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Main content
Current time:0:00Total duration:7:50
AP.CALC:
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FUN‑7.A.1 (EK)

Video transcript

let's now introduce ourselves to the idea of a differential equation and as we'll see differential equations are super useful for modeling and simulating phenomena and understanding how they operate but we'll get into that later for now let's just think about or at least look at what a differential equation actually is so if I were to write so let's here's an example of a differential equation if I were to write that the second derivative of y plus 2 times the first derivative of Y is equal to 3 times y this right over here is a differential equation another way we could write it if we said that Y is a function of X we could write this in function notation we could write the second derivative of our function with respect to X plus 2 times the first derivative of our function is equal to 3 times our function or if we wanted to use the the live knits notation we could also write we could also write the second derivative of Y with respect to X plus 2 times the first derivative of Y with respect to X is equal to 3 times y all three of these I guess equations are really representing the same thing they're saying ok can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself so just to be clear these are all essentially saying the same thing and you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions it's not just a value or a set of values so the solution here so the solution to a differential equation is a is a function is a function or a set of functions or a class of functions it's important to contrast this relative to a traditional equation so let me write that down so traditional equation I guess I could say maybe I shouldn't say traditional equation differential equations have been around for a while so let me write this as a may be an algebraic equation that you're familiar with algebraic an algebraic equation might look something like and I'll just write up a simple quadratic say x squared x squared plus 3x plus 2 is equal to 0 the solutions to this algebraic equation are going to be numbers or a set of numbers we can solve this is going to be X plus 2 times X plus 1 is equal to 0 so X could be equal to negative 2 or X could be equal to negative 1 the solutions here are numbers or a set of our set of values or set of values that satisfy that satisfy the equation here it's a relationship between a function and stir it is and so the solutions or the solution is going to be a function or a set of functions now how let's make that a little bit more tangible what would a solution to something like any of these 3 which really represent the same thing what would a solution actually look like and actually let me move over to the let me move this over a little bit move this a little bit so we can take a look at what some of these solutions could look like and we erase this little so stuff that I have right over here so and I'm going to kind of give you examples of solutions here we'll verify that these indeed are solutions for I guess this is really just one differential equation represented in different ways but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution so one solution to this differential equation and I'll just write it as our first one so one solution I'll call it y1 and I could even write it as y1 of X to make it explicit but it is a function of X one solution is y1 of X is equal to e to the negative 3x and I encourage you to pause this video right now and find the first derivative of y1 and the second derivative of y1 and verify that it does indeed satisfy this differential equation so I'm assuming you've had a go at it so let's work through this together so that's why one so the first derivative of y1 well this is going to be you'll see we just have to do the chain rule here derivative of negative 3x with respect to X is just negative 3 and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x and if we take the second derivative Y of y1 this is equal to the same exact idea derivative of this is negative 3 times negative 3 is going to be 9 e to the negative 3x and now we could just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function so let's verify that so let me so we first have what the second derivative of Y so that's that term right over there so we have 9 e to the negative 3x plus plus 2 times the first derivative so that's going to be 2 times this right over here so it's going to be net -6 I'll just write plus negative 6 e to the negative 3 X notice I just took this 2 times the first derivative 2 times the first derivative is going to be equal to or needs to be equal to if this indeed does satisfy if y1 does it does indeed satisfy the different the differential equation this needs to be equal to 3 times y well 3 times y is 3 times e to the negative 3x 3 e to the negative 3x and let's see if that indeed is true so these two terms right over here 9 e to the negative 3x essentially minus 6 e to the negative 3x that's going to be 3 e to the negative 3x which is indeed which is indeed equal to 3 e to the negative 3 X so y1 is indeed a solution to this differential equation but as we'll see it is not the only solution to this differential equation for example for example let's say let's say a y2 is equal to e e to the X is also a solution to this differential equation and I encourage you to pause the video again and verify that it's a solution so assuming you've had a go at it so the first derivative of this this is pretty straightforward is e to the X second derivative is one of the profound things of the exponential function the second derivative here is also e to the X and so if I have so the second derivative let me just do in those same colors so the second derivative it's going to be e to the X e to the X plus two times e to the X plus two times e to the X is indeed going to be equal to three times e to the x three times e to the X this is absolutely going to be true e to the X plus 2 DX is three e to the X so y 2 is also a solution to this differential equation so that's a start in the next few videos we'll explore this more we'll start to see what the solutions look like what classes of solutions are techniques for solving them visualizing solutions to differential equations and a whole toolkit for for kind of digging in and deeper