- [Instructor] Welcome to a lesson on exact first-order
differential equations.
A differential equation M of x comma y d x plus
N of x comma y d y equals zero, or the equation
in this form here is an exact differential equation
in a region R in the xy-plane if the left side
of the equation, this side here, corresponds to
the differential of some function f of x comma y.
And remember, if our function f of x comma y equals
a constant c, then its differential here must equal zero.
So the partial derivative of f with respects to x
must equal N of x comma y and the partial derivative
of f with respects to y must equal N of x comma y.
And we can use this information to find our function
f of x comma y and therefore solve
this differential equation if it's exact.
So the first step on these types of problem
is to determine if the given differential equation
is exact, the test for exactness is to see if
the partial derivative of M with respects to y
is equal to the partial derivative of N with respects to x.
So if this is true, we have an exact differential equation
and therefore solve it by finding f of x comma y equals c
by using this connection here.
So to summarize this, if we know the differential equation
is exact, then we can find the function f of x comma y
by solving these two equations here.
Let's take a look at an example.
Our first step is to identify our function M,
which in this case will be two x y,
and our function N, which will be x squared minus one.
And now we'll test to see if this is an exact
differential equation by finding the partial derivative of M
with respects to y and the partial derivative of N
with respects to x.
If these are equal, then the differential equation is exact.
So treating x as a constant here, the derivative
with respects to y would just be two x
and the derivative of x squared minus one with respects to x
would also be two x.
So this means we do have an exact differential equation,
which means our solution will be our function f of x comma y
equals a constant c and we can find our function
f of x comma y using these two equations here.
So if we know the partial derivative of f
with respects to x, it must equal M of x comma y.
Then if we integrate both sides of this equation
with respects to x, we'll recover the x part
of our function f of x comma y,
which means f of x comma y
must be equal to the integral of M of x comma y,
which is two x y
with respects to x.
Now keep in mind when we integrate here,
because our function f is a function of two variables
we'll only be recovering the x part of this function.
So f of x comma y is going to be equal to integrating
with respects to x we would have two times x squared
divided by two, so the twos would simplify out,
so we'd have x squared y,
but then to show we've only recovered the x part
of the function, not the y part, we'd have plus
a function of y, let's call it h of y.
So now if we can find h of y, we'll have f of x comma y
and therefore we can find our solution
to the differential equation.
So now we'll use this second equation here
where we know the partial derivative of f with respects to y
must equal N of x comma y.
So what we'll do now is we'll find the derivative of f
with respects to y and set it equal to N of x comma y
which we know is x squared minus one.
So here's our function f and then we'll find
the derivative with respects to y so we'll treat
x as a constant so here we would have x squared
plus the derivative of h of y, which would just be
h prime of y, and then again this must equal N,
which is x squared minus one.
So from here we should be able to conclude that
since this is x squared and this is x squared,
h prime of y must equal negative one.
So now we can recover h of y if we integrate
both sides of this equation with respects to y.
So we'll integrate here and here with respects to y.
So here the integration and derivative will undo each other,
so we have h of y must equal, this would just be negative y.
Now we have all the information we need to find the solution
to this differential equation,
by using this information here and knowing h of y
is equal to negative y.
So if our solution is going to be in this form here,
then we know that x squared y
and then plus h of y where h of y is negative y
or minus y, equals c, would be the general solution
to the given differential equation, which is
an exact differential equation.
So this is a family of solutions and now we can actually
show some of the solutions by selecting constants for c.
And to show this graphically, I've selected
the following values of c and then graphed the functions.
So each color here corresponds to one of the many solutions
to the exact differential equation
and we'll take a look at a second example in the next video.