, simply add
and
together
34
3.6
Cauchy-Euler DEs
A Cauchy-Euler DE is a linear, nonhomogeneous, th order differential equation of the form:
( )
Where
(
are constants,
, and
)
( )
.
To solve this DE, we once again must solve it in two parts: and as the homogeneous and particular
solutions respectively. We can calculate using variation of parameters, but solving for requires a
new technique.
Start by defining the following substitution:
Where
represents the roots we wish to determine.
Next, find as many derivatives of the substitution as necessary to substitute it in for all of the
the DE:
(
)
terms in
etc.
Once you’ve calculated the necessary substitutions, substitute them into the original DE. I’m going to
use a 2nd order DE for an example:
(
)
Simplify the equation into a polynomial that you can find the roots of:
(
)
(
)
(
)
Once you’ve found the roots of your polynomial, you’ll have three potential cases to deal with:
Unique real roots
give the solutions:
Duplicate real roots
give the solutions:
( )
, ( )-
, ( )Continued on next page
35
Complex roots
give the solutions:
(
( ))
In the event of duplicate complex roots, multiply by
(
( ))
( ) just like with duplicate real roots.
For example:
Start by solving for
:
Substitute in
and its derivatives:
(
)
Solve for the roots:
(
)
(
)
(
)(
Now that we have the roots, use them to form
Once
)
:
has been found, variation of parameters to find
|
:
|
|
|
|
|
Continued on next page
36
∫
∫
Now add
and
together to get the final answer:
tl;dr:

( )
(
)
Cauchy-Euler DEs have the form
( )
To solve:
o To solve for , start by defining the substitution
and as many of its derivatives
as needed to substitute into the homogeneous equation
o Substitute
and its derivatives into the homogeneous equation
o Simplify the equation to solve for the roots
o Unique real roots produce solutions of the form
( )),
o Complex roots
yield solutions of the form
(
( ))
(
o Duplicate roots (real or complex) are multiplied by ( ) as many times an necessary to
make them linearly independent
Once you’ve solved for , solve for using variation of parameters

Add


and
together to find the final solution to the DE
37
3.8
Applications of 2nd Order Linear DEs
Mass-Spring Systems
A mass-spring system can be modelled using the following differential equation:
( )
is the spring constant, and ( ) is the
Where is the object’s mass, is the damping coefficient,
function representing the forced motion.
Mass-spring system equations can be solved just like a normal nonhomogeneous linear DE, there’s no
special way of solving them. The one thing that may be troublesome is figuring out the values of the
parameters from the question:


is usually given as “a mass of #kg”, so that one’s fairly easy. It could also be given as “the mass
is attached to a spring with constant #N/m. The spring is stretched #m”
is usually given as “the damping force is numerically equivalent to (half, double, triple, etc.)
the instantaneous velocity”, which means is equal to


is usually given as “a force of #N stretches the spring #m”, giving you the distance and force
needed to calculate
( ) is usually given as “the mass is driven by a force ( )
”
LRC Circuits
An LRC circuit can be modelled using the following differential equation:
( )
Where is the inductance,
a function of .
is the resistance,
is the capacitance, and ( ) is the impressed voltage as
LRC circuit equations can be solved just like a normal nonhomogeneous linear DE, no special rules.
tl;dr:

