 So last time we determined if z equals y to power minus n, a first-order linear differential equation in z corresponds to a first-order non-linear equation of this form. And this is known as a Bernoulli equation. After Jaco Bernoulli, the eldest member of a family of Swiss mathematicians of the early 18th century, sibling rivalry and intergenerational jealousy caused many members of the Bernoulli family to become quite prominent in mathematics and physics. The Bernoulli family put the functional in dysfunctional. Now since the substitution z equal y to power minus n gives us this non-linear differential equation, this means that we can use a substitution to rewrite a differential equation of this form as a first-order linear differential equation. But what substitution? And here's an important thing. When dealing with differential equations, it's tempting to treat it like a cookbook. Do this when you see this. Unfortunately, that doesn't apply to very many differential equations, and so the idea is don't memorize formulas and rules, understand processes. And in this case the important process is that we made some substitution that turned a first-order linear differential equation into this thing. So let's think about this problem a little bit. In our differential equation, if n is equal to zero, we have... And that's not actually very interesting because this is a separable differential equation. And we already know how to solve those. If n is equal to negative one, we have... And while this might or might not be separable depending on what f of t and g of t are, this is a linear differential equation. And remember the main effect of our transformation will be to make a linear differential equation. So if we're starting with one, we probably wouldn't want to bother with the transformation. So we'll assume that n is not equal to zero or negative one. So let's have a differential equation like this. So we'll make our substitution z equals y to power minus n. And we'll find dz dt. Then solve for dy dt. And we'll make our substitutions and simplify. Now if we want this to be first-order in z, we have to get rid of this y variable. And so that means we need three minus n, the exponent, to be equal to zero. And so we make n equal to three, and this gives us the equation. And if it's not written down, it didn't happen. Making the substitution z equals y to power minus three gives us this differential equation, which we see is a first-order linear differential equation. Let's see if we can solve this first-order linear differential equation. So we'll multiply by negative three to get rid of those fractions. We'll rearrange things a little bit. We'll factor and separate. And since we separated the variables, we can find the antiderivative of both sides, which gives us a solution. Since our original differential equation was in terms of y, we can't actually leave a solution in terms of z. So we'll need to solve for y. So first, we'll solve our equation for z. And since we wrote down that z equals y to power minus three, we can recover y, which gives us our solution.