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We hopefully know at this point what a differential

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equation is, so now let's try to solve some.

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And this first class of differential equations I'll

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introduce you to, they're called separable equations.

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And I think what you'll find is that we're not learning

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really anything new.

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Using just your first year calculus derivative and

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integrating skills, you can solve a separable equation.

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And the reason why they're called separable is because

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you can actually separate the x and y terms, and integrate

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them separately to get the solution of the

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differential equation.

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So that's separable.

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Separable equations.

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So let's do a couple, and I think you'll get the point.

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These often are really more of exercises in algebra than

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anything else.

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So the first separable differential equation is: dy

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over dx is equal to x squared over 1 minus y squared.

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And actually, this is a good time to just review our

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terminology.

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So first of all, what is the order of this

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differential equation?

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Well, the highest derivative in it is just the first

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derivative, so the order is equal to 1.

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So it's first order.

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It's ordinary, because we only have a regular derivative, no

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partial derivatives here.

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And then, is this linear or non-linear?

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Well, first you say, oh well, you know, this looks linear.

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I'm not multiplying the derivative

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times anything else.

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But if you look carefully, something

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interesting is going on.

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First of all, you have a y squared.

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And y is a dependent variable.

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y is a function of x.

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So to have the y squared, that makes it non-linear.

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And even if this was a y, if you were to actually multiply

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both sides of this equation times 1 minus y, and get in

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the form that I showed you in the previous equation, you

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would have 1 minus y squared.

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This is actually the first step of what we have to do

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anyways, so I'll write it down.

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So if I'm just multiplying both sides of this equation

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times 1 minus y squared, you get 1 minus y squared times dy

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dx is equal to x squared.

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And then you immediately see that even if this wasn't a

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squared here, you'd be multiplying the y times dy dx,

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and that also makes it non-linear because you're

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multiplying the dependent variable times the

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derivative of itself.

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So that also makes this a non-linear equation.

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But anyway, let's get back to solving this.

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So this is the first step.

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I just multiply both sides by 1 minus y squared.

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And the real end goal is just to separate the y's and the

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x's, and then integrate both sides.

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So I'm almost there.

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So now what I want to do is I want to multiply both sides of

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this equation times dx, so I have a dx here and get rid of

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this dx there.

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Let me go here, I don't want to waste too much space.

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So you get 1 minus y squared dy is equal to x squared dx.

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I have separated the x and y variables and the

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differentials.

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All I did is I multiplied both sides of this equation times

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dx to get here.

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Now, I can just integrate both sides.

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So let's do that.

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Whatever you do to one side of the equation you

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have to do the other.

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That's true with regular equations or

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differential equations.

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So we're going to integrate both sides.

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So what's the integral of this expression with respect to y?

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Let's see.

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The integral of 1 is y, the integral of y squared, well

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that's minus y to the third over 3.

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And I'll write the plus c here just to kind of show you

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something, but you really don't have to write a plus c

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on both sides.

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I'll call the plus the constant due to y.

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The y integration.

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You'll never see this in a calculus class, but I just

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want to make a point here.

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I just want to show you that our plus c has never

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disappeared from when we were taking our traditional

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antiderivatives.

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And what's the derivative of this?

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Well that's x to the third over 3.

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And this is also going to have a plus c

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due to the x variable.

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Now, the reason why I did this magenta one in magenta and I

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labelled it like that, is because you really just have

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to write a plus c on one side of the equation.

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And if that doesn't make a lot of sense, let's subtract this

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c from both sides, and we get y minus-- let me scroll down a

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little bit, my y looks like a g.

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y minus y to the third over 3 is equal to x to the third

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over 3 plus the constant when we took the antiderivative of

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the x, minus the constant of the antiderivative when we

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took the y.

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But these two constants, they're just constants.

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I mean, we don't know what they are.

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There are arbitrary constants.

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So we could just write a general c here.

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So you could have just-- you have to have a constant, but

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it doesn't have to be on both sides of the equation, because

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they're arbitrary.

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cx minus cy, well, that's still just another constant.

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And then if we want to simplify this equation more,

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we can multiply both sides of this by 3, just

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make it look nicer.

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And you get 3y minus y to the third is equal to x to the

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third plus-- well, I could write 3c here.

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But once again, c is an arbitrary constant.

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So 3 times an arbitrary constant, that's just another

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arbitrary constant.

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So I'll write the c there.

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And there you have it.

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We have solved this differential equation.

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Although it is in implicit form right now, and it's

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fairly hard to get it out of implicit form.

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We could put the c on one side, so the solution could be

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3y minus y to the third minus x to the third is equal to c.

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Some people might like that little bit better.

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But that's the solution.