Mass-spring systems can be modelled using

LRC circuits can be modelled using

( )
( )
nd
Both of these models can be solved like any other 2 order linear DE
38
4.1
Laplace Transform
The Laplace transform of a function ( ) is defined as:
* ( )+
( )
∫
( )
*Note: and must have opposite units, e.g. if is in seconds, will be in sec-1 (Hz).
One very important thing to remember about the Laplace transformation is that it’s linear, just like
integration. This means that constants can be put in front of the transform rather than inside, and the
transform of a sum is equal to the sum of the transforms.
*
+
* +
* +
Now here’s a generic example of a Laplace transform to give you a feel for the process:
*
+
∫
*
+
∫
*
*
)
(
+
+
,
*
(
+
*
(
)
|
)
(
(
)
)
-
+
*
+
Here are some common Laplace transforms that will be useful to memorize:
* +
*
+
*
( )+
*
( )+
*
( )+
*
( )+
39
4.2
Inverse Laplace Transform and Transform of Derivatives
If a function ( ) is the result of the Laplace transform of ( ), then ( ) is the inverse Laplace
transform of ( ):
( )
* ( )+
Unfortunately, there’s no way for us to convert this into a derivative or some other operation that we
know how to do, so when presented with a problem that requires you to perform an inverse Laplace
transform you’ll have to convert ( ) into one of the forms below:
{
}
{
}
{
}
( )
2
3
( )
{
}
( )
2
3
( )
Here’s an example of this in action using the following:
{ }
Start by matching the given function up with a known transform that you can use. In this example, the
inverse transform is most similar to
2
3. By comparing the exponent, we can see that
.
Once you’ve found the inverse transform you want to use, figure out how you need to change the
equation to make use of the predefined inverse transform. Remember the properties of the Laplace
transform, you’ll need them to do this part.
In this example, we need
to become , so we multiply the entire equation by :
{ }
{ }
{ }
Now that the inverse transform is in a form that we know the result of, we can convert it back into ( ):
{ }
40
Larch’s Theorem
Larch’s Theorem (which you probably don’t need to remember, but is here for its result) states that:
If ( ) and
( ) are both piecewise continuous and of exponential order, then:
* ( )+
* ( )+
( )
The important thing to take from this is that when you extend the theorem to an th order derivative:
{
( )
* ( )+
( )}
( )
(
( )
)
( )
This consequence of the theorem can be used to solve initial value problems with linear DEs if you really
love doing Laplace transformations. For example:
( )
( )
To start, use Larch’s Theorem to take the Laplace transform of the entire equation:
*
,
( )
+
* +
( )
* +
,
( )-
( )
*
+
( )-
( )
Next, isolate for ( ):
( )
( )
(
( )
) ( )
(
)(
( )
) ( )
(
)(
)(
)
Then take the inverse Laplace transform of ( ) to solve for ( ):
( )
( )
8
{
}
(
)(
9
)
)(
{
}
{
}
( )
Continued on next page
41
tl;dr:



Whenever you need to perform an inverse Laplace transformation, you’ll have to manipulate
the given equation until the inverse transform matches the form of one of the predefined
inverse transforms in the table above
The Laplace transform of an th order derivative is equal to:
(
)
* ( )+
( )
( )
( )
{ ( ) ( )}
This can be used to solve initial value problems of the form:
( )

(
)
( )
To solve:
o Take the Laplace transform of the entire equation
o Isolate for ( )
o Take the inverse Laplace transform of the entire equation to get ( )
42
4.3
Translation of Laplace Transforms
Translation Along the s Axis
To translate a Laplace transform of a function
function by
:
(
units along the axis, simply multiply the original
)
( )+
*
Translation along the axis can be used to easily solve Laplace transforms that fit the form given above.
For example:
(
*
)
+
Using translation along the axis, we can turn this into a simpler Laplace transform:
*
Where
+
* +
indicates that we must later replace with
.
Now that it has been simplified, calculate the transform:
* +
|
And finally make the substitution to get the transform we were looking for:
*
+
(
)
Translation Along the t Axis
Translating a Laplace transform along the axis requires the use of the Heaviside/Unit Step function:
(
)
2
The Laplace transform of which looks like:
* (
To translate ( )
)+
units along the axis, you must do the following:
( )
* (
) (
)+
Continued on next page
43
Translation along the axis can be useful when trying to determine the inverse transform of
For example, finding the following inverse:
8
( ).
9
We start by comparing it to the translation formula:
8
From this we can see that
and ( )
( )+
*
9
. We then use the other half of the translation formula
to build our solution:
(
) (
)
Now we need to find (
), which can be done by first finding ( ) from the inverse transform of
( ) and then replacing with
:
8
9
8
8
} ( )|
{
( )|
9
9
(
)
(
)
tl;dr:






