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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 7

Lesson 6: Finding general solutions using separation of variables- Separable equations introduction
- Addressing treating differentials algebraically
- Separable differential equations
- Separable differential equations: find the error
- Worked example: separable differential equations
- Separable differential equations
- Worked example: identifying separable equations
- Identifying separable equations
- Identify separable equations

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# Worked example: separable differential equations

Two worked examples of finding general solutions to separable differential equations.

## Want to join the conversation?

- I don't understand what Sal did with the constant in the second example.

The anti-derivative of sin(x) is -cos(x) + C. After that he multiplies it by -1, and he gets cos(x) + C. Shouldn't that be cos(x) - C?(16 votes)- Remember, 𝐶 = ℝ. That is, the constant of integration can be
*any*real number. Writing +𝐶 is the same as writing -𝐶 because 𝐶 encapsulates all real numbers. That is, if we fix 𝐶₁ as one possible value of 𝐶, then since {𝐶₁ ∪ -𝐶₁} ∈ ℝ = 𝐶, the sign of the constant of integration does not matter.

For example,*an*antiderivative of sin 𝑥 is -cos 𝑥 + 5. Multiplying this by -1 for whatever reason gives us cos 𝑥 – 5. However, another antiderivative of sin 𝑥 is -cos 𝑥 – 5. Multiplying this by -1 gives us cos 𝑥 + 5. So cos 𝑥 ± 5 are both valid results from multiplying antiderivatives of sin 𝑥 by -1. This same logic applies to all real numbers (𝐶) so whether or not we distribute the negative to 𝐶 is irrelevant. It still encapsulates ℝ no matter what its sign is.

Comment if you have questions!(40 votes)

- Would it be fine if I simplify my answer to y = sec(x) + C?(5 votes)
- Tempting but no. That would only work if the solution were:

y = (1 / cos x) + C

But the answer was

y = 1 / (cos x + C)

and you can't pull the Constant out of the denominator.(16 votes)

- At 1 :48 why Sal is putting c on right hand side only but not with y^2/2. Please explain.(5 votes)
- Greetings!

A few videos back, Sal mentioned that if you put C on both sides, then you subtract C from both sides, you simply end up with C on the right side. So, it is a shortcut to simply leave the C off of the Y (left side) of the equation as once you solve for Y, you will only end up with +C on the right side anyway.(8 votes)

- I didn't get the idea behind the "general solution" and "Particular solution"

Could someone help me please(5 votes)- that + C at the end after integrating casues it to be a general solution, since we don't know what the exact solution is. Or maybe another way to think of it is that C can be any number, and if you differentiated it you'd get the exact same answer. think x^2+1, x^2+2 and even just x^2. They all differentiate to 2x.

It isn't until we have some initial conditions we can figure out what that C is. initial conditions need to be when x is something then y is something else. Then you can plug these into the equation once you have things in terms of y and use algebra to figure out what C is. Does that help?(7 votes)

- That was strangely fun Sal, im ecstatic over here(5 votes)
- yeah, calc is fun :D(2 votes)

- I've heard people say "At3:45, it was A-OK for Sal to keep C as C when multiplying by -1, because C is still just an arbitrary constant."

That I understand.

This I do not understand:

At4:02, Sal takes the reciprocal of both sides. And he puts the C in the denominator. But if you have and arbitrary constant C and you raise it to the -1 power, it'll still be an arbitrary constant!! It's domain hasn't changed.(3 votes)- We cannot take the reciprocal of C if C=0. Keeping C in the denominator reflects that.(5 votes)

- Isn't y = 0 also a solution for the 2nd one? It's not given by the general solution he gave.(2 votes)
- It is a solution, and it's not given because Sal started by dividing both sides by y. Therefore, he implicitly assumed that y≠0, and missed that solution. After solving, Sal should've gone back and checked the y=0 case.(3 votes)

