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Worked example: separable differential equation (with taking exp of both sides) | Khan Academy


3m read
·Nov 11, 2024

What we're going to do in this video is see if we can solve the differential equation: the derivative of y with respect to x is equal to x times y. Pause this video and see if you can find a general solution here.

So, the first thing that my brain likes to do when I see a differential equation is to say, hey, is this separable? And when I say separable, can I get all the y's and dy's on one side and all the x's and dx's on the other side? You can indeed do that if we treat our differentials like if we could treat them like algebraic variables, which is fair game when you're dealing with differential equations like this.

We could multiply both sides by dx. So, multiply both sides by dx and we could divide both sides by y. Let me move this over a little bit so we have some space. So, we could also multiply both sides by 1 over y, 1 over y. And what that does is these dx's cancel out, and this y and 1 over y cancels out.

We are left with, let me write all the things in terms of y on the left-hand side in blue. So, we have 1 over y dy is equal to, and then I'll do all this stuff in orange. We have: is equal to... we're just left with an x and a dx, x dx. And then we’ll want to take the indefinite integral of both sides.

Now, what's the antiderivative of 1 over y? Well, if we want it in the most general form, this would be the natural log of the absolute value of y, and then this is going to be equal to the antiderivative of x, which is x squared over 2. And then we want to put a constant on either side; I'll just put it on the right-hand side plus c. This ensures that we're dealing with the general solution.

Now, if we want to solve explicitly for y, we could raise e to both sides power. Another way to think about it is if this is equal to that, then e to this power is going to be the same thing as e to that power.

Now, what does that do for us? Well, what is e to the natural log of the absolute value of y? Well, I'm raising e to the power that I would have to raise e2 to get to the absolute value of y. So, the left-hand side here simplifies to the absolute value of y, and we get that as being equal to...

Now, we could use our exponent properties. This over here is the same thing as e to the x squared over 2 times e to the c. I am just using our exponent properties here. Well, e to the c we could just view that as some other type of constant; this is just some constant c.

So, we could rewrite this whole thing as c e e to the x squared over 2. Hopefully, you see what I'm doing there. I just use my exponent properties: e to the sum of two things is equal to e to the first thing times e to the second thing.

And I just said, well, e to the power of some constant c we could just relabel that as, let's call that our blue c. And so, this simplifies to blue c times e to the x squared over 2.

Now, we still have this absolute value sign here, so this essentially means that y could be equal to... We could write it this way: y could be equal to plus or minus c e e to the x squared over 2.

But once again we don't know what this constant is; I didn't say that it was positive or negative. So, when you say a plus or minus of a constant here, you could really just subsume all of this. I'll just call this with red c, so we could say that y is equal to... I’ll just rewrite it over again for fun: y is equal to red c, not the red c, but a red z times e to the x squared over 2.

This right over here is the general solution to the original separable differential equation.

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