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.