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Let's solve another 2nd order linear homogeneous

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differential equation.

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And this one-- well, I won't give you the details before I

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actually write it down.

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So the differential equation is 4 times the 2nd derivative

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of y with respect to x, minus 8 times the 1st derivative,

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plus 3 times the function times y, is equal to 0.

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And we have our initial conditions y of

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0 is equal to 2.

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And we have y prime of 0 is equal to 1/2.

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Now I could go into the whole thing y is equal to e to the

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rx is a solution, substitute it in, then factor out e to

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the rx, and have the characteristic equation.

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And if you want to see all of that over again, you might

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want to watch the previous video, just to see where that

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characteristic equation comes from.

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But in this video, I'm just going to show you, literally,

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how quickly you can do these type of problems mechanically.

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So if this is our original differential equation, the

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characteristic equation is going to be-- and I'll do this

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in a different color-- 4r squared minus 8r plus 3r is

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equal to 0.

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And watch the previous video if you don't know where this

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characteristic equation comes from.

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But if you want to do these problems really quick, you

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just substitute the 2nd derivatives with an r squared,

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the first derivatives with an r, and then the function

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with-- oh sorry, no.

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This is supposed to be a constant--

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And then the coefficient on the original function is just

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a constant, right?

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I think you see what I did.

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2nd derivative r squared.

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1st derivative r.

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No derivative-- you could say that's r to the 0, or just 1.

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But this is our characteristic equation.

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And now we can just figure out its roots.

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This is not a trivial one for me to factor so, if it's not

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trivial, you can just use the quadratic equation.

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So we could say the solution of this is r is equal to

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negative b-- b is negative 8, so it's positive 8-- 8 plus or

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minus the square root of b squared.

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So that's 64, minus 4 times a which is 4,

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times c which is 3.

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All of that over 2a.

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2 times 4 is 8.

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That equals 8 plus or minus square root of 64 minus--

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what's 16 times 3-- minus 48.

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All of that over 8.

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What's 64 minus 48?

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Let's see, it's 16, right?

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Right.

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10 plus 48 is 58, then another-- so it's 16.

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So we have r is equal to 8 plus or minus the square root

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of 16, over 8, is equal to 8 plus or minus 4 over 8.

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That equals 1 plus or minus 1/2.

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So the two solutions of this characteristic equation--

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ignore that, let me scratch that out in black so you know

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that's not like a 30 or something-- the two solutions

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of this characteristic equation are r is equal to--

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well 1 plus 1/2 is equal to 3/2-- and r is equal to 1

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minus 1/2, is equal to 1/2.

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So we know our two r's, and we know that, from previous

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experience in the last video, that y is equal to c times e

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to the rx is a solution.

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So the general solution of this differential equation is

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y is equal to c1 times e-- let's use our first r-- e to

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the 3/2 x, plus c2 times e to the 1/2 x.

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This differential equations problem was literally just a

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problem in using the quadratic equation.

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And once you figure out the r's you have

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your general solution.

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And now we just have to use our initial conditions.

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So to know the initial conditions, we need to know y

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of x, and we need to know y prime of x.

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Let's just do that right now.

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So what's y prime?

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y prime of our general solution is equal to 3/2 times

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c1 e to the 3/2 x, plus-- derivative of the inside-- 1/2

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times c2 e to the 1/2 x.

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And now let's use our actual initial conditions.

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I don't want to lose them-- let me rewrite them down here

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so I can scroll down.

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So we know that y of 0 is equal to 2, and y prime of 0

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is equal to 1/2.

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Those are our initial conditions.

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So let's use that information.

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So y of 0-- what happens when you substitute x

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is equal to 0 here.?

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You get c1 times e to the 0, essentially, so that's just 1,

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plus c2-- well that's just e to the 0 again, because x is

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0-- is equal to-- so this is, when x is equal to 0, what is

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y? y is equal to 2.

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Y of 0 is equal to 2.

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And then let's use the second equation.

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So when we substitute x is equal to 0 in the derivative--

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so when x is 0 we get 3/2 c1-- this goes to 1 again-- plus

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1/2 c2-- this is 1 again, e to the 1/2 half times 0 is e to

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the 0, which is 1-- is equal to-- so when x is 0 for the

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derivative, y is equal to 1/2, or the derivative is 1/2 at

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that point, or the slope is 1/2 at that point.

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And now we have two equations and two unknowns, and we could

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solve it a ton of ways.

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I think you know how to solve them.

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Let's multiply the top equation-- I don't know--

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let's multiply it by 3/2, and what do we get?

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We get-- I'll do it in a different color-- we get 3/2

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c1 plus 3/2 c2 is equal to-- what's 3/2 times 2?

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It's equal to 3.

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And now, let's subtract-- well, I don't want to confuse

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you, so let's just subtract the bottom from the top, so

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this cancels out.

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What's 1/2 minus 3/2?

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1/2 minus 1 and 1/2.

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Well, that's just minus 1, right?

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So minus c2 is equal to-- what's 1/2 minus 3?

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It's minus 2 and 1/2, or minus 5/2.

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And so we get c2 is equal to 5/2.

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And we can substitute back in this top equation.

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c1 plus 5/2 is equal to 2, or c1 is equal to 2, which is the

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same thing as 4/2, minus 5/2, which is equal to minus 1/2.

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And now we can just substitute c1 and c2 back into our

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general solution and we have found the particular solution

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of this differential equation, which is y is equal to c1-- c1

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is minus 1/2-- minus 1/2 e to the 3/2 x plus c2-- c2 is

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5/2-- plus c2, which is 5/2, e to the 1/2 x, and we are done.

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And it might seem really fancy.

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We're solving a differential equation.

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Our solution has e in it.

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We're taking derivatives and we're doing

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all sorts of things.

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But really the meat of this problem was solving a

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quadratic, which was our characteristic equation.

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And watch the previous video just to see why this

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characteristic equation works.

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But it's very easy to come up with the characteristic

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equation, right?

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I think you obviously see that y prime turns into r squared,

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y prime turns into r, and then y just turns into 1,

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essentially.

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So you solve a quadratic.

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And then after doing that, you just have to take one

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derivative-- because after solving the quadratic, you

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immediately have the general solution-- then you take its

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derivative, use your initial conditions.

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You have a system of linear equations which is Algebra I.

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And then you solve them for the two constants, c1 and c2,

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and you end up with your particular solution.

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And that's all there is to it.

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I will see you in the next video.

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