Original subtitles

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Hey everyone.

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And this video we have a differential equation y prime plus two X Y squared equals zero.

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So this is a non linear differential equation because of the Y squared.

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And we're told that a one parameter

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family of solutions so family of solutions

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is given by y equals one over X squared plus C and the questions are to find

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the solutions or find the solution rather that passes through the given point.

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So through the given point and find the interval of definition the interval of definition

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part a let's go through this very carefully.

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This is a really key example.

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So maybe the solution passes through zero negative 1 So again the question is to find a solution that

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passes through this point and find the interval of definition.

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So solution so in this case x is 0 and Y is negative 1.

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So x is 0 and Y is negative 1.

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So all we have to do is we have to find c.

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So we're gonna put the negative one where the y is right here and that's equal to 1 over 0 squared plus

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C which is putting the zero here where the X is.

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So that means that we have negative 1 equals 1 oversee to solve this perceived just multiply by C so

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you get negative C equals 1.

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So c is equal to negative 1.

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We're almost there.

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All we have to do is plug the C back into our 1 parameter family so we get Y equals 1 over x squared

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minus 1.

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So that is the solution to the D E that passes through this point.

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You can think of this as an initial condition.

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This is why of zero equals negative one.

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OK so now we just have to find the interval of definition.

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So to do that what you can do is you can graph this craft this solution wrap this function rather.

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So how do you graft this function.

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Well skill.

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So this function has two vertical asymptote.

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I'll be brief one in negative one so there's 1 and then here's negative 1.

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This is one and this is negative one and it has a horizontal asymptote at zero.

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That's because the degree in the denominator is bigger than the degree and the numerator.

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So it has a horizontal asymptote but zero.

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And if you plug in zero you get negative one.

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In other words it passes through zero come a negative 1.

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So it looks something like this and it looks something like this.

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You can use your calculator to it's pretty easy to plug in numbers and figure out like you plug in 5.

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If you plug in 5 you'll notice it's up here.

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So it has to look like that.

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If you plug a negative 10 you'll notice the y values up here so you know it has to look like that.

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OK so what is the interval of definition.

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Well we talked about this before in a previous video.

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The domain here would be negative infinity to negative 1 negative 1 to 1 1 to infinity with union symbols

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between them.

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We would exclude negative 1 to 1 but an interval can't have that right.

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Interval cannot have gaps.

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So the answer is going to be in this case negative 1 to 1 right here.

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And the reason that that's the answer is because that is where the point is right.

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The graph passes through zero negative 1.

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So we want to pick this.

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So this here is blue graph that I'm drawing.

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That's the graph of the solution.

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Right.

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And so where is that defined.

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On negative 1 to 1.

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Let's do another example.

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So if it had been like over here or something if the point had been here then we would have picked one

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to infinity.

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Check this one out.

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Let's say it passes through to one third so a solution.

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The same thing your your X is 2 and your Y is one third.

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So you get one third equals one over two squared plus C plugging it into our R function and so we get

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one third equals one over four plus C then we can cross multiply so we get.

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So this goes here and this goes here.

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So we get 4 plus C equals 3.

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Subtract 4.

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So we get C equals negative 1 Hurrah.

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So we get exactly the same thing.

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It's totally rigged.

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So we already have the graph.

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And in this case I'll read I'll redraw that well I'll just use this graph.

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Here's 2 1.

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To one third right.

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There's two one third so this time we use this piece here and the purple piece.

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So now the interval of definition would be 1 to infinity.

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So you have to pick the interval from where the point is in the first example it was here.

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So we pick negative 1 to 1 and the second example it was here.

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So we picked one to infinity.

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If it had been over here on the left we would pick negative infinity to negative 1.

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Let's do one more one more really really tricky one.

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What if we're told it passes through five comma zero then again that's just like before we do the same

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thing.

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This is our X. This is our y.

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So we get 0 equals 1 over 5 squared plus C..

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So zero is equal to one over twenty five plus C then multiply by twenty five policy.

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So you get right.

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You would you would put that here like this and do that here like this.

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And then what happens is these cancel so you get zero equals 1.

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So there is no solution.

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All right there is no c.

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So does that mean there is no answer.

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No.

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No it does not.

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The question is to find the solution that passes through the point someone to rewrite the D here.

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The D was my prime plus 2 x y squared.

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That was our original question and our one parameter family was won over X squared plus C we tried to

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find the solution that passed this point and we got something ridiculous.

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So what is the solution.

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Well now we have to think Is there another solution to this differential equation.

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Right there is y equals zero because if you let Y equal to zero then y prime is also zero.

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And you can check that this indeed does satisfy the differential equation plugging it in.

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You get 0 plus 2 x 0 squared so zero zero zero.

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So it checks.

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So y equals zero is a solution.

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So the answer is y equals zero.

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That is a solution.

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And it does indeed pass through five comma zero Y.

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Well if you graph it y equals zero as a horizontal line and there is five comma zero so it certainly

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passes through five comma zero.

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So y equals zero is the solution to this differential equation.

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And it does pass through five comma zero.

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So as it is the answer to this question what about the interval of definition.

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Well this thing is defined everywhere so it would just be negative infinity to infinity.

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Hope that made sense.