)
( )+
Translation along the axis is defined as (
*
It’s used to find the transform of functions that match the form given above
To solve:
o Remove the
part from the transform and add a note to later replace with
o Solve the Laplace transform as you normally would
o Replace all instances of with
( )
) (
)+
Translation along the axis is defined as
* (
It’s used to find the inverse transform of functions that have been multiplied by
To solve:
o Remove the
from the inverse transform and add a note to later replace with
o
o
o
Solve the inverse Laplace transform as you normally would
Multiply the result by ( )
Replace all instances of with
44
4.4
Additional Laplace Transform Properties and Operations
Derivatives of Transforms
The th derivative of a Laplace transform is equivalent to the transform of ( ) multiplied by
*
( )+
(
)
( )
( )
This can be used to solve transforms of functions that have been multiplied by
*
*
( )+
( )+
(
)
(
. For example:
)
This technique can be applied to solve initial value problems such as the one below:
( )
( )
( )
,
Start by taking the Laplace transform of the differential equation and isolating for ( ):
(
) ( )
( )
(
)
Now we take the inverse transform of the equation:
( )
{
}
{
(
)
}
Manipulate the equation to get the inverse transforms into forms we can solve:
( )
{
From here, we can apply the transform of
( )
}
{
(
)
}
( ) to the second inverse transform to solve:
( )
( )
:
45
Convolution
Convolution is a special product between two piecewise continuous functions denoted with a star:
∫ ( ) (
The convolution operation is commutative so
two functions is equal to the product of their transforms:
*
+
)
, and the transform of the convolution of
( ) ( )
This convolution formula can be used to solve inverse Laplace transforms that are made up of the
product of two different transforms. For example:
{
(
)
}
{
{
{
{
{
(
)
(
)
(
)
(
)
{
(
}
}
).
{(
(
)
∫
( )
( )
( )
( )∫
( )
( )∫
}
)
}
( )
∫
( ),
∫
}
}
(
( )[
{
(
}
(
( )
)
( )
( )
( )-
( )
( )
( )
( )∫
( )
( )
∫
)
(
)
( )
/}
( )∫
)]|
( )[
( )
( )
( )
( )]|
46
Integral Equations
Integral equations are equations of the form:
( )
( )
∫ ( ) (
)
To solve this, we need to make use of Laplace transforms and convolution. Start by taking the Laplace
transform of the entire equation:
* ( )+
* ( )+
8∫ ( ) (
)
9
Using convolution, the transform of the integral can be simplified:
( )
( )
( ) ( )
( )
( ) ( )
( )
( ))
( )
Rearrange to solve for ( ):
( )(
( )
( )
( )
Use the inverse Laplace transform to find ( ):
( )
8
( )
9
( )
∫ ( )
(
Here’s an example equation:
( )
)
Take the Laplace transform:
* +
( )
8∫ ( )
(
)
* ( )+ *
( )+
9
Use convolution to simplify:
( )
* +
Continued on next page
47
Evaluate transforms:
( )
( )(
)
Rearrange for ( ):
( )
( )(
)
( )(
)
( )4
5
( )
4
5
( )
Take inverse transform:
* ( )+
{ }
( )
{
{ }
{
( )
}
}
( )
tl;dr:



The derivatives of a transform can be used to easily find the transform:
*
( )+ ( ) ( ) ( )
This special property can be applied when you come across inverse Laplace transforms of the
derivatives shown above
Convolution is another special property which equates the product of the transforms of two
functions to the transform of a special integral involving both functions:
( ) ( )

2∫
( ) (
)
3
Convolution can be used to solve inverse Laplace transforms of functions that are products of
two Laplace transforms
Continued on next page
48
tl;dr cont’d:

Integral equations are equations of the form: ( )

To solve:
o Take the Laplace transform of the entire equation
o Isolate for ( )
o Take the inverse transform of the equation to solve for ( )
( )
∫
( ) (
)
49
4.6
Systems of Linear Differential Equations
The Laplace transform can be used to solve systems of DEs with initial values, such as the one below:
( )
( )
,
( )
,
( )
,
Start by taking the Laplace transform of both equations. Larch’s theorem will be needed here.
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
Collect like terms and substitute in the initial values to simplify:
(
) ( )
( )
( )
(
) ( )
Now use whatever your preferred method is to solve for ( ) and ( ). Here I will use the Wronskian:
|
( )
|
|
|
|
( )
(
)(
(
)
|
|
(
|
)(
)
(
)
(
)
(
)
)
Manipulate the equation as necessary so you can take the inverse Laplace transform:
( )
( )
√
√
4
4
√
√
5
5
√
√
4
4
√
√
5
5
Finally, take the inverse Laplace transform to find ( ) and ( ):
( )
( )
√
√
(√ )
(√ )
√
√
(√
)
(√
)
Continued on next page
50
tl;dr:


