# Separable equationsÂ (old)

## Video transcript

We hopefully know at this point
what a differential equation is, so now let's
try to solve some. And this first class of
differential equations I'll introduce you to, they're called
separable equations. And I think what you'll find
is that we're not learning really anything new. Using just your first year
calculus derivative and integrating skills, you can
solve a separable equation. And the reason why they're
called separable is because you can actually separate the
x and y terms, and integrate them separately to get
the solution of the differential equation. So that's separable. Separable equations. So let's do a couple, and I
think you'll get the point. These often are really more of
exercises in algebra than anything else. So the first separable
differential equation is: dy over dx is equal to x squared
over 1 minus y squared. And actually, this is a good
time to just review our terminology. So first of all, what
is the order of this differential equation? Well, the highest derivative
in it is just the first derivative, so the order
is equal to 1. So it's first order. It's ordinary, because we only
have a regular derivative, no partial derivatives here. And then, is this linear
or non-linear? Well, first you say, oh well,
you know, this looks linear. I'm not multiplying
the derivative times anything else. But if you look carefully,
something interesting is going on. First of all, you have
a y squared. And y is a dependent variable. y is a function of x. So to have the y squared, that
makes it non-linear. And even if this was a y, if you
were to actually multiply both sides of this equation
times 1 minus y, and get in the form that I showed you in
the previous equation, you would have 1 minus y squared. This is actually the first step
of what we have to do anyways, so I'll
write it down. So if I'm just multiplying both
sides of this equation times 1 minus y squared, you get
1 minus y squared times dy dx is equal to x squared. And then you immediately see
that even if this wasn't a squared here, you'd be
multiplying the y times dy dx, and that also makes it
non-linear because you're multiplying the dependent
variable times the derivative of itself. So that also makes this
a non-linear equation. But anyway, let's get back
to solving this. So this is the first step. I just multiply both sides
by 1 minus y squared. And the real end goal is just
to separate the y's and the x's, and then integrate
both sides. So I'm almost there. So now what I want to do is I
want to multiply both sides of this equation times dx, so I
have a dx here and get rid of this dx there. Let me go here, I don't want
to waste too much space. So you get 1 minus y squared dy
is equal to x squared dx. I have separated the x and
y variables and the differentials. All I did is I multiplied both
sides of this equation times dx to get here. Now, I can just integrate
both sides. So let's do that. Whatever you do to one side
of the equation you have to do the other. That's true with regular
equations or differential equations. So we're going to integrate
both sides. So what's the integral of this
expression with respect to y? Let's see. The integral of 1 is y, the
integral of y squared, well that's minus y to the
third over 3. And I'll write the plus c here
just to kind of show you something, but you really don't
have to write a plus c on both sides. I'll call the plus the
constant due to y. The y integration. You'll never see this in a
calculus class, but I just want to make a point here. I just want to show you that
our plus c has never disappeared from when we were
taking our traditional antiderivatives. And what's the derivative
of this? Well that's x to the
third over 3. And this is also going
to have a plus c due to the x variable. Now, the reason why I did this
magenta one in magenta and I labelled it like that, is
because you really just have to write a plus c on one
side of the equation. And if that doesn't make a lot
of sense, let's subtract this c from both sides, and we get y
minus-- let me scroll down a little bit, my y
looks like a g. y minus y to the third over 3
is equal to x to the third over 3 plus the constant when we
took the antiderivative of the x, minus the constant of
the antiderivative when we took the y. But these two constants,
they're just constants. I mean, we don't know
what they are. There are arbitrary constants. So we could just write
a general c here. So you could have just-- you
have to have a constant, but it doesn't have to be on both
sides of the equation, because they're arbitrary. cx minus cy, well, that's still
just another constant. And then if we want to simplify
this equation more, we can multiply both sides
of this by 3, just make it look nicer. And you get 3y minus y to the
third is equal to x to the third plus-- well, I could
write 3c here. But once again, c is an
arbitrary constant. So 3 times an arbitrary
constant, that's just another arbitrary constant. So I'll write the c there. And there you have it. We have solved this differential
