Welcome to the video on the introduction to
differential equations.
What's a differential?
Well, we've been doing things like if I said that y is
equal to, I don't know, 3x.
If we took the derivative of both sides of this equation and
we said -- one of the notations we used is we said dy
over dx is equal to 3.
We took the derivative of both sides of that.
But this you could almost use differential notation.
dy is a differential and dx is a differential.
What does a differential mean?
A differential just means an infinitely small change in y,
in this case, so that's a differential in y, or an
infinitely small change in x.
So a differential it's like a difference -- when we learned
slope, we said slope is change in y over change in x.
Or you could say the difference in y over difference in x.
But once we started taking the slope of curves we had to make
this difference and this difference approach 0.
That's what a differential essentially is.
A differential is a really, really, really, really
infinitely small difference.
So what is a differential equation?
Well it's an equation that involves differentials.
So, for example, this is a differential equation.
This will be the first example that we solve.
dy over dx is equal to x squared plus 1.
So how do we solve this differential equation?
Well, there's a couple of ways you could say it.
You say oh, well this is the same thing as f prime of x.
You know, y to the function of x and you just take the
antiderivative of both sides.
But let's solve this properly as a differential equation.
So you can actually manipulate these differentials the same
way that you would manipulate numbers for the most part.
And so what could we do here?
Let's multiply both sides of this equation by dx.
So you get, for a very small change in y, dy, you get x
squared plus 1 times a very small change in the x.
Or another way to say it is for a very small change in x, if
you want to figure out how much does y change, you multiply
times x squared plus 1 wherever you are on the curve.
So what does that do for us?
Well let's take the antiderivative of both
sides or let's take the integral of both sides.
And if you watched all of the videos on integration and the
definite integral and area and a curve, you realize that an
integral is essentially a sum, it's kind of an infinite
sum of a bunch of these infinitely small dy's.
So if you take the sum of all of the changes in dy's,
you're left with a y on this side of this equation.
You might want to re-watch all the videos that we
did on integration.
Then what is this?
This is essentially just -- this is the indefinite
integral of x squared plus 1.
So it's the antiderivative of that.
And what's the antiderivative of x squared plus 1?
Well it's x to the third over 3.
We're just taking the derivative backwards.
And once again, watch the videos on the antiderivative.
There's I think eight or nine of them.
Plus x and then plus c.
And where does the c come from?
Well we know that when we take the derivative of a
constant it goes to 0.
So when you take the antiderivative, we're like
well, there could have been a constant there.
And that's where that plus c comes from.
So this is the general solution to this differential equation.
And that's something interesting.
So with traditional equations, the solution tended to
be a number, right?
If I just told you y is equal to 2y minus 1.
Then you would get minus y is equal to minus 1, you'd
get y is equal to 1.
So this is a traditional equation and your solution
was just a value -- you solved for the variable.
Differential equations are something different.
The solution is actually a function.
You're saying what function satisfies this
differential equation.
So that's something to keep in mind.
Right now we're doing very basic differential equations,
but that's something to keep in mind the whole time you learn
differential equations.
I think I'll eventually do a play list on essentially
an introductory course on differential equations that
you would take at college.
And that applies even when you start doing partial
differential equations, et cetera, et cetera.
The solution to a differential equation is not a number,
it is a function.
So anyway, this was the general solution to this
differential equation.
And if you want the particular solution, people normally give
you initial conditions or they give you points on the function
and then you can substitute back.
In this problem they said that dy dx equals x squared plus 1.
And they said that -- let me switch colors -- they said that
y is equal to 1 at x is equal to 1, or y is equal to 1
when x is equal to 1.
So we can use this information now to solve for c.
How do you do that?
Well it says y is equal to 1.
So 1 is equal to when x is equal to 1, is equal to 1/3,
right, 1 over 3, 1 to the third power over 3, plus 1 plus c.
And let's see what I can do.
Subtract one from both sides -- this comes 0.
Subtract 1/3 from both sides and you get c is
equal to minus 1/3.
So using these conditions, a point where this function
crosses through, we can now give you the particular
solution to this differential equation.
And that is y is equal to x to the third over 3 plus x minus
1/3, and then we just solve for the c.
And if you don't believe me, take this expression and
substitute it here and you will see that it equals, if you were
to take the derivative of y with respect to x, you would
see that it equals x squared plus 1.
Let's do another one.
So it tells us -- so this is a little bit more interesting --
it says dy dx is equal to x over y, and it has
the same conditions.
It says y equals 1 at x equals 1.
Or when x equals 1, y equals 1.
So let's find the function that satisfies this equation.
This one's interesting because we have an x and a y on
the right hand side of the equation.
So this kind of looks like something we got after we
did some type of implicit differentiation.
But let's see where we go.
So let's do the same thing.
Let's multiply both sides of this equation by dx until you
get dy is equal to x over y dx.
