Worked example: separable equation with an implicit solution | Khan Academy

We're given a differential equation right over here: cosine of y + 2, this whole thing times the derivative of y with respect to x is equal to 2x. We're given that for a particular solution, when x is equal to 1, y of 1 is equal to zero. We're asked, what is x when y is equal to π?

The first thing I like to look at when I see a differential equation is, is it separable? Can I get all the y's and dy on one side, and can I get all the x's and dx's on the other side? This one seems like it is. If I multiply both sides by dx, where you can view dx as the X differential of an infinitely small change in x, well then you get cosine of y + 2 * dy is equal to 2x * dx.

So just like that, I've been able to— all I did is I multiplied both sides of this times dx, but and I was able to separate the y's and the dy from the x's and the dx's. Now I can integrate both sides. So if I integrate both sides, what am I going to get?

The anti-derivative of cosine of y with respect to y is sine of y. Then the anti-derivative of two with respect to y is 2y. That is going to be equal to—well, the anti-derivative of 2x with respect to x is x^2. We can't forget that we could say a plus a different constant on either side, but it serves our purpose just to say plus C on one side.

So this is a general solution to this separable differential equation, and then we can find the particular one by substituting in when x is equal to 1, y is equal to 0. Let's do that to solve for C. So we get, or when y is equal to 0, x is equal to 1.

So sine of 0 + 2 * 0— all I did is I substituted in the zero for y— is equal to x^2. Well now, x is 1, so sine of 0 + C. Well, sine of 0 is 0, 2 * 0 is 0— all of that’s just going to be zero. So we get 0 is equal to 1 + C, or C is equal to -1.

So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here: sine of y + 2y is equal to x^2 - 1.

Now, what is x when y is equal to π? So sine of π + 2π is equal to x^2 - 1. Sine of π is equal to 0, and so we get—let's see, we can add one to both sides and we get 2π + 1 is equal to x^2.

Or we could say that x is equal to the plus or minus square root of 2π + 1. So I would write the plus or minus square root of 2π + 1, and we're done.