WEBVTT 00:00:00.085 --> 00:00:08.945 Find the equation of the curve which passes through the point negative eight, one given that the gradient of the tangent at any point is equal to two times the square of the 𝑦-coordinate. 00:00:09.245 --> 00:00:12.015 Let’s first interpret the information given by this question. 00:00:12.265 --> 00:00:15.535 We’re told about the gradient to the tangent of a curve at any point. 00:00:15.585 --> 00:00:18.365 This gradient can be represented by d𝑦 by dπ‘₯. 00:00:18.555 --> 00:00:21.775 We’re told that this is equal to two times the square of the 𝑦-coordinate. 00:00:21.925 --> 00:00:24.295 So in other words, is equal to two 𝑦 squared. 00:00:24.375 --> 00:00:27.205 Here, we recognize that we’ve been given a differential equation. 00:00:27.425 --> 00:00:30.315 Not only that, we’ve been given a separable differential equation. 00:00:30.535 --> 00:00:35.945 This is one that can be expressed in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. 00:00:36.215 --> 00:00:41.755 In our case, we’ll be taking 𝑔 of π‘₯ to be equal to the constant two and 𝑓 of 𝑦 to be equal to 𝑦 squared. 00:00:41.975 --> 00:00:45.665 Given an equation of this form, we can find the general solution in the following way. 00:00:45.895 --> 00:00:48.715 We first divide both sides of our equation by 𝑓 of 𝑦. 00:00:48.745 --> 00:00:50.165 In our case, this is 𝑦 squared. 00:00:50.425 --> 00:00:54.275 This gives us that one over 𝑦 squared d𝑦 by dπ‘₯ is equal to two. 00:00:54.605 --> 00:00:58.115 Next, we can integrate both sides of our equation with respect to π‘₯. 00:00:58.325 --> 00:01:02.455 Now, the chain rule tells us that d𝑦 by dπ‘₯ dπ‘₯ is equal to d𝑦. 00:01:02.565 --> 00:01:08.855 This gives us that the integral of one over 𝑦 squared with respect to 𝑦 is equal to the integral of two with respect to π‘₯. 00:01:08.985 --> 00:01:14.685 As a side note, you often see methods where before the integration, d𝑦 by dπ‘₯ is treated somewhat like a fraction. 00:01:14.845 --> 00:01:16.495 Both sides are multiplied by dπ‘₯. 00:01:16.495 --> 00:01:21.275 And the equivalent statement is obtained: one over 𝑦 squared d𝑦 is equal to two dπ‘₯. 00:01:21.515 --> 00:01:23.565 Both sides of the equation are then integrated. 00:01:23.595 --> 00:01:26.575 And we reach the same state as we did before, but with less working. 00:01:26.875 --> 00:01:30.735 Always remember that this is just a shorthand for the more rigorous steps that we’ve shown here. 00:01:30.965 --> 00:01:35.655 Returning to our method, one over 𝑦 squared can be expressed as 𝑦 to the power of negative two. 00:01:35.715 --> 00:01:45.565 You might be more comfortable performing the integration in this form to get negative one over 𝑦 plus the constant of integration 𝑐 one is equal to two π‘₯ plus some other constant of integration, 𝑐 two. 00:01:45.845 --> 00:01:53.895 Usually, we don’t worry about putting a constant on both sides of this equation because we just define a new constant, which in this case would be 𝑐 two minus 𝑐 one. 00:01:53.925 --> 00:01:55.845 And this could just go on one side of the equation. 00:01:56.225 --> 00:01:59.855 Now, at this stage, we have found the general solution to our differential equation. 00:02:00.005 --> 00:02:05.075 And it’s in implicit form, which means it’s not in the form where 𝑦 is equal to some function of π‘₯. 00:02:05.385 --> 00:02:12.355 We found the general solution for all lines where the gradient of the tangent at any point is equal to two times the square of the 𝑦-coordinate. 00:02:12.595 --> 00:02:17.145 Now, our next thought might be to convert our implicit solution into an explicit solution. 00:02:17.425 --> 00:02:19.305 But as it turns out, this isn’t necessary. 00:02:19.725 --> 00:02:22.775 This is because the general solution is not our final destination. 00:02:22.955 --> 00:02:25.195 Rather, we are seeking a particular solution. 00:02:25.475 --> 00:02:28.685 We want the line which passes through the point negative eight, one. 00:02:28.905 --> 00:02:31.225 This information is an initial value. 00:02:31.395 --> 00:02:34.485 And we’ll be finding the particular solution which matches in the following way. 00:02:34.765 --> 00:02:40.245 We know that if negative eight, one lies on the line, when π‘₯ is equal to negative eight, 𝑦 must be equal to one. 00:02:40.485 --> 00:02:44.565 We can substitute these known values into the equation for our general solution. 00:02:44.705 --> 00:02:46.445 We then perform some simplification. 00:02:46.585 --> 00:02:49.425 And we find that our constant 𝑐 three is equal to 15. 00:02:49.585 --> 00:02:54.335 We can now substitute this value into our general solution which we’ll turn it into a particular solution. 00:02:54.525 --> 00:02:57.655 Negative one over 𝑦 is equal to two π‘₯ plus 15. 00:02:57.835 --> 00:03:05.285 Now, if we wanted, we could convert this to an explicit solution first by multiplying by negative one and then raising both sides to the power of negative one. 00:03:05.675 --> 00:03:09.905 To recap, we used the information given in the question to form a separable differential equation. 00:03:10.145 --> 00:03:17.505 The particular solution to this which passes through the point negative eight, one is 𝑦 equals negative one over two π‘₯ plus 15. 00:03:17.675 --> 00:03:19.435 And this is the answer to our question.