Video: Finding the Equation of the Curve That Passes through a Given Point given Its Slope as a Function of 𝑦 Using Differential Equations

Find the equation of the curve which passes through the point (βˆ’8, 1) given that the gradient of the tangent at any point is equal to 2 times the square of the 𝑦-coordinate.

03:19

Video Transcript

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.

Let’s first interpret the information given by this question. We’re told about the gradient to the tangent of a curve at any point. This gradient can be represented by d𝑦 by dπ‘₯. We’re told that this is equal to two times the square of the 𝑦-coordinate. So in other words, is equal to two 𝑦 squared. Here, we recognize that we’ve been given a differential equation. Not only that, we’ve been given a separable differential equation. This is one that can be expressed in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. 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.

Given an equation of this form, we can find the general solution in the following way. We first divide both sides of our equation by 𝑓 of 𝑦. In our case, this is 𝑦 squared. This gives us that one over 𝑦 squared d𝑦 by dπ‘₯ is equal to two. Next, we can integrate both sides of our equation with respect to π‘₯. Now, the chain rule tells us that d𝑦 by dπ‘₯ dπ‘₯ is equal to d𝑦. 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‘₯. As a side note, you often see methods where before the integration, d𝑦 by dπ‘₯ is treated somewhat like a fraction. Both sides are multiplied by dπ‘₯.

And the equivalent statement is obtained: one over 𝑦 squared d𝑦 is equal to two dπ‘₯. Both sides of the equation are then integrated. And we reach the same state as we did before, but with less working. Always remember that this is just a shorthand for the more rigorous steps that we’ve shown here. Returning to our method, one over 𝑦 squared can be expressed as 𝑦 to the power of negative two. 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.

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. And this could just go on one side of the equation. Now, at this stage, we have found the general solution to our differential equation. And it’s in implicit form, which means it’s not in the form where 𝑦 is equal to some function of π‘₯. 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.

Now, our next thought might be to convert our implicit solution into an explicit solution. But as it turns out, this isn’t necessary. This is because the general solution is not our final destination. Rather, we are seeking a particular solution. We want the line which passes through the point negative eight, one. This information is an initial value. And we’ll be finding the particular solution which matches in the following way. We know that if negative eight, one lies on the line, when π‘₯ is equal to negative eight, 𝑦 must be equal to one. We can substitute these known values into the equation for our general solution. We then perform some simplification. And we find that our constant 𝑐 three is equal to 15.

We can now substitute this value into our general solution which we’ll turn it into a particular solution. Negative one over 𝑦 is equal to two π‘₯ plus 15. 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. To recap, we used the information given in the question to form a separable differential equation. The particular solution to this which passes through the point negative eight, one is 𝑦 equals negative one over two π‘₯ plus 15. And this is the answer to our question.

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