Main content

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

Current time:0:00Total duration:5:38

# Worked example: finding a specific solution to a separable equation

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

- Let's now get some practice
with separable differential equations, so let's say I have
the differential equation, the derivative of Y with
respect to X is equal to two Y-squared, and let's say
that the graph of a particular solution to this, the graph
of a particular solution, passes through the point one
comma negative one, so my question to you is, what is Y,
what is Y when X is equal to three for this particular
solution, so the particular solution to the differential
equation that passes through the point one comma negative
one, what is Y when X is equal to three, and I encourage
you to pause the video, and try to work through it on your own. So I'm assuming you had a
go at it, and the key with a separable differential
equation, and that's a big clue that I'm even calling it a
separable differential equation, is that you separate the Xs from the Ys. Or all the Xs and the
DXs from the Ys and DYs. So how do you do that here? Well, what I could do,
let me just rewrite it. So it's gonna be DY DX is
equal to two Y-squared, is equal to two Y, equal to two Y-squared. So let's see, we can
multiply both sides by DX, and let's see, so then we're
gonna have, that cancels with that if we treat it as just
a value, or as a variable. We're gonna have DY is
equal to two Y squared DX. Well, we're not quite done yet. We gotta get this two Y
squared on the left hand side. So we can divide both
sides by two Y-squared. So if we divide both sides by
two Y-squared, two Y-squared, the left hand side, we could
rewrite this as 1/2 Y to the negative two power, is going
to be equal to DY, DY is equal to DX, and now, we
can integrate both sides. So we can integrate both sides. Let me give myself a
little bit more space. And so, what is, what is this
left hand side going to be? Well, we increment the exponent, and then divide by that value,
so Y to the negative two, if your increment is
Y to the negative one, and then divide by negative one, so this is going to be -1/2
Y to the negative one power, and we could do a plus C like
we did in the previous video, but we're gonna have a
plus C on both sides, and you could subtract,
or you know, you have different arbitrary
constants on both sides and you could subtract them from each other, so I'm just gonna write the
constant only on one side. So you have that is equal to,
well if I integrate just DX, that's just going to give me
X, that's just gonna give me X. So this right over here is X, and of course I can have
a plus C over there, and If I want I can, I can
solve for Y if I multiply, let's see, I can multiply
both sides by negative two, and then I'm gonna have, the
left hand side you're just gonna have Y to the negative
one, or 1/Y is equal to, if I multiply the right hand
side times negative two, I'm gonna have negative
two times X plus, well it's some arbitrary constant,
it's still going to, it's gonna be negative two
times this arbitrary constant but I could still just call it
some arbitrary constant, and then if we want we can take
the reciprocal of both sides, and so we will get Y is equal
to, is equal to 1/-2X+C. And now we can use, we can use
the information they gave us right over here, the fact
that our particular solution needs to go through this
point to solve for C. So, when X is negative one,
so when X is negative one. Oh sorry, when X is one, when
X is one, Y is negative one, so we get negative one
is equal to 1/-2+C, or we could say C minus two, we
could multiply both sides times C minus two, if then we will
get, actually let me just scroll down a little bit, so
if you multiply both sides times C minus two, negative
one times C minus two is going to be negative C plus two or
two minus C is equal to one. All I did is I multiplied C
minus two times both sides, and then, let's see, I can
subtract two from both sides, so negative C is equal to negative
one, and then if I multiply both sides by negative one,
we get C is equal to one. So our particular solution
is Y is equal to 1/-2X+1. And we are almost done,
they didn't just ask for, we didn't just ask for
the particular solution, we asked, what is Y when
X is equal to three. So Y is going to be equal
to one over, three times negative two is negative six
plus one, which is equal to negative, is going to be
equal to 1/-5, or -1/5. And we are done.

AP® is a registered trademark of the College Board, which has not reviewed this resource.