- Not a question but is it just me or is this one of the easiest things I've learned in calculus so far? It's really straightforward and makes a lot of intuitive sense(2 votes)
- Separable differential equations are probably the easiest DEs to solve. If you take a DE course, you'll stumble upon linear DEs and homogeneous DEs, which are generally harder to solve.(2 votes)

- isn't the antiderivative of dy, y(2 votes)
- The antiderivative of dy is y(1 vote)

- at3:10, why is the antiderivative of y^-2 not (2y)(ln|y|)?(1 vote)
- If you were to take the derivative of that, you would use the product rule. You have to make sure you recognize that when taking an integral because the derivative of (2y)(ln|y|) is actually equal to

2*ln|y|+2 which is evidently not y^-2, the answer would be y^-3/-3 using the power rule for integrals. If you are unsure, you can always take the derivative of your answer to check if it is correct(2 votes)

## Video transcript

- [Instructor] What we're
gonna be doing in this video is get some practice
finding general solutions to separable differential equations. So let's say that I had the
differential equation DY, DX, the derivative of Y with
respect to X, is equal to E to the X, over Y. See if you can find
the general solution to this differential equation. I'm giving you a huge hint. It is a separable differential equation. All right, so when we're dealing with a separable differential
equation, what we wanna do is get the Ys and the
DYs on one side, and then the Xs and the DXs on the other side. And we really treat these
differentials kind of like variables, which
is a little hand-wavy with the mathematics. But that's what we will do. So let's see. If we
multiply both sides times Y, so we're gonna multiply
both sides times Y, what are we going to get? We're gonna get Y times a derivative of Y, with respect to X, is equal to E to the X, and now we can multiply both
sides by the differential, DX; multiply both of them
by DX; those cancel out. And we are left with Y times
DY is equal to E to the X, DX. And now we can take the
integral of both sides. So let us do that. So what is the integral of Y, DY? Well here we would just
use the reverse power rule. We would increment the exponent,
so it's Y to the first, but so now when we take
the anti-derivative, it will be Y squared, and
then we divide by that incremented exponent, is equal to, well the exciting thing about
E to the X is that it's anti-derivative, and its
derivative, is E to the X, is equal to E to the X,
plus is equal to E to the X plus C. And so we can leave it
like this if we like. In fact this right over
here is, this isn't an explicit function. Y here isn't an explicit function of X. We could actually say Y is
equal to the plus or minus square root of two times
all of this business, but this would be a pretty
general relationship, which would satisfy this
separable differential equation. Let's do another example. So let's say that we
have the derivative of Y with respect to X is equal
to, let's say it's equal to Y squared times sine of X. Pause the video and see if you can find the general solution here. So once again, we wanna
separate our Ys and our Xs. So let's see, we can
multiply both sides times Y to the negative two power,
Y to the negative two, Y to the negative two, these become one, and then we could also
multiply both sides times DX. So if we multiply DX here,
those cancel out, and then we multiply DX here,
and so we're left with Y to the negative two
power times DY is equal to sine of X, DX, and now we
just can integrate both sides. Now what is the anti-derivative
of Y to the negative two? Well, once again we use
the reverse power rule. We increment the exponents,
so it's gonna be Y to the negative one, and
then we divide by that newly incremented exponent. So we divide by negative one. Well that would just
make this think negative. That is going to be equal to... So, what's the
anti-derivative of sine of X? Well, it is, you might recognize it if I put a negative there,
and a negative there. The anti-derivative of negative sine of X, well that's cosine of X. So this whole thing is gonna
be negative cosine of X, or another way to write this:
I could multiply both sides times a negative one, and so these would both become positive, and so I could write one over Y is equal to
cosine of X, and actually let me write it this
way, plus C; don't wanna forget my plus Cs. Plus C, or I could take the
reciprocal of both sides if I wanna solve explicitly
for Y, I could get Y is equal to one over cosine of X plus C as our general solution. And we're done. That was strangely fun.