WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome to www.educator.com, welcome back to AP Calculus.
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Today, we are going to talk about a technique for actually solving any particular differential equation that you are faced with.
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The technique is called separation of variables.
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It is a very simple technique.
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It does not apply to every differential equation but of course to the ones that it does, it is very straightforward.
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You just literally separate the x and y variables and you integrate.
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It is that simple, so let us get started.
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Let us look at the following differential equation.
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I will go ahead and work with blue maybe.
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Let us look at the following differential equation.
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Let us look at y' is equal to y² × cos(x).
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Or if you want to put it in terms of dy dx, it is dy dx is equal to y² cos(x).
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For separation of variables, you definitely want to use the dy dx.
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You will see why in just a second.
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The idea is this, if we can rearrange a differential equation
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such that all symbols containing x's are on one side of the equality sign
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and all symbols containing y are on the other, we call this a separable equation or a separable differential equation.
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We can apply this idea of separation of variables.
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Separable equations are very easy to solve.
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They are easy in the sense that all you do is integrate.
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It is as easy as the integration is.
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You just integrate both sides.
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In this particular case, we have dy dx is equal to y² cos x.
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We are going to separate the variables, if we can.
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dx, I’m going to move it over to the right side.
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I get dy = y² cos x dx.
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Now everything that involves of y, I'm going to bring over to this side.
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I’m going to divide everything by y².
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I’m going to end up with dy/ y² = cos x dx.
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Everything involving a y is on the left, everything involving an x is on the right.
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We were able to separate the variables.
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We have separated the variables.
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Now we just integrate both sides.
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Let me write the equation again.
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We have dy/ y² is equal to cos x dx.
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You have an equality.
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Anything you do to the left side of the equality, as long as you do the right side, the equality is maintained.
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We just integrate both sides.
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It is literally that simple.
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What you end up with is, the integral, I’m going to rewrite this y⁻² dy = the integral of cos x dx.
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This one here becomes –y⁻¹ + let us just call it c1, constant, these are indefinite integrals.
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It is equal to sin x + c2.
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C1 and c2 are just constants.
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I’m just going to put them together.
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I’m going to combine c1 and c2 into just a single constant and put it on whichever side I want, in this case, on the right side.
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I’m going to write this as -1/y = sin x + c.
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Now I just solve for y.
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Y is equal to -1/ sin x + c.
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There you go, that is your differential equation.
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This is your function of x.
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In this case, we are actually able to solve explicitly for y = some function of x.
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You would not be always be able to do that.
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You might have to leave it in implicit form because the y and x might be a little too mixed up.
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We will see some examples of that, it is not a problem.
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It is literally that simple.
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C is the constant and it can be anything.
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This is the general solution.
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Remember, a general solution, you have a constant.
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For different values of the constant, you are going to get different curves in the xy plane.
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It is that simple, literally, it is that simple.
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If you can separate the variables, you can integrate.
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Hopefully, you can integrate.
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It is that simple.
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You separate variables, if you can, and two, you integrate.
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You are going to solve the different equation for y, some function y that your looking for.
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That is your unknown.
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Let us take a look at what this looks like graphically, for different values of c.
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Here you have it, here are the different values of c.
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The orange, black, you have a purple, and you have a green.
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When c = 1, you have the black graph, that is this one right here.
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Let me do this in red.
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When c = 1, you get this black curve right here, that is the solution.
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This is one particular number that goes on.
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When c = 2, we have the green graph.
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Now it changes a little bit, now it is this way, that is that one.
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When c = -2, that is the purple graph.
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This time it is above the x axis.
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When c = -1/2, we have the orange graph.
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That is all that is going on here.
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This is the family of solutions for different values of c.
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Once again, in this particular case, this was y =,
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y as a function of x is equal to -1/ sin(x) + c, for different values of c.
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Once you pick a c, it is going to be one particular curve.
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Let us go ahead and do an example here.
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Solve the following initial value problem.
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We remember, an initial value problem is the differential equation and some initial value.
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The x valued does not always have to be 0, it can be any number.
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In this particular case, it happens to be 0.
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They tell me that the curve, when x is equal to 0, it passes through the point 4.
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It passes through the point 0,4.
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I know this much and I know that this is the relationship, that dy dx = x²/ y³.
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Let us see if this is possible to solve.
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Here we have, y’, I'm going to write that as, let me do this in blue.
