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