 The easiest type of non-separable differential equation to solve is a first-order linear. This is a differential equation where the highest derivative present is the first, and the solution function only appears as a first power. For example, y' plus 12y equals 15 is a first order because we only have the first derivative and linear because the only time our solution function y appears is as a first power. It's important to understand that the linearity applies to the solution function, so dy dx plus e to the power ty equals sine t is also first-order linear, first-order because it has a first derivative, and even though e to the t and sine t are not linear functions of t, we don't care because what's important is that y only appears to the first power. With any luck, we only ever encounter first-order linear differential equations. We are not that lucky. However, it might be possible to transform a nonlinear equation into a linear equation through some transformation of the variables, and so we'll consider the following problem. Suppose we have a nice first-order linear differential equation in z, and z is some function of y. What's the corresponding differential equation in y? Well, that's the easy problem because that just requires us to use the chain rule. However, it does give us insight onto the harder problem, which is the only really one to solve. Suppose we have a nonlinear differential equation in y. Can we find z as some function of y, which will give us a linear differential equation in z? And there's only one way to find out the answer to this question. Well, actually you could sit around waiting for someone to tell you the answer to the question. But that's not as interesting, and you don't learn nearly as much. So there really is only one real way to find the answer to this question, which is to try it out and see what happens. Now one of the problems we'll face is that we may get buried in a mountain of notation, and so it's useful to remember that any function applied to a constant of anti-differentiation is going to produce a constant. And this allows us to reduce any expression involving our constants to a single constant. So no matter how horrible the expression looks, since all we're doing over here is applying functions to constant values, we can absorb all of this work into a single constant, where we recycle the constant name. The analogous situation with functions is that any composition of functions of one variable is a function of one variable. So maybe I have some horrible composition of functions, but because these are all functions of one variable, we can absorb all of these functions into a single function of t, where we recycle the function name. So again, back to our problem. Suppose we have a first-order linear differential equation, and z is some function of y. What's the corresponding differential equation in y? We might begin with a simple linear substitution, z equals ay plus bt plus c, where a, b, and c are constants. So if we differentiate, we get... this transforms our first-order linear differential equation in z into... and we'll simplify by merging constants and functions of t. So if we divide everything by a algebraically, we get... but b divided by a is just some constant, and we'll recycle the constant name. A function of t divided by a constant is just a function of t, so f of t over a and g of t over a will absorb this division by a. If we expand the product we get algebraically, but multiplying a function of t by a gives us a function of t, and multiplying a function of t by another function of t gives us a function of t, which we'll introduce a new name for. And if I add a constant to a function of t, it's just a function of t. To put our differential equation in something looking like a standard form, we'll subtract h of t from both sides. But again, over on the right-hand side, this is a difference of two functions of t, which is going to be a function of t. And so we can simplify our differential equation by merging constants and functions of t into... However, this is just a first-order linear differential equation. So while the substitution may make the solution easier to find, it won't help us solve different types of equations. But it might be useful elsewhere, so I'd make a note of it. Now in retrospect, we should never expect it too much, since z equals ay plus bt plus c is still a linear function of y. So what if we choose a nonlinear function of y? Well, since there are so many ways we could be nonlinear, let's try to limit our possibilities. So we might view the transformation z equals f of y, where f is a nonlinear as a graph transformation. So given some graph y equals g of x, we'll transform the graph by applying f to our function y. And since the graph tells us about the derivative, similar graphs lead to similar differential equations. So what transformations might we consider? Well, let's consider. If our transformation is z equals log y, our original graph is going to become something like this, which is very different from what we started with, maybe too different. If we apply the transformation z equals sine y, our graph is very different again, and possibly too different. If our transformation is z equals y squared, we get a graph that's, well, very similar, and maybe that's too similar to be useful. If we try z equals one over y squared, our graph is different in parts and similar in parts. And so this graph transformation might be useful. So let's try transformations of the form z equals one over y to power n. So let's try to answer that first question. Let z equal y to minus n to what type of differential equations will first order linear differential equations in z correspond? So a first order linear differential equation in z is going to be dz dt plus some function of tz equals some function of t. If we make the substitution z equal y to power minus n, then we'll need to make a substitution on dz dt. So let's find that. If z equal y to power minus n, then dz dt is going to be equals means replaceable, so I can replace dz dt and z in our original differential equation with what they're equal to. Now, this is rather messy, so let's do a little bit of algebra. One of the things we can do is we can multiply every term by y to power n plus one to clear up some of our exponents. So if we do that, we get and let's use our power of absorbing constants and functions. So I have this minus n here. If we divide everything by minus n, we can let that division by minus n get absorbed into the functions f of t and g of t and reduce, reuse, recycle. We'll call them the same thing, even though they are slightly different from what we started with. So notice that we have a first order differential equation because we have a first derivative, but because we have a power on y other than one, this is nonlinear. This type of first order nonlinear differential equation is important enough to have a name. It's called a Bernoulli equation. So let's take a look at how to solve Bernoulli equations next.