Amplitude-Phase Form Proof

Again, I pulled formulas out of thin air when I showed amplitude-phase form. Your professor may want to see justification, so I will show an example here.

Say you have solved a differential equation, and you got an answer in the form y = c₁e^βxcos(γx) + c₂e^βxsin(γx). Now you're supposed to convert to amplitude phase form. As I stated in my previous post, this answer will be in the following form:  y(x) = Ae^βxcos(γx - θ). But why?

First, we will start by defining the constant A, which is equal to √(c₁² + c₂²). Now on the right-hand side of your equation, multiply and divide by A. This is the same as multiplying by 1, and does not change the actual problem. You'll get:

y = c₁√(c₁² + c₂²)/√(c₁² + c₂²)e^βxcos(γx) + c₂√(c₁² + c₂²)/√(c₁² + c₂²)e^βxsin(γx)

Now visualize a right triangle in which the legs have length c₁ and c₂. By the Pythagorean theorem, the hypotenuse will be √(c₁² + c₂²).

So you can replace several pieces of the above equation in terms of θ.

y = √(c₁² + c₂²)cos(θ)e^βxcos(γx) + √(c₁² + c₂²)sin(θ)e^βxsin(γx).

And what's left is equal to A:

y = Acos(θ)e^βxcos(γx) + Asin(θ)e^βxsin(γx).

Pull out common factors:

y = Ae^βx(cos(θ)cos(γx) + sin(θ)sin(γx)).

Now use a trigonometic identity to clean it up: cos(A)cos(B) + sin(A)sin(B) = cos(B-A)

y = Ae^βx(cos(γx - θ)

And that's the proof!

Second Order Linear Answer Proofs

A few posts ago, I explained how to get answers for second order linear diff EQs. However, I just provided formulas. This isn't enough for some professors! They might ask that you justify where that formula came from. To justify, you'll have to show a proof.

In lecture, your professor will probably show you a proof using variables for every term. This can get messy and confusing, plus there is no point in me repeating exactly what you've already been shown. So, I am going to show proofs on specific example problems.

As outlined in the previous post on this topic, there are three scenarios you will encounter. Which scenario you are in will depend on the results of your characteristic equation.

Scenario 1: b²-4ac > 0: 2 real roots
Example problem: 6y'' + y' - y = 0

As I already explained, the answer in this situation will take the following form: y(x) = c₁e^α₁x + c₂e^α₂x. But why?

Well, start with a guess. Guess that your answer will take the form y = e^αx where α is a constant. Now take the first derivative: y' = αe^αx. And the second: y'' =  α²e^αx. Now substitute these values in for y, y', and y'' in the original equation:

6α²e^αx + αe^αx - e^αx = 0

Now you can factor out a common term of e^αx:

e^αx(6α² + α - 1) = 0

Now, e^αx can never equal zero. It is a positive number raised to some power, and in this situation you will never receive zero for an answer. This means that the only way to make the equation hold true is to make the part in parenthesis equal 0.

(6α² + α - 1) = 0
α = -1/2, 1/3

And that's where your characteristic equation comes from. Now you can see why it is that when you solve the characteristic equation for its roots, you are finding solutions to the differential equation. Since this characteristic equation is a second order equation (the highest exponent is 2), it will have two solutions. This is why a second order differential equation also has two solutions.

We assumed at the beginning that the answer was in the form y = e^αx, and we were able to show that this assumption can lead us to answers that satisfy the equation. So it is valid to take the values of α you find from the characteristic equation, and substitute them into  y = e^αx.

The Superposition Principle says that any answer to a differential equation multiplied by a constant is still an answer to the equation, and any two answers added together are also an answer. This is why we take the two solutions to the characteristic polynomial and put constants (c₁ and c₂) in front, and why we add them together to get the final solution.

Scenario 2: b²-4ac < 0: imaginary roots
Example problem: y'' + 2y' + 2y = 0

I stated in my previous post that the answer will end up being in the following form: y(x) = c₁e^αxcos(βx)+ c₂e^αxsin(βx). Now let's prove it.

