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## Differential equations

### Unit 1: Lesson 4

Separable equations- Separable equations introduction
- Addressing treating differentials algebraically
- Worked example: identifying separable equations
- Worked example: finding a specific solution to a separable equation
- Worked example: separable equation with an implicit solution
- Particular solutions to separable differential equations
- Separable equations (old)
- Separable equations example (old)

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# Worked example: identifying separable equations

AP.CALC:

FUN‑7 (EU)

, FUN‑7.D (LO)

, FUN‑7.D.1 (EK)

, FUN‑7.D.2 (EK)

Separable equations can be written in the form dy/dx=f(x)g(y). See how we analyze various differential equations to see if they are separable.

## Video transcript

- [Sal] Which of the differential
equations are separable? And 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, and if when I solve for the derivative, if I get dy, dx is equal to some function of y times some other function of x then I say okay, this is separable. 'Cause I could rewrite this as, I could divide both sides by g of y, and I get one over g of y, which is itself a function of y, times dy is equal to h of x dx. You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx. And 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 gonna get x times, I'll write y prime is the derivative of y with respect to x, is equal to 3 minus y. So I subtracted y from both sides. Let's see if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually I'm gonna write it this way. I'm gonna write it 3
minus y times 1 over x. And so it's clear I'm able to write the derivative as a product of a function of y, and a function of x. So this indeed is separable. And I could show you, I could multiply both sides by dx and I could divide both
sides by 3 minus y now, and I would get 1 over 3 minus y dy is equal to 1 over x dx. So clearly, this one right
over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a slightly different colour so we don't get all of our
math all jumbled together. So in this second one, let's see, if I subtract the 2x, the 2y from both sides, so actually, lemme just do, whoops, lemme do a couple things at once. I'm gonna subtract 2x from both sides. I am going to subtract 2y from both sides. So I'm gonna subtract 2y from both sides. I'm gonna add one to both sides, so I'm gonna add one to both sides. And then what am I going
to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have 2 times the derivative of y with respect to x is equal to negative 2x minus 2y plus 1. And now see I can divide
everything by two; I would get the derivative of y with respect to x is equal to, and actually, yeah I would get, I'm just gonna divide by two, so I'm gonna get negative x minus y and then I'm going to get plus 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 gonna, I'm gonna say is not 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 dy dx. If I multiply both sides by dx and divide both sides
by this right over here, I would get one over y squared plus y dy is equal to x squared plus 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
gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y, is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y. And here my algebraic tool kit of how do I separate x and y so 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.