Skip to main content

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 - Basic Concepts: Definitions


Differential Equation


The first definition that we should cover should be that of differential equation. A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives.

There is one differential equation that everybody probably knows, that is Newton’s Second Law of Motion. If an object of mass m is moving with acceleration a and being acted on with force F then Newton’s Second Law tells us.
(1)F=ma
To see that this is in fact a differential equation we need to rewrite it a little. First, remember that we can rewrite the acceleration, a, in one of two ways.
(2)a=dvdtORa=d2udt2
Where v is the velocity of the object and u is the position function of the object at any time t. We should also remember at this point that the force, F may also be a function of time, velocity, and/or position.
So, with all these things in mind Newton’s Second Law can now be written as a differential equation in terms of either the velocity, v, or the position, u, of the object as follows.
(3)mdvdt=F(t,v)(4)md2udt2=F(t,u,dudt)
So, here is our first differential equation. We will see both forms of this in later chapters.
Here are a few more examples of differential equations.
(5)ay+by+cy=g(t)(6)sin(y)d2ydx2=(1y)dydx+y2e5y(7)y(4)+10y4y+2y=cos(t)(8)α22ux2=ut(9)a2uxx=utt(10)3u2xt=1+uy

Order


The order of a differential equation is the largest derivative present in the differential equation. In the differential equations listed above (3) is a first order differential equation, (4)(5)(6)(8), and (9) are second order differential equations, (10) is a third order differential equation and (7) is a fourth order differential equation.
Note that the order does not depend on whether or not you’ve got ordinary or partial derivatives in the differential equation.
We will be looking almost exclusively at first and second order differential equations in these notes. As you will see most of the solution techniques for second order differential equations can be easily (and naturally) extended to higher order differential equations and we’ll discuss that idea later on.

Ordinary and Partial Differential Equations


A differential equation is called an ordinary differential equation, abbreviated by ode, if it has ordinary derivatives in it. Likewise, a differential equation is called a partial differential equation, abbreviated by pde, if it has partial derivatives in it. In the differential equations above (3) - (7) are ode’s and (8) - (10) are pde’s.
The vast majority of these notes will deal with ode’s. The only exception to this will be the last chapter in which we’ll take a brief look at a common and basic solution technique for solving pde’s.

Linear Differential Equations


linear differential equation is any differential equation that can be written in the following form.
(11)an(t)y(n)(t)+an1(t)y(n1)(t)++a1(t)y(t)+a0(t)y(t)=g(t)
The important thing to note about linear differential equations is that there are no products of the function, y(t), and its derivatives and neither the function or its derivatives occur to any power other than the first power. Also note that neither the function or its derivatives are “inside” another function, for example, y or ey.
The coefficients a0(t),,an(t) and g(t) can be zero or non-zero functions, constant or non-constant functions, linear or non-linear functions. Only the function,y(t), and its derivatives are used in determining if a differential equation is linear.
If a differential equation cannot be written in the form, (11) then it is called a non-linear differential equation.
In (5) - (7) above only (6) is non-linear, the other two are linear differential equations. We can’t classify (3) and (4) since we do not know what form the function F has. These could be either linear or non-linear depending on F.

Solution


solution to a differential equation on an interval α<t<β is any function y(t) which satisfies the differential equation in question on the interval α<t<β. It is important to note that solutions are often accompanied by intervals and these intervals can impart some important information about the solution. Consider the following example.


Example 1 Show that y(x)=x32 is a solution to 4x2y+12xy+3y=0 for x>0.

We’ll need the first and second derivative to do this.
y(x)=32x52y(x)=154x72Plug these as well as the function into the differential equation.
4x2(154x72)+12x(32x52)+3(x32)=015x3218x32+3x32=00=0So, y(x)=x32 does satisfy the differential equation and hence is a solution. Why then did we include the condition that x>0? We did not use this condition anywhere in the work showing that the function would satisfy the differential equation.
To see why recall that
y(x)=x32=1x3In this form it is clear that we’ll need to avoid x=0 at the least as this would give division by zero.
Also, there is a general rule of thumb that we’re going to run with in this class. This rule of thumb is : Start with real numbers, end with real numbers. In other words, if our differential equation only contains real numbers then we don’t want solutions that give complex numbers. So, in order to avoid complex numbers we will also need to avoid negative values of x.

So, we saw in the last example that even though a function may symbolically satisfy a differential equation, because of certain restrictions brought about by the solution we cannot use all values of the independent variable and hence, must make a restriction on the independent variable. This will be the case with many solutions to differential equations.

In the last example, note that there are in fact many more possible solutions to the differential equation given. For instance, all of the following are also solutions
y(x)=x12y(x)=9x32y(x)=7x12y(x)=9x32+7x12
We’ll leave the details to you to check that these are in fact solutions. Given these examples can you come up with any other solutions to the differential equation? There are in fact an infinite number of solutions to this differential equation.