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I’m going to write that as dy dx is equal to x²/ y³ and we try to separate variables.
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We are going to end up with dy = x² dx/ y³.
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Now I’m going to multiply by y³.
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I end up with y³ dy = x² dx.
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Yes, in this particular case, we were able to separate the variables.
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Now we just integrate both sides.
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Integrate this side, integrate this side, and I end up with y⁴/ 4 is equal to x³/ 3 + c.
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This is my general solution.
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If you want to, you can solve explicitly for y.
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Let me just say, you are welcome to leave it in this form.
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Excuse me, you are welcome to leave it in this implicit form.
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In other words, not y = some function of x.
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Again, sometimes it is not going to be possible, sometimes it is.
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Or if possible, you can solve explicitly for y.
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In this particular case, it ends up being, I multiply by 4.
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I end up with y⁴ = 4/3 x³ + 4c.
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I get y = 4/3 x³ + 4c¹/4.
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This is the explicit, it does not matter, it is a personal choice, however far you want to take it.
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This is the general solution, the one that involves the constant.
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Now let us deal with the initial value problem.
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y(0) = 4 or y(0) = 2, it does not matter.
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I think on my paper I have a different value but it does really matter.
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For the initial value, y(0) is equal to 4.
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Now we have y⁴/ 4, we will write our equation down.
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It is equal to x³/ 3 + c.
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Now we are looking for this particular c.
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We put these in, y is equal to 4.
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It is going to be 4⁴/ 4 is equal to 0³/ 3.
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This is x, this is y, + c.
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C is going to equal 4³.
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4 × 4 is 16, 4 × 16 is 64, c = 64.
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Therefore, our final solution is y⁴/ 4, our particular solution is equal to x³/ 3 + 64.
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That is it, it is that simple.
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This is what the family of solutions looks like for different values of c.
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That is it, like that.
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For a particular value of c, you are going to get a curve that looks like that, one of those curves.
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Let us take a look at another example.
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Solve the following initial value problem.
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This time we have y’ = y sin x/ y³ + 2 with initial value y(0) = -2.
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Let us see if we can separate the variables here.
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We are going to write this as dy dx is equal to y × sin(x)/ y³ + 2.
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When I separate the variables, I'm going to end up with the following.
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I’m going to end up with y³ + 2/ y dy is equal to sin x dx.
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Bring this over here, bring this down here, bring the dx up there.
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I was able to separate the variables.
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Everything with a y on one side, everything with an x on the other side.
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The variables are separated, now we integrate both sides.
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We integrate this side and we integrate that side.
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This integral over here becomes the integral of y² dy, because y³/ y is y².
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I’m going to separate this fraction, + the integral of 2/ y dy is going to equal the integral sin x dx.
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Now I take care of the integrals.
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This one becomes y³/ 3.
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This one becomes 2 × natlog of the absolute value of y.
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This one = - cos(x) + a constant c.
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This is our general solution.
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Now we deal with the initial value y(0) = -2.
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I put these values into here, x is 0, y is 2.
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This is the y, this is my x.
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I put these into here and I solve for c.
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I have y is -2, this is going to be -2³.
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I should do that actually, /3 + 2 × natlog of y absolute value of -2 = -cos(0) + c.
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I end up with -8/3 + 2 ln 2 = -1 + c.
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I move the 1 over, I get c = -5/3 + 2 ln 2.
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This is my value of c.
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Put that back in there and I get my particular solution.
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Our particular solution, in other words, our solution to the initial value problem,
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our particular solution is y³/ 3 + 2 × ln of the absolute value of y = -cos x + 2 × natlog of 2 ln 2 - 5/3.
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There you go, that is it, this is the solution.
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In this particular case, we definitely want to leave it in implicit form.
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In fact, we have no choice because separating the y here and we have a y here as the argument of the logarithm function.
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This is going to be very difficult to separate, if it is possible at all.
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I do not ever think it is possible to separate the y from the x.
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This is a perfectly good, perfectly valid, completely acceptable solution to that particular initial value problem.
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Once again, separate the variables, if possible, integrate.
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And then, put the values in for the initial value problem, if there is an initial value.
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Or else just leave it with the constant c for the general solution.
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Let us see what this one looks like.
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This looks like this.
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This is the particular solution, here is the equation.
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This is what it looks like.