And let's get this y over onto the dy side, because
it will be easy to take the antiderivative then.
So we get y dy is equal to x dx.
Now we can take the integral of both sides of this equation, or
take the antiderivative on this side with respect to y on
the side with respect to x.
I don't know why I did it in brown.
So what's the antiderivative here?
Well, it is y squared over 2.
And I could say plus c, some constant.
Let me do that.
There could be a constant here.
I'll call it c1.
We don't know what that constant is.
And that equals x squared over 2 plus c2, right?
Or some other constant.
Maybe it's the same number, I don't know.
But we don't know what either of these are.
I could re-write it as -- let's see, what could I do?
Let me take the x out of that side.
So I could have y squared over 2 minus x squared over 2 is
equal to -- and let me subtract this from that side so I get
the constants all on the right hand side of the equation.
c2 minus c1.
I just took a c1, put it on the right hand side, took the x
squared over 2, put it on the left hand side, that's
why it's negative.
And we didn't know.
We said this could be any constant and this
could be any constant.
So the difference between two arbitrary constants could just
be a third arbitrary constant.
So I'll just re-write that as c.
So the general solution to this differential equation is y
squared over 2 minus x squared over 2 is equal to c.
Actually let's do something else just to clean it up a
little bit, because once again this could be any constant.
So let's multiply both sides of this equation by 2 and
you get y squared minus x squared equal to 2c.
Well now this is still any constant number, so we could
still write this as a c.
So we have y squared minus x squared is equal to c.
Now let's use our initial conditions to see what c is.
So when y is 1, so 1 squared minus when x is 1, 1 squared is
equal to c, this is 1 minus 1, so it's 0, right?
c is equal to 0.
So what is the particular solution to this
differential equation?
I'll do it in green.
It is y squared minus x squared is equal to 0.
Or we could add x squared to both sides of that.
We could also write it as y squared is equal to x squared.
Now you might be tempted to take the square root of both
sides of this and say that y is equal to x.
The reason why this would not be accurate is because here x
could be minus 2 and y could be plus 2 or vice versa.
So this would satisfy this equation, but it would not
satisfy this equation.
So be careful when you take that square root.
You have to worry about the plus or minus.
Let's do one more.
So this one says the derivative of y with respect to x is equal
to -- OK, so this is even a little bit more interesting.
This is equal to the square root of x over y.
And it says that y is equal to 4 when is equal to 1.
Yup, I think I'm reading that right.
We could do this one very similarly.
So let's just do the same step.
So multiply both sides of the equation times dx.
So you get dy is equal to -- and just to skip a step, I'm
going to re-write square root of x over y is the square root
of x over the square root of y, and I multiply both
sides of that times dx.
Now let's multiply both sides of this equation
by the square root of y.
And I'm just going to re-write it as y to the 1/2 power.
That's the same thing as the square root of y.
dy is equal to x to the 1/2 power dx.
I just multiplied that there and re-wrote
it as y to the 1/2.
So what is the antiderivative of y to the 1/2 power?
Well it's just this plus 1, so it's y the 3/2 power, and then
times the inverse of this.
So times 2/3.
Or you could have said divided by this, either way.
And if you're not sure about this, because sometimes it is
confusing with the fractions and whatever else,
take the derivative.
3/2 times 2/3 is 1, and then you subtract this 1 from
the exponent and you get y to the 1/2.
So that works.
And of course, we kind of go through the same drill, plus c1
is equal to -- it's going to be the same thing on this side
-- 2/3 x to the 3/2 plus c2, and what can we do?
Well let's take the x to the left hand side of this
equation, so we get 2/3 y to the 3/2 minus 2/3 x to the 3/2
is equal to c2 minus c1, which we can just re-write as c.
And let's multiply both sides of this equation by 3/2, so
these two will become 1.
So you get y to 3/2 minus x to the 3/2 is equal to -- well
what's 3/2 times some constant c.
But we didn't even know what it was, we haven't solved for
it, so we can still write c.
I hope that doesn't confuse you.
3/2 c, we didn't know what it was.
We could call this c3 and now this is c4.
It's a different constant, but we still have to solve for it.
And now let's use our initial conditions.
So the initial conditions tell us that 4 -- and they don't
have to be initial conditions.
You can kind of say conditions or points where we know that
the particular solution to this differential equation
are satisfied.
So 4 to the 3/2 minus 1 to the 3/2 is equal to c.
What's 4 to the 3/2?
So 4 to the 1/2 is 2, and then that to the 3 power, that's 8.
Then 1 to any power, but especially in this case, 1 to
the square root of 1 is 1.
Then to the third power is 1.
So 8 minus 1 is equal to c.
And so c is equal to 7.
So the particular solution of this differential equation I
will do in a different color, and it is y to the 3/2 minus
x to the 3/2 is equal to 7.