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## Finding particular solutions using initial conditions and separation of variables

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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.