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That is it, for a particular value of x, now you know what the particular value of y is going to be.
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It is that simple.
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Let us move on to the next.
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Find an equation of the curve whose slope at the point xy is x² y, and which passes through the point 3/5, 5.
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We are looking for some y(x), again, some curve.
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This is the verbal description of an initial value problem.
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The initial value problem is the following.
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Find an equation of the curve whose slope of the point xy is x² y.
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What is the slope?
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The slope is the derivative.
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The derivative is dy dx.
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Dy dx, the slope at the point xy is equal to x² y.
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That is our differential equation, and which passes through the point 3/2, 5.
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The initial value is y of 3/2 = 5.
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This is the initial value problem that is represented by this word problem.
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Let us go ahead and see what we can do.
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We have dy dx = x² y.
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We separate variables, if we can.
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We divide by y, move the dx up here.
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We should have something like dy/y is equal to x² dx.
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Now we can integrate both sides.
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We are left with the natlog of the absolute value of y is equal to x³/ 3 + c.
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This is our general solution.
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It gives us a family of curves.
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Now let us go ahead and use the initial value.
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This is our x value, this is our y value.
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The natlog of the absolute value of 5.
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The absolute value 5 is 5, this becomes natlog of 5 is equal to x which is 3/2³/ 3 + c.
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And then, you just solve it from here, not a problem at all.
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This ends up giving the natlog of 5 is equal to 9/8 + c.
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Therefore, c is equal to the natlog of 5 - 9/8.
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Our solution is the natlog of the absolute value of y is equal to x³/ 3 + our value of c which is the natlog of 5 - 9/8.
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This is our particular solution.
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That is it, it is that simple.
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Let us express this another way, it is a logarithm.
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If you want to, you do not have to, this perfectly acceptable.
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If you want to, just to see what something else looks like.
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Let us express this, perhaps your teacher wants you to express it in terms of exponential,
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perhaps the test that you are taking wants you to express it, in terms of exponential, who knows.
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Perhaps you want to.
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Let us express this in another way.
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We have the natlog of the absolute value of y is equal to x³/ 3 + c.
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This is the natlog, just exponentiate both sides.
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What you are left with is the absolute value of y is equal to e ⁺x³/ 3 + c.
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We already found c.
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But if you did not find c, then if you just continued along with this, to put in the exponential form, you can find c.
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We already found c but we could also have come to this point and done,
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We said that y(3/2) = 5, this is our x value, this is our y value.
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Absolute value of 5 is 5.
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We have 5 = e x³/ 3, when I put 3/2 and I cube it and I divide by 3, 9/8 + c.
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I need to solve for c.
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I need to take the logarithm of both sides.
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I get ln of 5 = 9/8 + c which implies that c = ln 5 – 9/2, which is exactly what I got before.
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The exponential version of our solution becomes the absolute value of y is equal to e ⁺x³/ 3 + ln of 5 -9/8.
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This is the other version.
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I like to say something about the absolute value here, because absolute value scares the hell out of people.
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I do not understand why, it used to scare the hell out of me.
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It still does, from time to time.
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Let us just do a quick recollection here.
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Recall the meaning of absolute value.
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The absolute value of y, it means the following.
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It is y, if y happens to be bigger than 0.
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It is –y, if y happens to be less than 0.
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That is the definition of absolute value.
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We have the absolute value of y is equal to e ⁺x³/ 3 + c.
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For y greater than 0, the absolute value of y is just equal to y.
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Therefore, y is equal to just e ⁺x³ + c.
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For y less than 0, we have the absolute value of y is equal to –y.
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Therefore, -y is equal to e ⁺x³/ 3 + c.
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y is equal to -e ⁺x³/ 3 + c.
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That is what the absolute value means.
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Graphically, this is what is going to happen.
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The absolute value sign, the absolute value gives you graphs above the x axis, below the y axis.
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This is the positive, y = e ⁺x³ + c.
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This is negative, e ⁺x³ + c.
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What you have are different values for the constant c.
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In our particular case, which is this one right here.
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The blue, we said the absolute value of y is equal to x³/ 3 + ln 5 - 9/8, that is this blue curve right here.
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Notice how it passes through the point 3/2, 5, and this one here because absolute value.
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It is that simple, separate the variables and integrate.
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And then, solve the initial value problem for the value of the constant.
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If you need to, go ahead and graph it.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.