equation. Although it is in implicit
form right now, and it's fairly hard to get it out
of implicit form. We could put the c on one side,
so the solution could be 3y minus y to the third minus x
to the third is equal to c. Some people might like that
little bit better. But that's the solution. And notice, the solution, just
like when you take an antiderivative, the solution
is a class of implicit functions, in this case. And why is it a class? Because we have that
constant there. Depending on what number
you pick there, it will be another solution. But any constant there will
satisfy the original differential equation,
which was up here. This was the original
differential equation. And if you want to solve for
that constant, someone has to give you an initial condition. Someone has to say, well,
when x is 2, y is 3. And then you could
solve for c. Anyway, let's do another
one that gives us an initial condition. So this one's a little bit--
I'll start over. Clear image, different colors,
so I have optimal space. So this one is the first
derivative of y with respect to x is equal to 3x squared
plus 4x plus 2 over 2 times y minus 1. This is a parentheses, not
an absolute value. And they give us initial
conditions. They say that y of 0 is
equal to negative 1. So once we solve this
differential equation, and this is a separable differential
equation, then we can use this initial condition,
when x is 0, y is 1, to figure out the constant. So let's first separate
this equation. So let's multiply both sides
by 2 times y minus 1. And you get 2 times y minus 1
times dy dx is equal to 3x squared plus 4x plus 2. Multiply both sides times dx. This is really just an
exercise in algebra. And I can multiply this one out,
too, you get 2y minus 2, that's just this, dy. I multiplied both sides times
dx, so that equals 3x squared plus 4x plus 2 dx. I have separated
the equations. I've separated the independent
from the dependent variable, and their relative
differentials, and so now I can integrate. And I can integrate
in magenta. What's the antiderivative
of this expression with respect to y? Well, let's just see. It's y squared minus 2y. I won't write the plus
c, I'll just do it on the right hand side. That is equal to 3x squared. Well, the antiderivative is x to
the third, plus 2x squared, plus 2x plus c. And that c kind of takes care of
the constant for both sides of the equation, and hopefully
you understand why from the last example. But we can solve for c using the
initial condition y of 0 is equal to negative 1. So let's see. When x is 0, y is negative 1. So let's put y as negative 1,
so we get negative 1 squared minus 2 times negative 1, that's
the value of y, is equal to when x is equal to 0. So when x is equal to 0, that's
0 to the third plus 2 times 0 squared plus
2 times 0 plus c. So this is fairly
straightforward. All of these, this is all 0. This is, let's see, negative
1 squared, that's 1. Minus 2 times minus 1, that's
plus 2, is equal to c. And we get c is equal to 3. So, the implicit exact solution,
the solution of our differential equation-- remember
now, it's not a class, because they gave us an
initial condition-- is y squared minus 2y is equal to x
to the third plus 2x squared plus 2x plus 3. We figured out that's
what c was. And actually, if you want, you
could write this in an explicit form by completing
the square. This is just algebra
this point. You're done. This is an implicit form. If you wanted to make it
explicit, you could add 1 to both sides. I'm just completing
the square here. So y squared minus 2y plus 1. If I add 1 to that side, I have
to add 1 to this side, so it becomes x to the third plus
2x squared plus 2x plus 4. I just added 1 to both sides
of the equation. Why did I do that? Because I wanted this
side to be a perfect square in terms of y. Then I can rewrite this side as
y minus 1 squared is equal to x to the third plus 2x
squared plus 2x plus 4. Then I could say y minus 1 is
equal to the plus or minus square root of x to the
third plus 2x squared plus 2x plus 4. I can add 1 to both sides, and
then I can get y is equal to 1 plus or minus the square root
of x to the third plus 2x squared plus 2x plus 4. And it has plus or minus here,
and if we have to pick one of the two, we'd go back to
the initial condition. Well, our initial condition told
us that y of 0 is equal to negative 1. So if we put 0 here for x, we
get y is equal to 1 plus or minus 0 plus 4. So 1 plus or minus 4. So if y is going to be equal to
negative 1, so we get y is equal to 1 plus or
minus-- sorry, 2. If this is going to be equal to
negative 1, then this has to be 1 minus 2. So the explicit form that
satisfies our initial condition, and we're getting a
little geeky here, you could get rid of the plus, it's 1
minus this whole thing. That's what satisfies our
initial condition. And you could figure out where
it's satisfied, over what domain is it satisfied. Well, that's satisfied when this
term is positive, this becomes negative, and you get
it's undefined in reals, and all of that. But anyway, I've run
out of time. See you in the next video.