You can solve systems of linear DEs with initial values using Laplace transforms
To solve:
o Start by taking the Laplace transform of each equation in the system
o Collect like terms and simplify the equations in the system
o Use your preferred method to solve for ( ), ( ), etc. Substitution, elimination, the
Wronskian, and possibly other methods can all be used
o Manipulate the equations as needed so you can calculate the inverse transform
o Take the inverse Laplace transform of each equation solve for ( ), ( ), etc.
51
12.1 Orthogonal Functions
Two functions are orthogonal on the interval ,
equal to 0:
- if the integral of their product along that interval is
( ) ( )
∫
To show whether or not a set of functions is orthogonal, you must show that each member is orthogonal
to every other member. For example, we’ll use question 7 from the textbook:
*
( )
(
)
(
+
)
,
-
0
1
From the given set, we need to come up with a general form for any given member of the set. From the
set given above, we can find the general form as:
) )
((
Where
With this general form, you must show that for all
orthogonal to every other member:
∫
((
and
) )
such that
, every member of the set is
) )
((
Using the product identities, we can simplify the integral:
,
∫
(
)
( (
∫
( (
∫
(
) )
)-
( (
) )
∫
) )
( (
) )
Now we can solve the integral:
(
)
( (
) )|
(
)
( (
) )|
Since the integral evaluated to zero, the each function in the set is orthogonal to each other function.
52
Norm of Functions
The norm of a function or set of functions on an interval ,
√∫
‖ ( )‖
- is calculated as:
( )
For example, lets continue to use question 7 from the textbook:
*
( )
(
)
(
)
+
,
-
0
1
Start by coming up with the general form for any member of the set:
) )
((
Where
Use this general form with the equation of the norm to calculate the norm of each function in the set:
√∫
√ ∫
√
|
) )
((
( (
(
)
) )
( (
) )|
√
√
tl;dr:

If the integral ∫

To determine if a set of functions is orthogonal:
o Start by finding a general form for any member of the set
o Use the integral above to test the arbitrary cases of ( ) and
o If the integral evaluates to 0, the entire set is orthogonal
( ) ( )
evaluates to 0, ( ) and ( ) are orthogonal
( ) where
Continued on next page
53
tl;dr cont’d:
√∫

The norm of a function can be determined using: ‖ ( )‖

To determine the norm of every function in a set:
o Start by finding a general form for any member of the set
o Use the equation for the norm above with the general form
o Whatever comes out is the norm of any function in the set
( )
54
12.2 Fourier Series
A Fourier series of a periodic function ( ) can be defined as:
∑0
Where
.
are constants, and
/
where
.
/1
is the period of ( ).
The formulas for the constants are given below:
∫
( )
∫
( )
.
/
∫
( )
.
/
To evaluate these formulas and any others you might come across, the following integrals will be
invaluable:
∫
.
/
∫
2
∫
∫
.
/
.
/
.
/
.
/
/
.
/
∫
2
.
.
/
2
Most of the time, you will be given some function ( ) and told to find the Fourier series of ( ) on
some given interval. To do that, you just need to fill out the variables in the general form given at the
beginning of this section. For example, question 3 on page 664 of the textbook:
( )
2
To start, determine the period of the given function, which for a piecewise function is the same as the
interval it is defined on. In this example,
. Using we can calculate the value of :
Continued on next page
55
Now that we have a value for , we have enough information to determine
( )
∫
∫
∫
|
(
(
∫
(
)|
(
)
)
(
,(
(
)
( )
(
)
∫
(
)
∫
(
(
(
(
)
)|
∫
|
( )
∫
∫
:
)
(
(
)
)
(
))|
-
)
(
)
)
(
))|
Once you’ve solved for the constants, simply substitute them into the general Fourier series to finish:
( )
∑6
(
)
(
)
(
)
(
)7
Continued on next page
56
tl;dr:

The Fourier series of some function ( )



are all constants;
∑
where
0
.
/
.
/1
is the period of ( )
For a piecewise function, is equal to the length of the interval the function is defined on
To create the Fourier series of a function:
o Start by determining
o
Solve for
using
∫
( )
o
Solve for
using
∫
( )
o
Solve for
o
(
)
)
∫ ( ) (
Once you’ve solved for all of the constants, substitute them into the general form of the
Fourier series to obtain the Fourier series of ( )
using
57
Review and Other Useful Things
Cover-up Method of Partial Fractions
Partial fractions can be a pain to deal with sometimes, but the cover-up method can make them much
faster. With the cover-up method, you separate the rational expression into a sum of rational
expressions as normal, but instead of creating a system of equations to solve for the unknown
constants, you can take the original expression and multiply it by the denominator of the term you’re
solving for, then evaluate the result at whatever value makes the denominator of your term 0. I know it
sounds ugly when put into words, but it’s actually quite simple. Here’s an example:
(
)(
)(
)
Lets say we want to solve for first. As mentioned above,
multiplied by (
) and evaluated at
:
(
(
)(
)(
(
Now just evaluate the expression at
and
)
|
)
)(
|
)
)(
to solve for :
(
You can solve for
equivalent to the original expression
)(
)
in the same way:
(
(
(
(
)(
)(
)(
)(
)(
)(
)
|
)
)
|
)
(
)(
|
)
(
)(
|
)
This method can be used to quickly solve any partial fractions you come across.
58
Trig Identities
Pythagorean Identities
( )
( )
( )
( )
( )
( )
Negative Angle Identities
(
(
)
)
( )
(
( )
( )
Angle Sum/Difference Identities
(
)
( ) ( )
( )
( )
(
)
( )
( )
Double Angle Identities
( )
(
( )
( )
( )
Half Angle Identities
( )
)
( )
,
( )
(
( )
(
)-
)
( )
( )
( )
)
( )
( )
( )
( )
( )
(
,
)-
( )
( )
Sum/Difference Identities
( )
( )
( )
(
)
4
( )
( )
5
( )
(
)
( )
(
.
/
)
.
/
Product Identities
( )
( )
,
(
)
(
)-
( )
( )
,
(
)
(
)-
( )
( )
,
(
)
(
)-
( )
( )
,
(
)
(
)-
This whole thing was written by Tom Machado
59
Integration by Parts
Integration is used when you’re trying to integrate the product of two functions. The formula is shown
below:
∫
Where
is one of the functions, and
∫
is the other.
You can use IBP in two ways: to reduce one of the functions until the integral ∫
is the integral of a
single function, or – with functions that repeat like ( ) and
( ) – until the integral ∫
is the
same as ∫
, which allows us to rearrange. First, here is an example of the first use:
∫
In this case we want to choose to be the function that will get eventually disappear as we take its
derivative.
will then be the other function along with :
Now we need to solve for
and
by taking the derivative of
and the integral of
respectively:
Now that we have all of the components, we can fill out the IBP formula:
∫
∫(
)
We now have an expression that contains a single-function integral that we can solve:
∫
(
)
Now here’s an example of using IBP to solve a product of functions with repeating derivatives:
∫
In this case we want
along with :
( )
to be the function with a repeating derivative, and
to be the other function
( )
Continued on next page
60
Now find
and
by taking the derivative of
and the integral of
respectively:
( )
Now that you have all of the components, fill out the IBP formula:
∫
( )
( )
( )
∫
∫(
( )
∫
( ))
( )
Notice how we’ve ended up with another product of functions in the integral. We need to do IBP again:
( )
( )
∫
( )
( )
( )
,
∫
( )
-
Now that we’ve done that, the integral we started with has repeated itself! Now we can use algebra to
isolate for the integral we’re trying to solve:
∫
( )
( )
∫
( )
,
( )
( )
( )-
tl;dr:




Integration by parts can be used to solve integrals of the product of two functions
IBP formula: ∫
∫
To solve when one function has finite derivatives:
o Set to be the function that will eventually go to 1, set
to be the other function
o Find the derivative of and the integral of
o Fill out the IBP formula
o If the integral ∫
is still an integral of two functions, repeat IBP
o If the integral is of a single function, evaluate it to finish
To solve when one function has infinite repeating derivatives:
o Set to be the function with repeating derivatives, set
to be the other function
o Find the derivative of and the integral of
o Fill out the IBP formula
o If the integral ∫
doesn’t match your original integral, repeat IBP
o If the integral matches the original integral, use algebra to solve for the original integral