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Course: AP®︎ Calculus BC (2017 edition) > Unit 10
Lesson 2: Separable differential equations- Separable equations introduction
- Addressing treating differentials algebraically
- Separable differential equations
- Separable differential equations: find the error
- Worked example: separable differential equations
- Worked example: separable differential equation (with taking log of both sides)
- Worked example: separable differential equation (with taking exp of both sides)
- Separable differential equations
- Worked example: identifying separable equations
- Identifying separable equations
- Identify separable equations
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Worked example: separable differential equation (with taking exp of both sides)
Solving the equation dy/dx=x⋅y. This involves an extra step of raising 𝑒 by both sides of the resulting x-y equation, in order to isolate y.
Want to join the conversation?
- Hi! Isn't C atalways positive since we assumed earlier that it's equal to e^c which can't be negative? Then why do we consider the negative sign? 3:55(3 votes)
- We never assumed that C is positive, and e^c can't be negative but C can be negative since e^-x = 1/e^x. Also, we consider the negative sign because the left-hand side of the equation has |y|.(1 vote)
Video transcript
- [Tutor] What we're
going to do in this video is see if we can solve
the differential equation, the derivative of y with respect to x is equal to x times y, pause this video and see if you can find
a general solution here. So the first thing that
my brain likes to do, when I see a differential
equation is to say, hey, is this separable
and when I say separable, can I get all the ys and dys on one side and all the xs and dxs on the other side? And you can indeed do that, if we treat our differentials like, if we could treat 'em
like algebraic variables, which is fair game, when you're dealing with differential equations like this, we could multiple both sides by dx, so multiple both sides by dx and we could divide both sides by y, let me move this over a little
bit, so we have some space, so we could also multiple
both sides by one over y, one over y and so what that does is these dxs cancel out and this
y and one over y cancels out and we are left with, let
me write all the things in terms of y on the
left-hand side in blue, so we have one over you dy is equal to, and then I'll do all this stuff in orange, we have is equal to, we're just left with an x and a dx, x dx and then we'll wanna take the indefinite integral of both sides, now what's the
anti-derivative of one over y? Well, if we want it in
the most general form, this would be the natural log
of the absolute value of y and then this is going to be equal to the anti-derivative of x is x squared over two and then we wanna put a
constant on either side, I'll just put it on the
right-hand side, plus C, this ensures that we're dealing
with the general solution, now if we wanna solve explicitly for y, we could raise e to both sides' power, another way to think about it is, if this is equal to that, then e to this power is going to be the same
thing as e to that power, now, what does that do for us? Well, what is e to the natural log of the absolute value of y? Well, I'm raising e to the power that I would have to raise e too to to get to the absolute value of y, so the left-hand side here, it simplifies to the absolute value of y and we get that as being equal to, now, we could use our exponent properties, this over here is the same thing as e to the x squared over two times e to the C times e to the C, I am just using our
exponent properties here, well, e to the C, we could just view that as some other type of constant,
this is just sum constant C, so we could rewrite this whole thing as C e to the x squared x squared over two, hopefully
you see what I'm doing there, I just use my exponent properties e dah dah dah, sum of two things is equal to e to the first thing
times e to the second thing and I just said, well, e to
the power of sum constant C, we could just relabel that
as, let's call that our blue C and so this simplifies to blue C times e to the x squared over two, now we saw this absolute value sign here, so this essentially means that y could be equal to, we
could write it this way, y could be equal to plus or minus C, C e to the x squared over two, x squared over two, but once again, we don't
know what this constant is, I didn't say that it was
positive or negative, so when you see a plus or
minus of a constant here, you could really just subsume all of this, I'll just call this red C, so we could say that y is equal to, I'll just rewrite it over again for fun, y is equal to red C, not the Red Sea, but a red C (laughs) times e to the x squared over two, this right over here,
it's the general solution to the original, separable
differential equation.