Differential Equations - Basic Concepts: Thoughts

Before moving on to learning how to solve differential equations we want to give a few final thoughts. Any differential equations course will concern itself with answering one or more of the following questions.

Given a differential equation will a solution exist?Not all differential equations will have solutions so it’s useful to know ahead of time if there is a solution or not. If there isn’t a solution why waste our time trying to find something that doesn’t exist?
This question is usually called the existence question in a differential equations course.
If a differential equation does have a solution how many solutions are there?As we will see eventually, it is possible for a differential equation to have more than one solution. We would like to know how many solutions there will be for a given differential equation.
There is a sub question here as well. What condition(s) on a differential equation are required to obtain a single unique solution to the differential equation?
Both this question and the sub question are more important than you might realize. Suppose that we derive a differential equation that will give the temperature distribution in a bar of iron at any time $t$ . If we solve the differential equation and end up with two (or more) completely separate solutions we will have problems. Consider the following situation to see this.
If we subject 10 identical iron bars to identical conditions they should all exhibit the same temperature distribution. So only one of our solutions will be accurate, but we will have no way of knowing which one is the correct solution.
It would be nice if, during the derivation of our differential equation, we could make sure that our assumptions would give us a differential equation that upon solving will yield a single unique solution.
This question is usually called the uniqueness question in a differential equations course.
If a differential equation does have a solution can we find it?This may seem like an odd question to ask and yet the answer is not always yes. Just because we know that a solution to a differential equations exists does not mean that we will be able to find it.

In a first course in differential equations (such as this one) the third question is the question that we will concentrate on. We will answer the first two questions for special, and fairly simple, cases, but most of our efforts will be concentrated on answering the third question for as wide a variety of differential equations as possible.

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Digital Signal Processing - Basic Continuous Time Signals

To test a system, generally, standard or basic signals are used. These signals are the basic building blocks for many complex signals. Hence, they play a very important role in the study of signals and systems. Unit Impulse or Delta Function A signal, which satisfies the condition, δ ( t ) = lim ϵ → ∞ x ( t ) δ ( t ) = lim ϵ → ∞ x ( t ) is known as unit impulse signal. This signal tends to infinity when t = 0 and tends to zero when t ≠ 0 such that the area under its curve is always equals to one. The delta function has zero amplitude everywhere except at t = 0. Properties of Unit Impulse Signal δ(t) is an even signal. δ(t) is an example of neither energy nor power (NENP) signal. Area of unit impulse signal can be written as; A = ∫ ∞ − ∞ δ ( t ) d t = ∫ ∞ − ∞ lim ϵ → 0 x ( t ) d t = lim ϵ → 0 ∫ ∞ − ∞ [ x ( t ) d t ] = 1 Weight or strength of the signal can be written as; y ( t ) = A δ ( t ) y ( t ) = A δ ( t ) Area of the weighted impulse s...

Differential Equations - First Order: Bernoulli

In this section we are going to take a look at differential equations in the form, y ′ + p ( x ) y = q ( x ) y n y ′ + p ( x ) y = q ( x ) y n where p ( x ) p ( x ) and q ( x ) q ( x ) are continuous functions on the interval we’re working on and n n is a real number. Differential equations in this form are called Bernoulli Equations . First notice that if n = 0 n = 0 or n = 1 n = 1 then the equation is linear and we already know how to solve it in these cases. Therefore, in this section we’re going to be looking at solutions for values of n n other than these two. In order to solve these we’ll first divide the differential equation by y n y n to get, y − n y ′ + p ( x ) y 1 − n = q ( x ) y − n y ′ + p ( x ) y 1 − n = q ( x ) We are now going to use the substitution v = y 1 − n v = y 1 − n to convert this into a differential equation in terms of v v . As we’ll see th...

Differential Equations - Systems: Solutions

Now that we’ve got some of the basics out of the way for systems of differential equations it’s time to start thinking about how to solve a system of differential equations. We will start with the homogeneous system written in matrix form, → x ′ = A → x (1) (1) x → ′ = A x → where, A A is an n × n n × n matrix and → x x → is a vector whose components are the unknown functions in the system. Now, if we start with n = 1 n = 1 then the system reduces to a fairly simple linear (or separable) first order differential equation. x ′ = a x x ′ = a x and this has the following solution, x ( t ) = c e a t x ( t ) = c e a t So, let’s use this as a guide and for a general n n let’s see if → x ( t ) = → η e r t (2) (2) x → ( t ) = η → e r t will be a solution. Note that the only real difference here is that we let the constant in front of the exponential be a vector. All we need to do then is plug this into the d...

