Linear Differential Equations

Hey guys, this is Matt Kids 11, and today I'm going to be showing you the second math video on this channel.

Now, in the last math video I solved, I proved that the oil is identity by solving a differential equation. In that differential equation, like I just solved it by using a method called separation of variables. The differential equation written above cannot be ready to solve that way. That's why I wrote it. It is y prime instead of dy DX because that's not going to help us.

So, the first thing we're going to do is put this in standard form. Well, actually, the first thing you do is divide out the way, there's no coefficient to this term. But there already is no coefficient, so now we can subtract from both sides. We put it in standard form, so we have um, we have y prime minus 2xy is equal to x.

Now, you want to find something called the integration factor. So, what you do is you integrate the coefficient in front of the y term, and you want everything to be linear. You don't want any y squared or anything, ‘cause it's not going to work. And if you do have y squared, I'm going to show you how to sometimes make it linear.

So, we're going to take the integral of negative 2x DX, and now we get negative x squared. Now, it's negative x squared plus a constant, but we're not solving this equation. We're not integrative solving this integral. Basically, we're just finding something we can multiply both sides to, and that something is going to be e to the key to the integral of the coefficient of that term.

Okay, so now we have y prime e to the negative x squared minus 2xy e to the negative x squared is equal to x times e to the negative x squared. Now you might be looking at this and saying, “John, you do not make this any simpler." Like we went from there to there, but I'm not, trust me, it's simpler, and I'm going to show you why.

Now, if you have the derivative of a product of functions, that's going to be a prime B plus B prime A. So here we have the derivative of Y times this plus the derivative of that x y. You can write that as the derivative of something. So, that is almost specifically the derivative of Y times e to the negative x squared. The derivative of that is equal to x e to the negative x squared.

And here I mean prime, I don't need to be the one. Alright, now we're going to integrate both sides. So, the integral of the derivative is just the thing, so we can just cancel that out. We take it, y e to the negative x squared is equal to, and we will get a constant, but we're going to worry about the constant on that size; we don't worry about it on this side, worry.

And so it's going to be equal to the integral of x e to the negative x squared DX. So, now, all we need to do is take this integral and then we're basically done. So, to take this integral, what do I do? I'm going to do that in another color over here. Alright, we're going to let u equal negative x squared.

Then du is equal to negative 2x DX, so then this is equal to the integral of e to the U, and then we want instead of DX, it's going to be du over negative 2. Okay, and that is equal to negative 2 e to the U plus an arbitrary constant. Alright, but we want it in terms of X, not in terms of U.

So, it's equal to, this should be 1 over negative 2. Okay, so 1 over negative 2 e to the negative x squared plus a constant, right? 'Cause u was equal to negative negative x squared, and I plug that back in. So, I can erase all this unless you make that there. It's equal to 1 over negative 2 e to the negative x squared plus a constant.

Now, what you want to do is divide both sides by e to the negative x squared to isolate y and solve. So, you have Y is equal to, if you divide that by that, you just get 1 over negative 2 plus some constant times e to the negative x squared. And we're done, Arnaud!

Alright, so it's going to be some constant divided by e to the negative x squared C. I over both of those, but since it's negative, you could multiply it. So that's what I did, but I'm going to, I just put it here so it's clear. So, that would solve. You couldn't, if I had an initial condition, solve for the constant, but in this case, we don't.

Alright, the second thing I'm going to show you how to do is make a point linear when it's not. So, here it's linear because all of the powers of Y are one. So, let's say I have y squared. How would you solve that? Well, the first thing you want to do is make sure that if your power is higher than one, you can only have one of them. You also have a Y there that's going to help you cancel things out.

So, you're going to divide out by the highest power. So, you have dy prime over Y squared is equal to X plus 2XY divided by Y squared, which makes just over one. Okay, now we're going to let u be equal to, and by the way, this is known as a Bernoulli differential equation, so credits go to him. Ah, u equal one over Y because we have here, now u prime is equal to negative 1 over Y squared times y prime.

And then we see that we have that there. So, what within month negatives we have negative u. I'm going to bring that on to the other side. So, minus 2xy minus 2xu is equal to X. So, this is basically, it looks a lot like what we had last time.

Then you can solve it in terms of u, and then you can plug that again. So, I'm going to write here - you need to remember line. So, I'm just going to solve that quickly so I can review what I just taught you. Okay, so, and this should be a u prime. So, I'm going to divide both sides by negative 1 because we want this coefficient of e to be 1.

So, we have u prime plus 2xu is equal to negative x. Take the integral of this. So, the integral of 2x DX is equal to x squared, so multiply both sides by e to the x squared. So, we have e to the x squared u prime plus 2xu e to the x squared equals negative x e to the x squared.

Then we make that, we said that that's the derivative of e to the x squared times u prime is equal to negative x e to the x squared. Alright, we're going to take the integral of both sides. We have e to the x squared u is equal to the integral of negative x e to the x squared DX.

Which is equal to e to the x squared, and then ah, so we said u is equal to x squared. Do you or not? U, we're going to set on V equal to x squared. dv is equal to 2x DX. So since we have a 2x DX, it's going to be over 2 plus an arbitrary constant.

So now we have u is equal to, we're going to divide both sides by that. So we have 1 over 2 plus some constant divided by e to the x squared. Okay, now the equation was in terms of Y, not in terms of u. So I can replace u with 1 over Y, and then if I wanted to, I could isolate, you could take the reciprocal of both sides, use partial fractions, but I think you know how to do that.

So I'm going to end the video here. If you like the math series or you want to know how to solve more differential equations like homogeneous differential equations, systems of differential equations, just leave a comment below, or if you have any questions. Thanks for watching. Goodbye!