Video: Finding a Relative Maximum of a Function Which is a Solution to a Differential Equation

Consider the differential equation d𝑦/d𝑥 = −5 + 3𝑥² − 2𝑦. If 𝑦 = 𝑓(𝑥) is the solution to the differential equation, at what point does 𝑓 have a relative maximum?

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Video Transcript

Consider the differential equation d𝑦 by d𝑥 equals negative five plus three 𝑥 squared minus two 𝑦. If 𝑦 equals 𝑓 of 𝑥 is the solution to the differential equation, at what point does 𝑓 have a relative maximum? Is it a) negative three, 11, b) one, negative one, c) three, 11, or d) five, 35?

So, we have a differential equation whose solution is given as 𝑦 is equal to some function of 𝑥. We’re looking to find the location of a relative maximum, so we begin by recalling what we mean by a critical point. The critical points occur when the derivative of the function with respect to 𝑥, so d𝑦 by d𝑥, is equal to zero. Then, if the value of the second derivative, that’s d two 𝑦 by d𝑥 squared, is less than zero at any of these critical points, then we have a local maximum. We’ve been given an equation for d𝑦 by d𝑥. That’s the original differential equation in the question. But we don’t yet have an equation for d two 𝑦 by d𝑥 squared.

To find the second derivative, we’re going to differentiate our expression for d𝑦 by d𝑥. That’s negative five plus three 𝑥 squared minus two 𝑦 with respect to 𝑥. And we can actually do this term by term. The derivative of negative five or any constant is zero. Then, to differentiate three 𝑥 squared, we multiply that entire term by the exponent and then reduce the exponent by one. So, we get two times three 𝑥. The derivative of negative two 𝑦 is a little bit trickier.

To differentiate an expression for 𝑦 with respect to 𝑥, we use a special version of the chain rule. This is implicit differentiation. And the rule that we use is that the derivative of some function of 𝑦 with respect to 𝑥 is calculated by multiplying the derivative of that function with respect to 𝑦 by d𝑦 by d𝑥. So, we begin by differentiating negative two 𝑦 with respect to 𝑦. Well, that’s just negative two. And then, we multiply by d𝑦 by d𝑥. So, we get zero plus two times three 𝑥 minus two d𝑦 by d𝑥. And that simplifies quite nicely to six 𝑥 minus two d𝑦 by d𝑥.

Now, at this stage, what we could do is substitute the expression for d𝑦 by d𝑥 into this equation. However, we’re really looking to find the point where the second derivative is less than zero, so where six 𝑥 minus two d𝑦 by d𝑥 is less than zero. And we know that since we’re looking at a critical point, the derivative is zero. So, we’re looking to find really where six 𝑥 is less than zero. Now, it follows that for six 𝑥 to be less than zero, 𝑥 itself must be less than zero. And actually, if we go back to our original options, there’s only one point for which this is the case. It’s the point negative three, 11 that has an 𝑥-value of negative three and a 𝑦-value of 11.

Now, what we will do is just double check that the first derivative at this point is indeed equal to zero. We’ll substitute 𝑥 equals negative three and 𝑦 equals 11 into our expression for the first derivative. And we get negative five plus three times negative three squared minus two times 11, which is indeed zero. So, this first point satisfies all our criteria. At this point, the first derivative is equal to zero, so it is indeed a critical point. But also, the second derivative is less than zero, so this critical point is a local or a relative maximum.

And at this stage, what we could do is substitute all the other points into both of our equations, our equation for d𝑦 by d𝑥 and our equation for d two 𝑦 by d𝑥 squared. But of course, we know that even if d𝑦 by d𝑥 is equal to zero, six 𝑥 cannot be less than zero. And that’s because each of the 𝑥 values are themselves positive. So, the answer is a, negative three, 11.

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