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# Worked example: separable equation with an implicit solution

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.E (LO)
,
FUN‑7.E.1 (EK)
,
FUN‑7.E.2 (EK)
,
FUN‑7.E.3 (EK)

## Video transcript

- [Teacher] We're given a differential equation right over here, cosine of y plus two, this whole thing times the derivative of y with respect to x is equal to two x, and we're given that for a particular solution, when x is equal to one, y of one is equal to zero, and we're asked what is x when y is equal to pi? So the first thing I like to look at when I see a differential equation is: is it separable? Can I get all the ys and dys on one side, and can I get all the xs and dxs on the other side? And 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 plus two times dy is equal to two x times dx, so just like that I would've been able to, all I did is I multiplied both sides of this times dx, and I was able to separate the ys and the dys from the xs and the dxs, and now I can integrate both sides. So, if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y, is sine of y, and then the antiderivative of 2 with respect to y is 2y, and that is going to be equal to, well, the antiderivative of 2x with respect to x is x squared, and 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, and so this is a general solution to the separable differential equation, and then we can find the particular one by substituting in when x is equal to one, y is equal to zero, so let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one, so, sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero is zero, two times zero is zero, all of that's just gonna be zero, so we get zero is equal to one plus c, or c is equal to negative one. 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 plus two y is equal to x squared, and our constant is negative one, so minus one. And now what is x when y is equal to pi? So, sine of pi plus two times pi is equal to x squared minus one, see sine of pi is equal to zero, and so we get, let's see we can add one to both sides, and we get two pi plus one is equal to x squared, or we could say that x is equal to the plus or minus the square root of two pi plus one. So I would write the plus or minus square root of two pi plus one, and we're done.