Calculus III - 3-Dimensional Space: Equations of Lines

In this section we need to take a look at the equation of a line in R 3 R 3 . As we saw in the previous section the equation y = m x + b y = m x + b does not describe a line in R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a...

Differential Equations - First Order: Modeling - i

We now move into one of the main applications of differential equations both in this class and in general. Modeling is the process of writing a differential equation to describe a physical situation. Almost all of the differential equations that you will use in your job (for the engineers out there in the audience) are there because somebody, at some time, modeled a situation to come up with the differential equation that you are using. This section is not intended to completely teach you how to go about modeling all physical situations. A whole course could be devoted to the subject of modeling and still not cover everything! This section is designed to introduce you to the process of modeling and show you what is involved in modeling. We will look at three different situations in this section : Mixing Problems, Population Problems, and Falling Objects. In all of these situations we will be forced to make assumptions that do not accurately depict reality in most cases, but wi...

Digital Signal Processing - Miscellaneous Signals

There are other signals, which are a result of operation performed on them. Some common type of signals are discussed below. Conjugate Signals Signals, which satisfies the condition x ( t ) = x ∗ ( − t ) are called conjugate signals. Let x ( t ) = a ( t ) + j b ( t ) So, x ( − t ) = a ( − t ) + j b ( − t ) And x ∗ ( − t ) = a ( − t ) − j b ( − t ) By Condition, x ( t ) = x ∗ ( − t ) If we compare both the derived equations 1 and 2, we can see that the real part is even, whereas the imaginary part is odd. This is the condition for a signal to be a conjugate type. Conjugate Anti-Symmetric Signals Signals, which satisfy the condition x ( t ) = − x ∗ ( − t ) are called conjugate anti-symmetric signal Let x ( t ) = a ( t ) + j b ( t ) So x ( − t ) = a ( − t ) + j b ( − t ) And x ∗ ( − t ) = a ( − t ) − j b ( − t ) − x ∗ ( − t ) = − a ( − t ) + j b ( − t ) By Condition x ( t ) = − x ∗ ( − t ) ...

Differential Equations - Systems: Repeated Eigenvalues - i

This is the final case that we need to take a look at. In this section we are going to look at solutions to the system, → x ′ = A → x x → ′ = A x → where the eigenvalues are repeated eigenvalues. Since we are going to be working with systems in which A A is a 2 × 2 2 × 2 matrix we will make that assumption from the start. So, the system will have a double eigenvalue, λ λ . This presents us with a problem. We want two linearly independent solutions so that we can form a general solution. However, with a double eigenvalue we will have only one, → x 1 = → η e λ t x → 1 = η → e λ t So, we need to come up with a second solution. Recall that when we looked at the double root case with the second order differential equations we ran into a similar problem. In that section we simply added a t t to the solution and were able to get a second solution. Let’s see if the same thing will work in this case as well. We’ll see if → x = t e...