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And notice, the solution, just like when you take an

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antiderivative, the solution is a class of implicit

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functions, in this case.

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And why is it a class?

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Because we have that constant there.

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Depending on what number you pick there, it

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will be another solution.

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But any constant there will satisfy the original

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differential equation, which was up here.

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This was the original differential equation.

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And if you want to solve for that constant, someone has to

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give you an initial condition.

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Someone has to say, well, when x is 2, y is 3.

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And then you could solve for c.

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Anyway, let's do another one that gives

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us an initial condition.

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So this one's a little bit-- I'll start over.

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Clear image, different colors, so I have optimal space.

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So this one is the first derivative of y with respect

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to x is equal to 3x squared plus 4x plus 2 over 2

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times y minus 1.

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This is a parentheses, not an absolute value.

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And they give us initial conditions.

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They say that y of 0 is equal to negative 1.

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So once we solve this differential equation, and

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this is a separable differential equation, then we

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can use this initial condition, when x is 0, y is

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1, to figure out the constant.

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So let's first separate this equation.

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So let's multiply both sides by 2 times y minus 1.

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And you get 2 times y minus 1 times dy dx is equal to 3x

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squared plus 4x plus 2.

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Multiply both sides times dx.

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This is really just an exercise in algebra.

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And I can multiply this one out, too, you get 2y minus 2,

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that's just this, dy.

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I multiplied both sides times dx, so that equals 3x squared

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plus 4x plus 2 dx.

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I have separated the equations.

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I've separated the independent from the dependent variable,

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and their relative differentials, and so now I

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can integrate.

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And I can integrate in magenta.

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What's the antiderivative of this expression

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with respect to y?

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Well, let's just see.

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It's y squared minus 2y.

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I won't write the plus c, I'll just do it on

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the right hand side.

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That is equal to 3x squared.

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Well, the antiderivative is x to the third, plus 2x squared,

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plus 2x plus c.

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And that c kind of takes care of the constant for both sides

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of the equation, and hopefully you understand why from the

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last example.

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But we can solve for c using the initial condition y of 0

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is equal to negative 1.

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So let's see.

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When x is 0, y is negative 1.

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So let's put y as negative 1, so we get negative 1 squared

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minus 2 times negative 1, that's the value of y, is

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equal to when x is equal to 0.

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So when x is equal to 0, that's 0 to the third plus 2

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times 0 squared plus 2 times 0 plus c.

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So this is fairly straightforward.

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All of these, this is all 0.

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This is, let's see, negative 1 squared, that's 1.

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Minus 2 times minus 1, that's plus 2, is equal to c.

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And we get c is equal to 3.

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So, the implicit exact solution, the solution of our

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differential equation-- remember now, it's not a

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class, because they gave us an initial condition-- is y

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squared minus 2y is equal to x to the third plus 2x squared

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plus 2x plus 3.

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We figured out that's what c was.

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And actually, if you want, you could write this in an

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explicit form by completing the square.

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This is just algebra this point.

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You're done.

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This is an implicit form.

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If you wanted to make it explicit, you could add 1 to

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both sides.

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I'm just completing the square here.

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So y squared minus 2y plus 1.

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If I add 1 to that side, I have to add 1 to this side, so

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it becomes x to the third plus 2x squared plus 2x plus 4.

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I just added 1 to both sides of the equation.

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Why did I do that?

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Because I wanted this side to be a perfect

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square in terms of y.

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Then I can rewrite this side as y minus 1 squared is equal

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to x to the third plus 2x squared plus 2x plus 4.

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Then I could say y minus 1 is equal to the plus or minus

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square root of x to the third plus 2x squared

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plus 2x plus 4.

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I can add 1 to both sides, and then I can get y is equal to 1

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plus or minus the square root of x to the third plus 2x

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squared plus 2x plus 4.

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And it has plus or minus here, and if we have to pick one of

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the two, we'd go back to the initial condition.

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Well, our initial condition told us that y of 0 is equal

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to negative 1.

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So if we put 0 here for x, we get y is equal to 1 plus or

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minus 0 plus 4.

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So 1 plus or minus 4.

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So if y is going to be equal to negative 1, so we get y is

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equal to 1 plus or minus-- sorry, 2.

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If this is going to be equal to negative 1, then this has

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to be 1 minus 2.

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So the explicit form that satisfies our initial

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condition, and we're getting a little geeky here, you could

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get rid of the plus, it's 1 minus this whole thing.

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That's what satisfies our initial condition.

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And you could figure out where it's satisfied, over what

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domain is it satisfied.

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Well, that's satisfied when this term is positive, this

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becomes negative, and you get it's undefined in reals, and

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all of that.

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But anyway, I've run out of time.

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See you in the next video.