Worked example: exponential solution to differential equation | AP Calculus AB | Khan Academy

So we've got the differential equation: the derivative of y with respect to x is equal to 3 times y, and we want to find the particular solution that gives us y being equal to 2 when x is equal to 1.

So I encourage you to pause this video and see if you can figure this out on your own.

All right, now let's work through it together. Some of you might have immediately said, "Hey, this is the form of a differential equation where the solution is going to be an exponential," and you just got right to it. But I'm not going to go straight to that; I'm just going to recognize that this is a separable differential equation, and then I'm going to solve it that way.

When I say it's separable, that means we can separate all the y's (dy's) on one side and all the x's (dx's) on the other side. So what I could do is, if I divide both sides of this equation by y and multiply both sides by dx, I get 1 over y (dy) is equal to 3 (dx).

Now, on the left and right-hand sides, I have these clean things that I can now integrate. That's what people talk about when they say "separable differential equations."

Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the anti-derivative of 1 over y is going to be the natural log of the absolute value of y. I'm taking the anti-derivative with respect to y here. Now I could add a constant, but I'm going to add a constant on the right-hand side. So there's no reason to add two arbitrary constants on both sides; I can just add one on one side.

So that is going to be equal to the anti-derivative here, which is going to be 3x, and I'll add the promised constant plus c right over there. Now let's think about it a little bit.

Well, we can rewrite this in exponential form. We could say we could write that e to the (3x + c) is equal to the natural log of y. I could write the natural log of y is equal to e to the (3x + c). Now I could rewrite this as equal to e to the 3x times e to the c.

Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're going to be different values, but we're just trying to get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the 3x.

So another way of thinking about it is saying the absolute value of y is equal to this. This isn't a function yet; we're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the 3x, or y is equal to negative c e to the 3x.

Well, we've kept it in general terms; I haven't put any—we don't know what c is. So what we could do instead is just pick this one and then we can solve for c, assuming this one right over here, and so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative.

So let's do that. When y is equal to 2, I'm not going to solve for c to find the particular solution. x is equal to 1, or when x is equal to 1, y is equal to 2. So I could write it like that, and we get 2 is equal to c times e to the (3 times 1).

And so to solve for c, I can just divide both sides by e to the third, and so I could—or I could multiply both sides times e to the negative third, and I could get 2 e to the negative third power is equal to c.

And so let's now substitute it back in, and our particular solution is going to be y is equal to c, c is 2 e to the negative third power times e to the 3x. Now I have—I'm taking the product of two things with the same base; I can add the exponents.

So I could say y is equal to 2 times e to the (3x), and then I'll add the exponents to 3x minus 3. And there you go; this is one way that you could write the particular solution that meets these constraints for this separable differential equation.