Start out similarly to what we did last time: assume the answer will take the form y = e^αx. Take the first and second derivatives and plug into the original problem as before. Pull out the common term and you will get the characteristic equation:

(α² + 2α + 2) = 0

The problem now is that this equation does not have real roots. Use the quadratic formula to find its imaginary roots:

α = (-2 ± √(4 - 4*1*2))/2
α = 1 ± i

So you could plug in for α at this point and say that your solution is y = c₁e^(1+i)x + c₂e^(1-i)x. But those have imaginary numbers in them, and we don't like that.

To solve that problem, we use Euler's formula, which states that y = e^ix = cos(x) + isin(x). You don't need to prove this statement on your test.

So look at your solution as written right now: y = c₁e^(1+i)x + c₂e^(1-i)x.
Distribute: y = c₁e^1x+1ix + c₂e^1x-1ix.     (I am leaving the 1's in there to be clear about what is happening with your coefficients. They are preserved.)
Now remember a property of exponents: e^x+y = e^xe^y. So rewrite your equation using this rule:
y = c₁e^1xe^1ix  + c₂e^1xe^-1ix

Now you can apply Euler's formula on the terms containing i. You will get the following:

y = c₁e^1x(cos(1x) + isin(1x)) + c₂e^1x(cos(-1x) + isin(-1x))

Better, but there's still those pesky i's. We now use a property of cosine and sine to help us with the next step. Cosine is an even function. That means that cos(-x) = cos(x). Sine is an odd function, meaning that sin(-x) = -sin(x). We will use these rules to rewrite what is above:

y = c₁e^1x(cos(1x) + isin(1x)) + c₂e^1x(cos(1x) - isin(1x))

Great! Now we have like terms. Regroup the equation so that cosine and sine are together:

y = (c₁+ c₂)e^1xcos(1x) + i(c₁- c₂)e^1xsin(1x)

Now we just redefine our coefficients. c₁+ c₂ = c₃ and i(c₁- c₂) = c_4.

We can now write the equation as y = y = c₃e^1xcos(1x) + c₄e^1xsin(1x). All done!

There is a third scenario,  b²-4ac = 0: repeated roots. This proof is very similar to the first one, and I will leave it up to you! If you really want to see it, just shoot me a message.

Checking Your Answers

One of the great things about Diff EQs is you can always check your answer. The checks aren't necessarily super easy, though, so I'd recommend waiting until the end of an exam and checking if you still have time left over. Your professor will probably take off points for algebraic or differentiation errors, so checking is definitely a good idea!

Say you're solving the equation:

y'' - 10y' + 25y = 0; y(0) = 1, y'(0) = 0

And you've come up with the following solution:

y(x) = e^5x - 5xe^5x

First make sure your first initial condition holds: y(0) = 1

y(0) = e^5*0 - 5*0*e^5*0y(0) = 1 - 0 = 1

Now take the derivative of your equation:

y'(x) = 5e^5x - 5e^5x - 25xe^5x     (notice that you must use product rule on the second term)
y'(x) = -25xe^5x

Check your second initial condition: y'(0) = 0

y'(0) = -25*0*e^5*0y'(0) = 0

Now take the second derivative:

y''(x) = -25e^5x - 125xe^5x      (product rule again)

Now you have y'', y', and y - you can substitute these into the original equation and make sure it all equals zero.

y'' - 10y' + 25y = 0
-25e^5x - 125xe^5x - 10(-25xe^5x) + 25(e^5x - 5xe^5x) = 0
-25e^5x - 125xe^5x + 250xe^5x + 25e^5x - 125xe^5x = 0

Notice the 1st and 4th terms cancel, and the 2nd, 3rd, and 5th terms cancel, so the whole thing equals 0, and your solution checks out. Go ahead and try this anytime you're doubting yourself!

