Differential Equations Notes & Study Guide

This semester I am taking differential Equations and this is a study guide/cheatsheet I’m writing to help me better understand my notes and to help anyone else understand the topic who maybe in similar boat.

Textbook:

A First Course In DIFFERENTIAL EQUATIONS with Modeling Applications (11th Edition)

School:

Florida Polytechnic University

Note: I will try to not use textbook language wherever possible but I will include Book definition in some form to allow me to elaborate on what the book definition means.

Prerequisites:

Calc 2 skills such as

  • Logarithms and their properties
  • Complete the square
  • Partial Fraction decomposition

Table of contents:

i. Introduction

ii. Exam 1

iii. Exam 1 Summary

iv. Exam 2

v. Exam 2 Summary

vi. Exam 3

vii. Exam 3 Summary

viii. Final Exam

ix. Conclusion

Introduction:

Here is a video summary / Introduction to the topic

What is a Differential Equation?

Book Definition:

“An equation containing the derivatives of one or more unknown functions ( or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).”

The video above goes into detail about what this definition means but basically:

Differential Equations are equations written to express real life problems where things are changing and with ‘solutions’ to these equations being equations themselves. We are using math to represent real life scenarios, usually meaning a rate of change.

ie. beats per minute (bpm) , meters per second (m/s), or anything that isn’t exactly a finite value but rather a changing one.

Now to the goal in this class is to find the original function that has a derivative of the one in the equation you are given.

That may sound complex but this class is usually taken after Calculus 2 and that means Im assuming you have a simple understanding of Derivatives and integrals.

ie.

You’re given:

d/dx (x²) = 2x

and your job is to find:

integral of (2x)

which is x²+c

Where your answer is still a function but that’s very basic calculus, I will go back and write a Calculus 2 and 3 Article later but for now thats all you need to know.

Types of Differential Equations

Now the equations you’re given in differential equations problems will involve 1 or more derivatives of many unknown functions as defined below as ODE’s but eventually you will be doing Differential Equations with partial derivatives (PDE’s) from Calc 3 (More on this later)

  1. If a differential equation contains only ordinary derivatives of one or more unknown functions with respect to a single independent variable, it is said to be an ordinary differential equation (ODE)

2. An equation involving partial derivatives of one or more unknown functions of two or more independent variables is called a partial differential equation (PDE)

Initial Value Problems:

Book Definition:

“ An initial value problem is an ordinary differential equation together with a specified value, called the initial condition, of the unknown function at a given point in the domain of the solution.”

Here is an example of an IVP Seperable

To Solve the above:

y’ is equal to 𝑑𝑦/𝑑𝑥 in this case so:

𝑑𝑦/𝑑𝑥=𝑦+1

Then we move the y+1 and 𝑑𝑥 around

𝑑𝑦 /(𝑦+1)=𝑑𝑥

Then Integrate both sides to get

ln(𝑦+1)=𝑥+c

Now to remove the ln we need to make each side the base of e^x

hint: e^[ln(y+1)] is just y+1

𝑦+1 =𝑒^(𝑥+c)

𝑦= (𝑒^𝑥)(𝑒^c)-1

(𝑒^c) is still just C and we move the +1 to get:

𝑦= 𝑐𝑒^x -1

to arrive at the general solution of:

y = 𝑐𝑒^𝑥 -1

but the could be:

1𝑒^𝑥-1, 2𝑒^𝑥-1, 3𝑒^𝑥-1 etc.

The C represents a constant that we don’t know yet,

But we were given some initial condition in the problem y(0) = 5, then we can now use this to solve for our C value by plugging it in to our

5 = C𝑒⁰-1

5+1 = C𝑒⁰

5+1 = C(1)

and Then we get

6=C

which means C = 1

And this means we now have a single function answer of:

y = 6𝑒^𝑥 -1

as our solution, rather than a family of functions as a solution (The general solution)

We didn’t know that without knowing the initial condition but thats how we would solve for the C value given an initial condition but otherwise we just look for the general solution.

And thats only with there being one derivative which means the equation is of the first order.

First order differential equations:

The number of derivatives you have relates to the Order of the equation (equations with second derivates are said to be of the second order, equations with third derivative are said to be of the third order, etc)

for every extra derivative in your problem, you will get an extra C value which can be denoted

C₁+C₂…Cₓ

x being the order of derivative taken for that value

These Values are necessary for accuracy but eventually can be combined later so just make sure you never forget your +C

We will go into detailed examples but for now this is the basic theory behind the equations to help as it maybe a little tricky or obscure at first but we’ll get there.

