Worked example: identifying separable equations | AP Calculus AB | Khan Academy

Which of the differential equations are separable? I encourage you to pause this video and see which of these are actually separable.

Now, the way that I approach this is I try to solve for the derivative. If when I solve for the derivative, I get ( \frac{dy}{dx} = g(y) \cdot h(x) ), then I say, "Okay, this is separable," because I could rewrite this as I could divide both sides by ( g(y) ) and I get ( \frac{1}{g(y)} , dy = h(x) , dx ). You would go from this first equation to the second equation just by dividing both sides by ( g(y) ) and multiplying both sides by ( dx ). Then it's clear you have a separable equation; you can integrate both sides.

But the key is, let's solve for the derivative and see if we can put this in a form where we have the product of a function of ( y ) times a function of ( x ). So, let's do it with this first one here.

So, let's see if I subtract ( y ) from both sides. I'm just trying to solve for the derivative of ( y ) with respect to ( x ). I'm going to get ( x \cdot y' ) (where ( y' ) is the derivative of ( y ) with respect to ( x )) is equal to ( 3 - y ). So, I subtracted ( y ) from both sides.

Let's see; if I divide both sides by ( x ), I'm going to get ( y' = \frac{3 - y}{x} ). So, it's clear I'm able to write the derivative as the product of a function of ( y ) and a function of ( x ). So, this indeed is separable. I can show you: I can multiply both sides by ( dx ) and divide both sides by ( 3 - y ) now, and I would get ( \frac{1}{3 - y} , dy = \frac{1}{x} , dx ). So, clearly, this one right over here is separable.

Now, let's do the second one, and I'm going to just do the same technique. I'll do it in a slightly different color so we don't get all of our math jumbled together.

So in this second one, let's see if I subtract ( 2x + 2y ) from both sides. Actually, let me just do—oops, let me do a couple things at once. I'm going to subtract ( 2x ) from both sides. I am going to subtract ( 2y ) from both sides. So I'm going to subtract ( 2y ) from both sides. I'm going to add one to both sides. So I'm going to add ( 1 ) to both sides.

And then what am I going to get? If I do that, this is going to be zero; this is going to be zero; this is going to be zero. I'm going to have ( 2 \cdot \frac{dy}{dx} = -2x - 2y + 1 ).

Now, let's see. I can divide everything by ( 2 ). I would get ( \frac{dy}{dx} = \frac{-2x - 2y + 1}{2} ). Actually, yeah, I would get—I'm just going to divide by ( 2 ), so I'm going to get ( \frac{dy}{dx} = -x - y + \frac{1}{2} ).

So, it's not obvious to me how I can write this as a product of a function of ( x ) and a function of ( y ). So, this one does not feel—this one right over here is not separable. I don't know how to write this as a function of ( x ) times a function of ( y ), so this one I’m not going to say is separable.

Now this one, they've already written it for us as a function of ( x ) times a function of ( y ), so this one is clearly separable right over here. And if you want me to do the separating, I can rewrite this as well. This is ( \frac{dy}{dx} ). If I multiply both sides by ( dx ) and divide both sides by this right over here, I would get ( \frac{1}{y^2 + y} , dy = (x^2 + x) , dx ). So, clearly separable.

Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So, let's see if we were to unfactor the derivative. I'm just going to solve for ( \frac{dy}{dx} ). So, I'm going to factor it out. I'm going to get ( \frac{dy}{dx} \cdot (x + y) = x ).

Now, if I were to divide both sides by ( x + y ), I'm going to get ( \frac{dy}{dx} = \frac{x}{x + y} ). Here, my algebraic toolkit of how do I separate ( x ) and ( y ); I can write this as a function of ( x ) times a function of ( y )—not obvious to me here, so this one is not separable.

So, only the first one and the third one are separable.