Happy calculating,
Rachel

Amplitude-Phase Form

All the equations shown in my last blog post can be written in a different, equivalent form called amplitude-phase form. Converting to amplitude-phase form is not difficult, just take a few easy steps:

Find A. A is the amplitude, and it is equal to √(c₁² + c₂²).

Find θ. θ is the phase angle, and it can be found via its sine and cosine. cosθ = c₁/A, and sinθ = c₂/A. You can then use either arccosine or arcsine to find θ, but remember that both of those functions yield two answers, since all sines/cosines correspond to two different angles in two different quadrants, and your calculator won't necessarily give you the right one. So use the information about the sign of your cosine and sine to choose the angle in the correct quadrant. Remember that cosine is always positive in the quadrants 1 and 4 and negative in 2 and 3. Sine is positive in 1 and 2 and negative in 3 and 4.

Then assemble your terms like so:

y(x) = Ae^βxcos(γx - θ)

And that's the equation in amplitude-phase form. You're done!

Your professor might want more steps shown than what I have here. This is just the answer. The proof will be included in another blog post.

Let's do an example problem. Say you've found the following answer to an initial value problem and you want to put it in amplitude phase form:

y = cos(2x) - √3sin(2x)

First find A: your coefficients are 1 and √3, so A is: √(1² + (-√3)²) = √(1 + 3) = 2.

Now find θ. The two pieces of information we know about θ are that cosθ = 1/2 and sinθ = -√3/2.
This is where you have to remember some trigonometry. There are two "special" triangles that you ought to remember some information about. Here they are:

Right triangles whose angles are as shown will always have side lengths in proportion to the values shown. Back to our example problem, we are looking for an angle with cosine (adjacent over hypotenuse) of 1/2, and sine (opposite over hypotenuse) of -√3/2. See anything helpful above? If you guessed  π/3, you got it. Note that the angle labelled π/3 on the righthand triangle has got hypotenuse 2, opposite side √3, and adjacent side 1.

But the sine is negative for the angle we're looking for. π/3 is in the first quadrant, where both sine and cosine are positive. The only quadrant in which cosine is positive and sine is negative is the fourth quadrant. So the angle we're really looking for is one in the fourth quadrant that is π/3 away from the x axis. This can be found by 2π - π/3 = 5π/3. That is the value of θ.

Now that we have both A and θ, just arrange everything into the correct form. Your final answer is y = 2cos(2x - 5π/3).

Note that, when solving for θ, if your sine and cosine are not on those triangles, and they're not 0 or 1, your professor cannot possibly expect you to know the arcsines and arccosines required to find θ unless he allows you to use a calculator on your test. You can leave these alone, just might want to write down which quadrant as well to show that you know that.

I mentioned 0 and 1 because they're on the unit circle, which you should also try and remember from trigonometry:

Recall that cosine is x and sine is y. So, for example, if you've got a cosine of 0 and a sine of 1, look at the point (0, 1). The angle π/2 is there. And that's how you use the unit circle.

Happy calculating,
Rachel

Solving Second-Order Linear Homogeneous Constant Coefficient Ordinary Differential Equations

If that title seems like a mouthful, it's because our studies so far have taught us how to solve just one very specific kind of ordinary differential equation (ODE). If you're unsure of how to determine if your equation fits the bill, check out my last post on classifying ODEs. To put it shortly, your equation should take the form ay'' + by' + cy = 0, where a, b, c are constants.

To start, find the characteristic polynomial. This is just aα² + bα + c = 0. Your professor may have used r instead of alpha. Now, solve for α using the quadratic formula, α = (-b ± √(b² - 4ac))/(2a). An astute reader might notice at this point that our problem branches off into three possible situations.

First possibility is that b² - 4ac > 0. If this occurs, the term under the radical is positive, and you receive 2 real solutions for α, we'll call them α₁ and α₂. Your solution to the ODE will then take the form y(x) = c₁e^α₁x + c₂e^α₂x.

The second situation is that b² - 4ac = 0. In this case, the ± term in your quadratic equation vanishes, and you therefore get just one root, which we'll call α. Your general solution will now take the form y(x) = c₁e^αx + c₂xe^αx.