Exam 1 :

Sections 1.1 to 4.2

Our professor is testing us on Chapter 1 through chapter 3 section 1 (3.1)

Here’s the notes we’ve covered of the chapters covered (not skipped)

This first exam will be on ODE’s of any kind using the different techniques covered so I will go through each of the techniques and when they would be used to solve a ODE

To solve Any Differential equation, you must first start by deciding on which technique is most suitable for the equation in question.

The most effective way solve ODE’s is through this list of techniques, checking to see if the equation can be solved by each method in this order.

Note: Most problems tend to be solvable through multiple techniques but the optimal technique will make your solution much easier which makes it so important to pick the correct and best method.

The later techniques above often will turn your original equation into an equation solvable by the earlier methods (ex. A Bernoulli can sometimes create a separable equation which are much easier to solve)

1. Separable

A differential equation is said to be separable if the variables can be separated. A separable equation is one that can be written in the form dy/dx. Once this is done, all that is needed to solve the equation is to integrate both sides. Our example in the introduction is considered a separable equation as with the one below:

An example of a separable differential equation with solution
  • If you can split of the X and Y terms to be on opposite side of the equal sign, then it is a separable equation
  • Once separated, take integral of both sides

2. Homogenous

A differential equation can be homogeneous in either of two respects. A first order differential equation is said to be homogeneous if it may be written f(x,y)dy=g(x,y)dx, where f and g are homogeneous functions of the same degree of x and y. In this case, the change of variable y = ux leads to an equation of the form dx/x=h(u)du, which is easy to solve by integration of the two members.

homogeneous differential equation

Here is a light overview of how this type of equation is solved
  • Every term must be of the same degree

3. Substitution

A substitution problem is essentially a problem where you do a u-substitution for portion of the problem to make the problem easier to manipulate or solve.

Usually look for a term that when replaced with a u, makes the problem simpler

dy/dx =( some term)² could be u = some term

and hopefully you’d get

u = some term

and

dy/dx = du/dx + d/dx( some term)

You then substitute in your u term and new dy/dx term

and solve accordingly

u substitution method differential equations

  • Can be made separable

4. Linear (y or x)

To start a linear equation, you must first write it in standard form

This being:

y’ P(x) + y = Q(x)

Once this is done, the integrating factor is

I(x) = e^integral of P(x)dx

And then the general solution is now

y = 1/I(x)[integral of I(x)Q(x) dx + C]

Here is a great linear differential equation example

5. Bernoulli (y or x)

6. Exact

Here is how you solve an Exact Differential Equation

Basically, if in the form M(x,y) dx + N(x,y)dy = 0,

then you can separate the

M(x,y) and N(x,y) terms

Take the partial derivative of each of x and y and if they are equal,

your equation is exact.

If the equation is not exact then you can manipulate the equation to then use an integrating factor

7. Integrating Factor

Format your equation as such:

y’ + yP(x) = Q(x)

Integrating factor → I(x) = e^integral(

Use accordingly (Look at exact)

y = 1/I(x) * int[ I(x)*Q(x) + C ]

Linear Models

Growth & Decay

rate of change of x with respect to time

Modeling with First order Differential Equations

Now that you’ve practiced these different techniques for solving differential equations, Our next objective is to apply these differential equations to rate of change problems that we can now solve now that we’ve created am equation of the differential type.

Usually our differential equations are either separable or non-separable.

If separable, split up the variables and integrate

If not separabale, we must

Differential Equations — 11 — Modeling with 1st Order Diff. Eq’s (Tank Problem)

An example of modeling a situation using a differential equation
  1. Label the variables in the equation
  2. List any given constants or given initial conditions
  3. Write out equation in terms of words/variables
  4. Simplify
  5. Write equation in standard form
  6. Solve this differential equation
  7. Plug in the initial condition in to get C

Exam 1 Summary

The best way to go about first order differential equations

  1. Does the problem fall under SHEILDS — Separable, homogeneous, Exact, Linear, Direct Integration, or Substitution. (I know the order is different but shields is a good acrynonm

How to solve ANY differential equation

If Separable

  • Separate the variables into the dy and dx terms on the respective sides of the equal sign and then integrate and solve in terms of y

Else:

For Homogenous

  • Replace y = ux or x = vy
  • Find dy/dx and then substitute in y or x and dy/dx
  • Integrate the two sides

For Exact

  • Look at the equation and see if you can identify a M(x,y) and N(x,y) term which is set equal to 0
  • Find the partial derivative of M(x,y) and N(x,y) and see if they are equal

For Linear

  • Put the equation in standard form:

y’+yP(x) = Q(x)

