Linear homogeneous equations
2nd Order Linear Homogeneous Differential Equations 2 Let's find the general solution!
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- I've spoken a lot about second order linear homogeneous
- differential equations in abstract terms, and how if g
- is a solution, then some constant
- times g is also a solution.
- Or if g and h are solutions, then g
- plus h is also a solution.
- Let's actually do problems, because I think that will
- actually help you learn, as opposed to
- help you get confused.
- So let's say I have this differential equation, the
- second derivative of y, with respect to x, plus 5 times the
- first derivative of y, with respect to x, plus 6 times y
- is equal to 0.
- So we need to find a y where 1 times its second derivative,
- plus 5 times its first derivative, plus 6 times
- itself, is equal to 0.
- And now, let's just do a little bit of-- take a step
- back and think about what kind of function-- if I have the
- function and I take its derivative and then I take its
- second derivative, most times I get
- something completely different.
- Like, if y was x squared, then y prime would be 2x, and y
- prime prime would be 2.
- And then to add them together you'd say, well, how would my
- x terms cancel out so that you get 0 in the end?
- So draw back into your brain and think, is there some
- function that when I take its first and second derivatives,
- and third and fourth derivatives, it essentially
- becomes the same function?
- Maybe the constant in front of the function changes as I take
- the derivative.
- And if you've listened to a lot of my videos, you'd
- realize that it probably is what I consider to be the most
- amazing function in mathematics.
- And that is the function e to the x.
- And in particular, maybe e to the x won't work here-- or you
- can even try it out, right?
- If you did e to the x, it won't satisfy
- this equation, right?
- You would get e to the x, plus 5e to the x, plus 6e to the x.
- That would not equal to 0.
- But maybe y is equal to e to some constant r, times x.
- Let's just make the assumption that y is equal to some
- constant r times x, substitute it back into this, and then
- see if we can actually solve for an r that makes this
- equation true.
- And if we can, we've found the solution, or maybe we've found
- several solutions.
- So let's try it out.
- Let's try y is equal to e to the rx into this
- differential equation.
- So what is the first derivative of
- it, first of all.
- So y soon. prime is equal to what?
- Derivative chain rule.
- Derivative of the inside is r.
- And then derivative of the outside is still
- just e to the rx.
- And what's the second derivative?
- y prime prime is equal to derivative-- r is just a
- constant-- so derivative of the inside is r, times r on
- the outside, that's r squared, times e to the rx.
- And now we're ready to substitute back in.
- And I will switch colors.
- So the second derivative, that's r squared times e to
- the rx, plus 5 times the first derivative, so that's 5re to
- the rx, plus 6 times our function-- 6 times e to the rx
- is equal to 0.
- And something might already be surfacing to you as something
- we can do this equation to solve for r.
- All of these terms on the left all have an e to the rx, so
- let's factor that out.
- So this is equal to e to the rx times r squared, plus 5r,
- plus 6 is equal to 0.
- And our goal, remember, was to solve for the r, or the r's,
- that will make this true.
- And in order for this side of the equation to be
- 0, what do we know?
- Can e to the rx ever equal 0?
- Can you ever get something to some exponent and get 0?
- Well, no.
- So this cannot equal 0.
- So in order for this left-hand side of the equation to be 0,
- this term, this expression right here, has to be 0.
- And I'll do that in a different color.
- So we know, if we want to solve for r, that this, r
- squared plus 5r, plus 6, that has to be 0.
- And this is called the characteristic equation.
- This, the r squared plus 5r, plus 6, is called the
- characteristic equation.
- And it should be obvious to you that now this
- is no longer calculus.
- This is just factoring a quadratic.
- And this one actually is fairly
- straightforward to factor.
- So what is this?
- This is r plus 2, times r plus 3 is equal to 0.
- And so the solutions of the characteristic equation-- or
- actually, the solutions to this original equation-- are r
- is equal to negative 2 and r is equal to minus 3.
- So you say, hey, we found two solutions, because we found
- two you suitable r's that make this
- differential equation true.
- And what are those?
- Well, the first one is y is equal to e to
- the minus 2x, right?
- We can call that y1.
- And then the second solution we found, y2 is e to the--
- what is this?-- r is minus 3x.
- Now my question to you is, is this the
- most general solution?
- Well, in the last video, in kind of our introductory
- video, we learned that a constant times a solution is
- still a solution.
- So, if y1 is a solution, we also know that we can multiply
- y1 times any constant.
- So let's do that.
- Let's multiply it by c1.
- That's a c1 there.
- This is also going to be a solution.
- And now it's a little bit more general, right?
- It's a whole class of functions.
- The c doesn't have to just be 1, it can be any constant.
- And then when you use your initial values, you actually
- can figure out what that constant is.
- And same for y2.
- y2 doesn't have to be 1 times e to the minus 3x, it has to
- be any constant.
- And we learned that in the last video, that if
- something's a solution, some constant times
- that is also a solution.
- And we also learned that if we have two different solutions,
- that if you add them together, you also get a solution.
- So the most general solution to this differential equation
- is y-- we could say y of x, just to hit it home that this
- is definitely a function of x-- y of x is equal to c1e to
- the minus 2x, plus c2e to the minus 3x.
- And this is the general solution of this
- differential equation.
- And I won't prove it because the proof is fairly involved.
- I mean, we just tried out e to the rx.
- Maybe there's some other wacko function that would have
- worked here.
- But I'll tell you now, and you kind of have to take it as a
- leap of faith, that this is the only general solution.
- There isn't some crazy outside function there that would have
- also worked.
- And so the other question that might be popping in your brain
- is, Sal, when we did first order differential equations,
- we only had one constant.
- And that was OK, because we had one set of initial
- conditions and we solved for our constants.
- But here, I have two constants.
- So if I wanted a particular solution, how can I solve for
- two variables if I'm only given one initial condition?
- And if that's what you actually thought, your
- intuition would be correct.
- You actually need two initial conditions to solve this
- differential equation.
- You would need to know, at a given value of x,
- what y is equal to.
- And, maybe at a given value of x, what the
- first derivative is.
- And that is what we will do in the next video.
- See
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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