So, given that there are an infinite number of solutions to the differential equation in the last example (provided you believe us when we say that anyway….) we can ask a natural question. Which is the solution that we want or does it matter which solution we use? This question leads us to the next definition in this section.

Initial Condition(s)


Initial Condition(s) are a condition, or set of conditions, on the solution that will allow us to determine which solution that we are after. Initial conditions (often abbreviated i.c.’s when we’re feeling lazy…) are of the form,
y(t0)=y0and/or y(k)(t0)=yk
So, in other words, initial conditions are values of the solution and/or its derivative(s) at specific points. As we will see eventually, solutions to “nice enough” differential equations are unique and hence only one solution will meet the given initial conditions.
The number of initial conditions that are required for a given differential equation will depend upon the order of the differential equation as we will see.

Example 2 y(x)=x32 is a solution to 4x2y+12xy+3y=0y(4)=18, and y(4)=364.

As we saw in previous example the function is a solution and we can then note that
y(4)=432=1(4)3=18y(4)=32452=321(4)5=364and so this solution also meets the initial conditions of y(4)=18 and y(4)=364. In fact, y(x)=x32 is the only solution to this differential equation that satisfies these two initial conditions.


Initial Value Problem


An Initial Value Problem (or IVP) is a differential equation along with an appropriate number of initial conditions.


Example 3 The following is an IVP.
4x2y+12xy+3y=0y(4)=18,y(4)=364
Example 4 Here’s another IVP.
2ty+4y=3y(1)=4
As we noted earlier the number of initial conditions required will depend on the order of the differential equation.

Interval of Validity


The interval of validity for an IVP with initial condition(s)
y(t0)=y0and/or y(k)(t0)=yk
is the largest possible interval on which the solution is valid and contains t0. These are easy to define, but can be difficult to find, so we’re going to put off saying anything more about these until we get into actually solving differential equations and need the interval of validity.

General Solution


The general solution to a differential equation is the most general form that the solution can take and doesn’t take any initial conditions into account.

Example 5 y(t)=34+ct2 is the general solution to2ty+4y=3
We’ll leave it to you to check that this function is in fact a solution to the given differential equation. In fact, all solutions to this differential equation will be in this form. This is one of the first differential equations that you will learn how to solve and you will be able to verify this shortly for yourself.


Actual Solution


The actual solution to a differential equation is the specific solution that not only satisfies the differential equation, but also satisfies the given initial condition(s).

Example 6 What is the actual solution to the following IVP?2ty+4y=3y(1)=4

This is actually easier to do than it might at first appear. From the previous example we already know (well that is provided you believe our solution to this example…) that all solutions to the differential equation are of the form.
y(t)=34+ct2All that we need to do is determine the value of c that will give us the solution that we’re after. To find this all we need do is use our initial condition as follows.
4=y(1)=34+c12c=434=194So, the actual solution to the IVP is.
y(t)=34194t2
From this last example we can see that once we have the general solution to a differential equation finding the actual solution is nothing more than applying the initial condition(s) and solving for the constant(s) that are in the general solution.

Implicit/Explicit Solution


In this case it’s easier to define an explicit solution, then tell you what an implicit solution isn’t, and then give you an example to show you the difference. So, that’s what we’ll do.

An explicit solution is any solution that is given in the form y=y(t). In other words, the only place that y actually shows up is once on the left side and only raised to the first power. An implicit solution is any solution that isn’t in explicit form. Note that it is possible to have either general implicit/explicit solutions and actual implicit/explicit solutions.


Example 7 y2=t23 is the actual implicit solution to y=ty,y(2)=1
At this point we will ask that you trust us that this is in fact a solution to the differential equation. You will learn how to get this solution in a later section. The point of this example is that since there is a y2 on the left side instead of a single y(t)this is not an explicit solution!
Example 8 Find an actual explicit solution to y=ty,y(2)=1.

We already know from the previous example that an implicit solution to this IVP is y2=t23. To find the explicit solution all we need to do is solve for y(t).
y(t)=±t23Now, we’ve got a problem here. There are two functions here and we only want one and in fact only one will be correct! We can determine the correct function by reapplying the initial condition. Only one of them will satisfy the initial condition.
In this case we can see that the “-“ solution will be the correct one. The actual explicit solution is then
y(t)=t23
In this case we were able to find an explicit solution to the differential equation. It should be noted however that it will not always be possible to find an explicit solution.
Also, note that in this case we were only able to get the explicit actual solution because we had the initial condition to help us determine which of the two functions would be the correct solution.


We’ve now gotten most of the basic definitions out of the way and so we can move onto other topics.

Comments

Popular posts from this blog

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 - 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 ⁡ (...