4.5.1

Instructor: Example 1, put each of the following first order linear differential equations into standard form. Identify p of x and q of x for each equation. But in order to do that, we first need to read over our definition for what a first order linear differential equation is, and also what makes that standard form. Because that's what we're being asked to do. 

The standard form is going to be in this form. Y prime equals b of x over a of xy equals c of x over a of x. Or change of errors, which is what we're going to be using. Y prime plus p of x times y equals q of x. 

So notice that each of these functions are in terms of x. But we are going to have our dy dx, our slope term, something in terms of x times y, and then some other function in terms of x. So let's see if we can begin to get this first one in the right form. 

The first thing it looks like we need to do is get our y prime and our y term on the left side. So this would necessitate that this is y prime plus 4y equals 3x. Now, that's it actually for this one because we have a function in terms of x, our p of x is 4, which, I mean, I know that's a constant function. But we could say that's in terms of x. So that would be our p. 

And then our q of x is 3x. Again, that is going to come from this right side of the equation. That's all we need. All right, let's look at part B. 3xy prime over 4y minus 3 equals 2. And we're assuming that x is greater than 0. 

All right, well, we need to isolate y prime. That is a given. So let's multiply both sides by 4y minus 3. To answer that, this would be 3xy prime equals 8y minus 6. 

And then we'll need to divide by 3x to get it in standard form so that we have y prime equals 8y over 3x minus 6 over 3x. Of course, we'll reduce just a bit. And we want to write this where we have functions in terms of x. 

So this second term will be minus 2 over x. And this one right here will be 8 over 3x times y, y prime. But this is not in standard form because I need this term on the left-hand side of the equation. So let's go and write this as y prime minus 8 over 3x times y equals negative 2 over x. That makes our p of x equal to 8 over 3x, negative 8 over 3x and q of x equal to negative 2 over x. 

Part C, y equals 3y prime minus 4x squared plus 5. Again, we want our y terms together. So let's move the-- actually, you know what? Let's do this. Let's add 4x squared and subtract 5. 

And I also want you in the same breath to subtract y from both sides. So that would leave me with 4x squared. That's 5 on the left side and 3y prime minus y on the right side, which for ease of working, let's go ahead and rearrange that. 

And we need y prime isolated. It needs to have a coefficient of 1. So we will divide by 3 everywhere, which would be y prime minus 1/3y equals 4/3x squared minus 5/3. And it looks like we now have a p of x of minus 1/3 in terms of x again. And q of x will be 4/3x squared minus 5/3.