Your integrating factor is now

I(x) = e^integral(P(x))

You then multiply the equation by your integrating factor

I(x)* y =integral ( I(x) * Q(x) dx

For Direct Integration

  • Just integrate the equation, shouldn’t be too complicated

For Substitution

  • If the equation can be made simpler, substitute in a u = some term and then find du and apply that to the equation

Modeling

When given a modeling word problem, you need to memorize a couple of different problem types and their associated differential equation type so that you can solve for their general solution and for C if given the initial conditions

Bacterial Growth, Half-Life of Plutonium, Age of a Fossil

dx/dt = kx or dA/dt = kA

Cooling of a Cake

dT/dt = k(T-70)

Mixture of Two Salt Solutions

dA/dt = (input rate of salt) -(output rate of salt) =>R in — R out

Exam 2 :

Sections 4.2 to 7.1

Reduction of order

Homogeneous Linear Equations with Constant Coefficients

Undetermined Coefficients — Superposition Approach*

Undetermined Coefficients — Annihilator Approach

Auxiliary Equation:

y’’ + y’ + … = 0

y c: complementary function

y p: particular solution

y p = u1(x)cosx + u2(x)sinx

u1(x) = negative integral of y2 f(x) / w(x) dx

u1(x) = integral of y1 f(x) / w(x) dx

w(x) is the Wronskian

To find wronskian we do a crossproduct of |y1 and y2|

y = yc + yp

Exam 2 Summary

At the present moment in time, classes were now moved completely online meaning we did receive a lecture video review which I will be breaking down into sections, steps and explanations as best I understand.

The video that was made to cover our second exam material is from sections 4.1 to 4.8 and it is mainly solving homogenous and non-homogenous linear differential equations.

I took hand written notes that can be found here

I will write out verbal explanations with light equations due to the inability to type the mathematics properly with sub notation but I will do my best to describe steps in full

Topics and Examples

Prove that given solutions are indeed solutions to the homogenous, linear, second order differential equation

Ex.

Show that [e^x , e^-3x] is a fundamental set of solutions of the equation

y’’ + y’ — 6y=0

Ans

First we must check if the given solutions are linearly independent or not

so we use each of the provided solutions as inputs for y and take their respective first and second derivatives to see if they satisfy the equation.

In this problem the y’s and their derivatives satisfy the differential equation so now we must check if they are linearly linearly independent by finding the general equation and since the equation is a second order differential equation so there are only two solutions.

so e^(2x) & e^(3x) are our f(x)1 and f(x)2

To find the general solution we need to find the wronskian which is

the cross product of the f(x)’s and their derivatives.

If the final solution is never 0 then the solutions are linearly independent

In this case e^(2x)•-3e^(-3x) + -2^(2x)•e^(-3x) = -3e^(-x) — 2e^(-x)

  • This equals -5e^-x so they are linearly independent
  • Therefore both of the provided solutions are the fundamental set of solutions

Reduction of order formula

Ex.

Given the equation: x²y’’ — 3xy’+4y=0 and the solution: y1 = x² ,

Find the other solution

Ans

This is a second order linear differential equation that has a non-constant coefficient so we take the original equation and divide by the term in front of the y’’ to get it by itself

In this case we divide by x² to get

y’’ — 3y’/x + 4/x²y = 0

The coefficient of y’ is P(x)

The coefficient of y is Q(x)

Now at this point we need to use the reduction of order formula

Then using this formula we can plug in and solve for the the second solution the to the differential equation.

Find the general solution of the homogenous linear equation

Ex.

Ans

Solving non-homogenous differential equations using the method of undetermined coefficients

Ex.

Ans

Solving non-homogenous linear differential equations with constant coefficients using the method of variation of parameters

Ex.

Ans

Final Summary

1.1: Definitions and Terminology

1.2: Initial-Value Problems

1.3: Differential Equations as Mathematical Models

2.2: Separable Equations

2.3: Linear Equations

2.4: Exact Equations

2.5: Solutions by Substitutions

3.1: Linear Models

4.1: Preliminary Theory — Linear Equations

4.2: Reduction of Order

Exam 1

4.3: Homogeneous Linear Equations with Constant Coefficients

A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants (i.e., not functions), all terms are linear, and the entire differential equation is equal to zero (i.e., it is homogeneous).

4.4: Undetermined Coefficients — Superposition Approach*

4.6: VARIATION OF PARAMETERS

4.7:

7.1:

7.2:

7.3:

Exam 2

7.4

7.5

7.6

6.1

6.2

Thanks for reading! Buy me coffee? https://ko-fi.com/jonngan

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