Finally, you might run into the situation that b² - 4ac < 0, which means that the equation has imaginary roots equal to -b/(2a) ± i√(4ac - b²)/(2a). An imaginary number has two distinct parts, the real part and the imaginary part, written in the form β+iγ (i is √-1). In this case,  β = -b/(2a), and γ = √(4ac - b²)/(2a). And the general solution takes the form y(x) = c₁e^βxcos(γx)+ c₂e^βxsin(γx).

c₁ and c₂ represent any constant, and these cannot be found specifically without being given initial conditions - that is, a value for y(0) and y'(0). If you are given initial conditions, you must take some additional steps to determine c₁ and c₂. Plug in 0 for x and set the resulting expression equal to y(0). This will give you one equation involving c₁ and c₂. Then differentiate y(x), plug in 0 for x, and set the resulting expression equal to y'(0). You will then have a system of two equations involving c₁ and c₂. Use your choice of methods (substitution or elimination) to find c₁ and c₂.

These are just the answers! Your professor may want you to show many more steps than what I have shown here. And as always, if you have any questions, come to MASH. And if there's anything specific that you'd like to see on the blog, leave a comment. You don't need to have a blogspot account to comment.

Happy calculating,
Rachel

Classifying Ordinary Differential Equations

Ordinary differential equations (ODEs) have several basic properties that you must be aware of.

One of the first important ones is its order. The order of an ODE is simply equal to the highest order derivative found in the equation. So, for example, the equation

y''' + 2y'' - 6xy' + x²y + y⁴ = sin(x)

is a third order equation, because y''' is a third order derivative, and is the highest order derivative found in the equation. Please note the important difference between terms such as y''' and y⁴; the latter is not a derivative and so is not considered when determining order of an ODE.

A second very important characteristic of an ODE is linearity. An ODE is linear if none of the y terms have exponents or involve any other "fancy" functions (like cos(y), or e^y). Squared x's are okay, but squared y's, if present, mean the equation is nonlinear. So, the equation

y''' + 2y'' - 6xy' + x²y + y⁴ = sin(x)

is nonlinear, because of the y⁴. But the equation

y'' -  x²y' + e^xy = sin(x)

is linear because the only y terms are y derivatives and y itself.

A linear equation is always capable of being arranged into the following form:

a₁(x)y + a₂(x)y' + a₃(x)y'' + ... = f(x)

That's kind of icky to look at, so let me break it down. Each term on the left contains y or a y derivative. a_n(x) are just the coefficients of each term. Note that these coefficients can be functions of x, but they can also just be constants, or even 0. There can be any number of terms, with increasing orders of derivatives, and all these terms are summed. Any terms that do not contain a y or a y derivative are written on the righthand side, which here is represented as f(x). f(x) is called the forcing function, and it can be a function of x, a constant, or 0.

This leads me to the last two properties. These last properties ONLY apply to linear ODEs. If your ODE is nonlinear, it cannot be classified any more specifically. If it is linear, further analysis must be done.

I said above that the coefficients a_n(x) can be functions of x or constants. This is another one of the properties. If ALL coefficients are constant, a linear ODE can be further classified as a constant coefficient ODE. If even one coefficient contains an x, the ODE is said to have variable coefficients. So the equation

y'' + 6y' - 11y = sin(x)

has constant coefficients, and the equation

y'' + 3xy' - 7y = 0

has variable coefficients.

Finally, like I mentioned before, the function f(x) on the righthand side of the generic linear equation listed above can be a function of x, a constant, or 0. If it is zero, the equation is said to be homogeneous. If it's not zero, the equation is nonhomogeneous. Another way to think about this is, if there is a term anywhere in the equation that does not contain y or a derivative of y, the equation is nonhomogeneous. So the equation

y'' + 5xy' + y = 7

is nonhomogeneous, while

y'' + sin(x)y = 0

is homogeneous.

