PROFESSOR: Welcome to this session on separable equations.
So in this problem, you're asked in the first question
to solve the initial value problem
dy/dx equals y square with the initial condition y of zero
equals 1.
In the second part of the problem,
you're asked to find the general solution
where no initial condition is imposed.
So here you need to remember your method of separation
of variables to tackle the first question.
And then in the second part of the problem,
remember all the types of solutions and conditions
that you applied the first part to recover the lost solutions.
So why don't you take a minute, pause the video,
work through questions a and b.
And then we'll continue together when I come back.
Welcome back.
So in the first part of the problem,
we'll be solving the equation dy/dx equals y squared.
So here the method of separation of variables
tells us that we should regroup the variables of the same kind,
so all the y variables on one side and dx variable
on the other side of the equation
and then integrate from this point.
So here for this step, notice that I divided by y
squared, which means that we need
to impose the condition y not equal to zero from now on.
So from this step, we just use indefinite integrals
to integrate both sides of the equation.
So the left-hand side is the integral dy over y squared.
So integral of this gives us minus 1 over y.
And the right-hand side, integral dx, is just x.
Both sides would give us constant of integrations,
but we only need one because this
is a first-order differential equation.
And so we group them together on the right-hand side
with constant c.
So from this point, given that we're interested in variable y,
we need just to invert the expression.
And that gives us partial answer,
which is y of x equals minus 1 over x plus c.
So now we need to use our initial condition, y of zero
equals to 1, to determine the value
of c for this particular initial value problem.
So our initial condition was y of zero equals 1.
So if we substitute this in the expression
that we just obtained, we just have
zero plus c, which then only gives us c equal to 1.
So we end up with the value for our constant of integration,
c equals minus 1.
And so the solution to this problem is
y of x equals 1 over 1 minus x.
So if you examine this expression,
you see right away that we have a problem for x
equals to 1 because at x equals to 1,
we have 1 over 0 which means that, then, the solution blows
up and we have a vertical asymptote.
So let me draw this here.
So we're going to have an asymptote at x equals
1 and a solution that passes through our initial condition,
y equals 1, going to infinity when approaching x equals 1.
But then on the right side of the value x equals 1,
we have another part of the solution that goes
to zero as x goes to infinity.
And that diverges to minus infinity when x approaches 1.
So by convention, the solutions of differential equations
are defined on one single interval.
So we need here to realize that the solution we had
is basically two parts, the parts
on the left of the asymptote and the part
on the right of the asymptote.
So the solution to our initial value
problems needs to be the solution that
passes through the imposed initial condition, which
was y of zero equals to 1.
So it needs to be this solution.
So now, if we move on to the solution of the second part
of the problem, b, we were asked to find the general solution
of the problem, which means that we need to account now
for all the solutions, regardless
of their initial condition.
So we already answered this partially
during the solution of part a where
we solved using indefinite integrals
and arrived to the solution minus 1
over x plus c, where here we had,
basically, an undetermined constant of integration.
So this is one general solution.
But remember that we need to give all
the solutions of the problem.
So when we arrived at the solution,
we excluded the solution y equals
zero, which was basically a lost solution because we
had to impose the condition y not equal to zero.
So we need, when we give the general solution
to this differential equation, to recover the lost solution.
And then we basically have one kind of solution, minus 1
over x plus c, that excludes y equals to zero
and another kind of solution that
is simply the zero solution.
So to summarize, the important points
of this problem is to remember the separation of variables
and how to use it and the fact that using it
imposes conditions that require us to recover lost solutions
at the end of the problem if we're
asked to give general solution.
Another point to remember is that even a simple ODE
can lead to relatively complex behavior which
drives the presence of this vertical asymptote
that you need to then know how to deal with and determine
which part of the solutions that you've obtained
is the real solution to the initial value problem
that you're given.
So I hope that you are OK with this problem
and you will use these approaches many times
for the rest of the course.