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# Worked example: finding a specific solution to a separable equation

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.E (LO)
,
FUN‑7.E.1 (EK)
,
FUN‑7.E.2 (EK)
,
FUN‑7.E.3 (EK)
Solving a separable differential equation given initial conditions. In this video, the equation is dy/dx=2y² with y(1)=1.

## Video transcript

- Let's now get some practice with separable differential equations, so let's say I have the differential equation, the derivative of Y with respect to X is equal to two Y-squared, and let's say that the graph of a particular solution to this, the graph of a particular solution, passes through the point one comma negative one, so my question to you is, what is Y, what is Y when X is equal to three for this particular solution, so the particular solution to the differential equation that passes through the point one comma negative one, what is Y when X is equal to three, and I encourage you to pause the video, and try to work through it on your own. So I'm assuming you had a go at it, and the key with a separable differential equation, and that's a big clue that I'm even calling it a separable differential equation, is that you separate the Xs from the Ys. Or all the Xs and the DXs from the Ys and DYs. So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be DY DX is equal to two Y-squared, is equal to two Y, equal to two Y-squared. So let's see, we can multiply both sides by DX, and let's see, so then we're gonna have, that cancels with that if we treat it as just a value, or as a variable. We're gonna have DY is equal to two Y squared DX. Well, we're not quite done yet. We gotta get this two Y squared on the left hand side. So we can divide both sides by two Y-squared. So if we divide both sides by two Y-squared, two Y-squared, the left hand side, we could rewrite this as 1/2 Y to the negative two power, is going to be equal to DY, DY is equal to DX, and now, we can integrate both sides. So we can integrate both sides. Let me give myself a little bit more space. And so, what is, what is this left hand side going to be? Well, we increment the exponent, and then divide by that value, so Y to the negative two, if your increment is Y to the negative one, and then divide by negative one, so this is going to be -1/2 Y to the negative one power, and we could do a plus C like we did in the previous video, but we're gonna have a plus C on both sides, and you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other, so I'm just gonna write the constant only on one side. So you have that is equal to, well if I integrate just DX, that's just going to give me X, that's just gonna give me X. So this right over here is X, and of course I can have a plus C over there, and If I want I can, I can solve for Y if I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left hand side you're just gonna have Y to the negative one, or 1/Y is equal to, if I multiply the right hand side times negative two, I'm gonna have negative two times X plus, well it's some arbitrary constant, it's still going to, it's gonna be negative two times this arbitrary constant but I could still just call it some arbitrary constant, and then if we want we can take the reciprocal of both sides, and so we will get Y is equal to, is equal to 1/-2X+C. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for C. So, when X is negative one, so when X is negative one. Oh sorry, when X is one, when X is one, Y is negative one, so we get negative one is equal to 1/-2+C, or we could say C minus two, we could multiply both sides times C minus two, if then we will get, actually let me just scroll down a little bit, so if you multiply both sides times C minus two, negative one times C minus two is going to be negative C plus two or two minus C is equal to one. All I did is I multiplied C minus two times both sides, and then, let's see, I can subtract two from both sides, so negative C is equal to negative one, and then if I multiply both sides by negative one, we get C is equal to one. So our particular solution is Y is equal to 1/-2X+1. And we are almost done, they didn't just ask for, we didn't just ask for the particular solution, we asked, what is Y when X is equal to three. So Y is going to be equal to one over, three times negative two is negative six plus one, which is equal to negative, is going to be equal to 1/-5, or -1/5. And we are done.