So, quiz yourself: is the following equation homogeneous?

y'' + 7x²y' + y² = 0

Trick question! That equation isn't linear, so it cannot even be classified as homogeneous or nonhomogeneous. Those categories only apply to linear equations.

Okay, here's a real question. Is the following equation homogeneous?

y'' + (8/x)y' + √(x)y + 7 = 0

The answer is... no! Did I fool you? Don't get lazy and just look for a zero on the right side of the equation. There is a term (7) that does not contain y or a derivative of y, so the equation is inhomogeneous. This equation isn't shown in the standard form of a linear equation. If it were, that 7 would have been subtracted over to the righthand side, and this equation wouldn't be so deceptive.

One final question, to provide an overview of everything in this post. No tricks in this one, I promise. Tell the order of the following equation, whether it is linear or nonlinear, and, if applicable, whether it has constant/variable coefficients and is homogeneous/nonhomogeneous.

y''' + 2xy'' sin(x)y = 1/x

The answers are (highlight to read): third order, linear, variable coefficients, and nonhomogeneous.

As always, come to MASH if you want more help!

Happy calculating,
Rachel

Precalculus Toolkit

I thought I'd start by compiling a list of some of the useful algebraic and trigonometric rules that come up while solving differential equations. Many times, when students get stuck on a problem, it's not because they were unable to do the calculus. It's because they had trouble with the algebraic acrobatics that is sometimes necessary to get an answer into the proper form.

Sometimes, performing an operation on one representation of an equation would be very difficult or impossible using the methods you learn in this class. However, if you represent the equation in a different, equivalent way, the task might become much easier.

The list below is meant to refresh you on some of these methods, so that you won't spend hours beating your head against the wall to solve a problem that should be able to be solved easily. Also keep in mind that you might have to use the rules backwards, too!

Exponents:

1. Multiplying same base, different exponent: add the exponents
   (x^u)(x^v) = x^u+v
   Example: (x^4y)(x^y+6) = x^{4y + y +6} = x^{5y + 6}


2. Multiplying different base, same exponent: multiply the bases
   (x^u)(y^u) = (xy)^u
   Example: (x^3z+4)(y^3z+4) = (xy)^3z+4
    Please note that this rule DOES NOT apply if you are not multiplying! You cannot "distribute" an exponent over addition.


3. Raising an exponent to an exponent: multiply the exponents
   (x^u)^v = x^u*v
   Example:  (x^3z+4)² = x^(3z+4)(2) = x^6z+8


4. The Nth root is the same as having an exponent of 1/N
   _n√(u) = u^1/n
   Example: √(x+1) = (x+1)^1/2

Logarithms: 

1. Log of a product: add the log of each factor
   ln(u*v) = ln(u) + ln(v)
   Example: ln((x+1)(x-3)) = ln(x+1) + ln(x-3)
    
2. Log of a quotient: subtract the log of the denominator from the log of the numerator
   ln(u/v) = ln(u) - ln(v)
   Example: ln((x+1)/(x-3)) = ln(x+1) - ln(x-3)
    
3. Log of an exponential: bring the exponent out front
   ln(u^v)  =  u*ln(v)
   Example: ln((x+4)^x/5) = x/5*ln(x+4) 

Trigonometry:

1. Trig identities equal to 1
   sin²(u) + cos²(u) = 1
   sec²(u) - tan²(u) = 1
   csc²(u) - cot²(u) = 1
   Example:  sin²(2x + 3) + cos²(2x + 3) = 1
    
2. Multiple angle and power reduction trig formulas
   There are way too many to list here, but Wikipedia has a great table. These are to be used when you don't know what to do with that sin(2x) or sin³(x), and other similar situations.

Feel free to leave a comment if you think of more I should have included. You don't need to have a blogspot account to leave a comment. And as always, feel free to stop by the ARC in the lobby of Daniel's hall during my sessions to get help in person, or to email me at rachel.wigell (at) wpi (dot) edu.

Happy calculating,
Rachel