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I've spoken a lot about second order linear homogeneous

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differential equations in abstract terms, and how if g

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is a solution, then some constant

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times g is also a solution.

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Or if g and h are solutions, then g

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plus h is also a solution.

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Let's actually do problems, because I think that will

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actually help you learn, as opposed to

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help you get confused.

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So let's say I have this differential equation, the

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second derivative of y, with respect to x, plus 5 times the

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first derivative of y, with respect to x, plus 6 times y

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is equal to 0.

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So we need to find a y where 1 times its second derivative,

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plus 5 times its first derivative, plus 6 times

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itself, is equal to 0.

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And now, let's just do a little bit of-- take a step

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back and think about what kind of function-- if I have the

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function and I take its derivative and then I take its

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second derivative, most times I get

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something completely different.

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Like, if y was x squared, then y prime would be 2x, and y

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prime prime would be 2.

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And then to add them together you'd say, well, how would my

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x terms cancel out so that you get 0 in the end?

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So draw back into your brain and think, is there some

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function that when I take its first and second derivatives,

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and third and fourth derivatives, it essentially

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becomes the same function?

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Maybe the constant in front of the function changes as I take

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the derivative.

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And if you've listened to a lot of my videos, you'd

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realize that it probably is what I consider to be the most

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amazing function in mathematics.

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And that is the function e to the x.

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And in particular, maybe e to the x won't work here-- or you

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can even try it out, right?

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If you did e to the x, it won't satisfy

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this equation, right?

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You would get e to the x, plus 5e to the x, plus 6e to the x.

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That would not equal to 0.

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But maybe y is equal to e to some constant r, times x.

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Let's just make the assumption that y is equal to some

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constant r times x, substitute it back into this, and then

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see if we can actually solve for an r that makes this

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equation true.

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And if we can, we've found the solution, or maybe we've found

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several solutions.

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So let's try it out.

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Let's try y is equal to e to the rx into this

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differential equation.

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So what is the first derivative of

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it, first of all.

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So y soon. prime is equal to what?

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Derivative chain rule.

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Derivative of the inside is r.

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And then derivative of the outside is still

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just e to the rx.

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And what's the second derivative?

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y prime prime is equal to derivative-- r is just a

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constant-- so derivative of the inside is r, times r on

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the outside, that's r squared, times e to the rx.

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And now we're ready to substitute back in.

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And I will switch colors.

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So the second derivative, that's r squared times e to

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the rx, plus 5 times the first derivative, so that's 5re to

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the rx, plus 6 times our function-- 6 times e to the rx

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is equal to 0.

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And something might already be surfacing to you as something

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we can do this equation to solve for r.

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All of these terms on the left all have an e to the rx, so

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let's factor that out.

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So this is equal to e to the rx times r squared, plus 5r,

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plus 6 is equal to 0.

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And our goal, remember, was to solve for the r, or the r's,

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that will make this true.

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And in order for this side of the equation to be

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0, what do we know?

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Can e to the rx ever equal 0?

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Can you ever get something to some exponent and get 0?

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Well, no.

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So this cannot equal 0.

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So in order for this left-hand side of the equation to be 0,

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this term, this expression right here, has to be 0.

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And I'll do that in a different color.

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So we know, if we want to solve for r, that this, r

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squared plus 5r, plus 6, that has to be 0.

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And this is called the characteristic equation.

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This, the r squared plus 5r, plus 6, is called the

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characteristic equation.

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And it should be obvious to you that now this

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is no longer calculus.

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This is just factoring a quadratic.

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And this one actually is fairly

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straightforward to factor.

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So what is this?

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This is r plus 2, times r plus 3 is equal to 0.

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And so the solutions of the characteristic equation-- or

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actually, the solutions to this original equation-- are r

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is equal to negative 2 and r is equal to minus 3.

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So you say, hey, we found two solutions, because we found

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two you suitable r's that make this

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differential equation true.

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And what are those?

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Well, the first one is y is equal to e to

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the minus 2x, right?

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We can call that y1.

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And then the second solution we found, y2 is e to the--

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what is this?-- r is minus 3x.

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Now my question to you is, is this the

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most general solution?

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Well, in the last video, in kind of our introductory

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video, we learned that a constant times a solution is

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still a solution.

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So, if y1 is a solution, we also know that we can multiply

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y1 times any constant.

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So let's do that.

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Let's multiply it by c1.

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That's a c1 there.

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This is also going to be a solution.

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And now it's a little bit more general, right?

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It's a whole class of functions.

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The c doesn't have to just be 1, it can be any constant.

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And then when you use your initial values, you actually

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can figure out what that constant is.

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And same for y2.

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y2 doesn't have to be 1 times e to the minus 3x, it has to

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be any constant.

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And we learned that in the last video, that if

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something's a solution, some constant times

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that is also a solution.

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And we also learned that if we have two different solutions,

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that if you add them together, you also get a solution.

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So the most general solution to this differential equation

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is y-- we could say y of x, just to hit it home that this

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is definitely a function of x-- y of x is equal to c1e to

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the minus 2x, plus c2e to the minus 3x.

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And this is the general solution of this

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differential equation.

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And I won't prove it because the proof is fairly involved.

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I mean, we just tried out e to the rx.

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Maybe there's some other wacko function that would have

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worked here.

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But I'll tell you now, and you kind of have to take it as a

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leap of faith, that this is the only general solution.

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There isn't some crazy outside function there that would have

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also worked.

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And so the other question that might be popping in your brain

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is, Sal, when we did first order differential equations,

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we only had one constant.

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And that was OK, because we had one set of initial

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conditions and we solved for our constants.

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But here, I have two constants.

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So if I wanted a particular solution, how can I solve for

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two variables if I'm only given one initial condition?

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And if that's what you actually thought, your

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intuition would be correct.

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You actually need two initial conditions to solve this

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differential equation.

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You would need to know, at a given value of x,

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what y is equal to.

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And, maybe at a given value of x, what the

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first derivative is.

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And that is what we will do in the next video.

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See

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