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Current time:0:00Total duration:4:57

AP.CALC:

FUN‑7 (EU)

, FUN‑7.D (LO)

, FUN‑7.D.1 (EK)

, FUN‑7.D.2 (EK)

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 I get dy/dx is equal to some function of Y times some other function of X then I say ok this is separable because I could rewrite this as I could divide both sides by G of Y and I get 1 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 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 the 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 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 now let's see if I divide both sides by X I'm going to get the derivative of Y with respect to X is equal to actually I'm gonna write it this way I'm going to write it 3 minus y times 1 over X and 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 and I could show you I can multiply both sides by DX and I can 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 going to just do the same technique I'll do it in a slightly different color so we don't get 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 let me just do whoops 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 1 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 0 this is going to be 0 this is going to be 0 I'm going to have 2 times the derivative of Y with respect to X is equal to negative 2x minus 2y plus 1 and now let's see I can divide everything by 2 I would get the derivative of Y with respect to X is equal to and actually yeah I would get I'm just going to drive divided by 2 so I'm going to 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 is this one right over here is not separable I don't know how to how to write this as a function of x times a function of Y so this one I'm not going to I'm going to say is not separable now this one they've already written it as 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 1 over y squared plus y dy is equal to x squared plus X DX so clearly separable all right now this last choice this is interesting they've essentially distributed the derivative right over here so let's see if we were to unfactored the derivative I'm just going to solve for dy DX so I'm going to factor it out I'm going to get dy x 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 going to get dy/dx is equal to x over X plus y and here I my algebraic toolkit how do I separate X&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

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