Differential Equations - Systems: Repeated Eigenvalues - ii

Example 3 Solve the following IVP. → x ′ = ( − 1 3 2 − 1 6 − 2 ) → x → x ( 2 ) = ( 1 0 ) x → ′ = ( − 1 3 2 − 1 6 − 2 ) x → x → ( 2 ) = ( 1 0 ) First the eigenvalue for the system. det ( A − λ I ) = ∣ ∣ ∣ ∣ − 1 − λ 3 2 − 1 6 − 2 − λ ∣ ∣ ∣ ∣ = λ 2 + 3 λ + 9 4 = ( λ + 3 2 ) 2 ⇒ λ 1 , 2 = − 3 2 det ( A − λ I ) = | − 1 − λ 3 2 − 1 6 − 2 − λ | = λ 2 + 3 λ + 9 4 = ( λ + 3 2 ) 2 ⇒ λ 1 , 2 = − 3 2 Now let’s get the eigenvector. ( 1 2 3 2 − 1 6 − 1 2 ) ( η 1 η 2 ) = ( 0 0 ) ⇒ 1 2 η 1 + 3 2 η 2 = 0 η 1 = − 3 η 2 ( 1 2 3 2 − 1 6 − 1 2 ) ( η 1 η 2 ) = ( 0 0 ) ⇒ 1 2 η 1 + 3 2 η 2 = 0 η 1 = − 3 η 2 → η = ( − 3 η 2 η 2 ) η 2 ≠ 0 → η ( 1 ) = ( − 3 1 ) η 2 = 1 η → = ( − 3 η 2 η 2 ) η 2 ≠ 0 η → ( 1 ) = ( − 3 1 ) η 2 = 1 Now find → ρ ρ → , ( 1 2 3 2 − 1 6 − 1 2 ) ( ρ 1 ρ 2 ) = ( − 3 1 ) ⇒ 1 2 ρ 1 + 3 2 ρ 2 = − 3 ρ 1 = − 6 − 3 ρ 2 ( 1 2 3 2 − 1 6 − 1 2 ) ( ρ 1 ρ 2 ) = ( − 3 1 ) ⇒ 1 2 ρ 1 + 3 2 ρ 2 = − 3 ρ 1 = − 6 − 3 ρ 2 → ρ = ( − 6 − 3 ρ 2 ρ 2 ) ⇒ → ρ = ( − 6 0 ) if ρ 2 = 0 ρ → ...

Differential Equations - Second Order: Repeated Roots

In this section we will be looking at the last case for the constant coefficient, linear, homogeneous second order differential equations. In this case we want solutions to a y ′′ + b y ′ + c y = 0 a y ″ + b y ′ + c y = 0 where solutions to the characteristic equation a r 2 + b r + c = 0 a r 2 + b r + c = 0 are double roots r 1 = r 2 = r r 1 = r 2 = r . This leads to a problem however. Recall that the solutions are y 1 ( t ) = e r 1 t = e r t y 2 ( t ) = e r 2 t = e r t y 1 ( t ) = e r 1 t = e r t y 2 ( t ) = e r 2 t = e r t These are the same solution and will NOT be “nice enough” to form a general solution. We do promise that we’ll define “nice enough” eventually! So, we can use the first solution, but we’re going to need a second solution. Before finding this second solution let’s take a little side trip. The reason for the side trip will be clear eventually. From the quadratic formula we know that the roots to the characteristic equation are, r 1 , 2 = ...

Differential Equations - Laplace Transforms: Table

f ( t ) = L − 1 { F ( s ) } f ( t ) = L − 1 { F ( s ) } F ( s ) = L { f ( t ) } F ( s ) = L { f ( t ) } 1 1 s 1 s e a t e a t 1 s − a 1 s − a t n , n = 1 , 2 , 3 , … t n , n = 1 , 2 , 3 , … n ! s n + 1 n ! s n + 1 t p t p , p > − 1 p > − 1 Γ ( p + 1 ) s p + 1 Γ ( p + 1 ) s p + 1 √ t t √ π 2 s 3 2 π 2 s 3 2 t n − 1 2 , n = 1 , 2 , 3 , … t n − 1 2 , n = 1 , 2 , 3 , … 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 1 ) √ π 2 n s n + 1 2 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 1 ) π 2 n s n + 1 2 sin ( a t ) sin ⁡ ( a t ) a s 2 + a 2 a s 2 + a 2 cos ( a t ) cos ⁡ ( a t ) s s 2 + a 2 s s 2 + a 2 t sin ( a t ) t sin ⁡ ( a t ) 2 a s ( s 2 + a 2 ) 2 2 a s ( s 2 + a 2 ) 2 t cos ( a t ) t cos ⁡ ( a t ) s 2 − a 2 ( s 2 + a 2 ) 2 s 2 − a 2 ( s 2 + a 2 ) 2 sin ( a t ) − a t cos ( a t ) sin ⁡ ( a t ) − a t cos ⁡ ( a t ) 2 a 3 ( s 2 + a 2 ) 2 2 a 3 ( s 2 + a 2 ) 2 sin ( a t ) + a t cos ( a t ) sin ⁡ ( a t ) + a t cos ⁡ ( a t ) 2 a s 2 ( s 2 + a 2 ) 2 2 a s 2 ( s 2 + a 2 ) 2 cos ( a t ) − a t sin ( a t ) cos ⁡ (...

Robert Karamagi

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Calculus III - 3-Dimensional Space: Equations of Lines