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# Worked example: equation from slope field

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.C (LO)
,
FUN‑7.C.1 (EK)

## Video transcript

we've already seen scenarios where we start with a differential equation and then we generate a slope field that describes the solutions to the differential equation that we use that to visualize those solutions what I want to do in this video is do an exercise that takes us the other way start with a slope field and figure out which differential equation is the slope field describing the solutions for and so I encourage you to look at each of these options and think about which of these the differential equations is being described by this slope field I encourage you to pause the video right now and try it on your own so I'm assuming you have had a go at it so let's work through each of them and the way I'm going to do it is I'm just going to find some points that seem to be easy to do arithmetic with and well we'll see if the if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field and I know just for simplicity maybe I'll do x equals 1 y equals 1 for all of these so I'm going to see what so when x equals 1 and Y is equal to 1 so this first differential equation right over here if X is 1 and Y is 1 then dy/dx would be negative 1 over 1 or negative 1 DUI DX would be negative 1 now is that depicted here when X is equal to 1 and y is equal to 1 our slope is a negative 1 our slope here looks positive so we can rule this one out now let's try the next one so if X is equal to 1 and Y is equal to 1 well then dy/dx would be equal to 1 minus 1 or 0 and once again I'm just I just picked 1 x equals 1 and y equals 1 for convenience I could have picked any other I could pick negative 5 and negative 7 this just makes the arithmetic a little easier but once again when you look at that point that we've already looked at our slope is clearly not 0 we have a positive slope here so we could rule that out once again for this magenta differential equation if x and y are both equal to 1 then 1 minus 1 is once again going to be equal to 0 and we've already seen the slope is not zero here so rule that one out and now here we have X plus y so when X is 1 Y is 1 our the derivative of Y with respect to X is going to be 1 plus 1 which is equal to 2 now this looks interesting it looks like this the slope right over here could be 2 this looks like 1 this looks like 2 I'd want to validate some other points but this looks like a really really good candidate and you could also see what's happening here at when dy/dx is equal to X plus y you would expect that as x increases for a given Y your slope would increase and as Y increases for a given X your slope increases and we see that if we were to just follow if we were to hold Y constant 1 but increase X along this line we see that the slope is increasing it's getting steeper and if we were to keep X constant and increase Y across this line we see that the slope increases and in general we see that the slope increases as we go to the top right and we see that it decreases as we go to the bottom left in both x and y become much much more negative so I'm feeling pretty good about this especially if we can knock this one out here if we can knock that one out so dy/dx is equal to x over Y well then when x equals 1 y equals 1 dy/dx would be equal to 1 and this slope looks larger than 1 it looks like 2 but since we're really just eyeballing it let's see if we can find something that more clearly where this more clearly falls apart so let's look at the situation when let's look at the situation when they both equal negative 1 so x equals negative 1 and Y is equal to negative 1 well in that case dy/dx should still be equal to 1 because you have negative 1 over 1 do we see that over here so when X is equal to negative 1 Y is equal to negative 1 our derivative here looks looks negative it looks like negative 2 which is consistent with this yellow differential equation the slope here is definitely not a positive 1 so we could rule this one out as well and so we should feel pretty confident that this is the differential equation being described and now that we've done it we can actually think about well okay what are the solutions for this differential equation going to look like well it depends where they start if you have or what points they contain if you have a solution that contains that point it looks like it might go looks like it might do something looks like it might do something like something like this if you had a solution that contained I don't know if you had a solution that contained at this point it might do something it might do something like that and of course it keeps going it looks like with asymptotes towards Y is equal to negative x this downward sloping this this this ain't this this essentially is the line y is equal to negative x actually no that's not the line equals y equals negative x this is the line y is equal to negative x minus 1 so that's this line right over here and it looks like if you had if you if you had you're in a solution if you had a if the solution contains say this point right over here you would actually that would actually be a solution to the differential equation y is equal to Y is equal to negative X whoops Y is equal to negative X minus 1 and you can verify that if Y is equal to negative X minus 1 then the X and negative x cancel out you're just left with dy/dx is equal to negative 1 which is exactly what's being described by this slope field anyway hopefully you found that interesting
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