### Video Transcript

Find the equation of the curve
which passes through the point negative eight, one given that the gradient of the
tangent at any point is equal to two times the square of the π¦-coordinate.

Letβs first interpret the
information given by this question. Weβre told about the gradient to
the tangent of a curve at any point. This gradient can be represented by
dπ¦ by dπ₯. Weβre told that this is equal to
two times the square of the π¦-coordinate. So in other words, is equal to two
π¦ squared. Here, we recognize that weβve been
given a differential equation. Not only that, weβve been given a
separable differential equation. This is one that can be expressed
in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of π¦. In our case, weβll be taking π of
π₯ to be equal to the constant two and π of π¦ to be equal to π¦ squared.

Given an equation of this form, we
can find the general solution in the following way. We first divide both sides of our
equation by π of π¦. In our case, this is π¦
squared. This gives us that one over π¦
squared dπ¦ by dπ₯ is equal to two. Next, we can integrate both sides
of our equation with respect to π₯. Now, the chain rule tells us that
dπ¦ by dπ₯ dπ₯ is equal to dπ¦. This gives us that the integral of
one over π¦ squared with respect to π¦ is equal to the integral of two with respect
to π₯. As a side note, you often see
methods where before the integration, dπ¦ by dπ₯ is treated somewhat like a
fraction. Both sides are multiplied by
dπ₯.

And the equivalent statement is
obtained: one over π¦ squared dπ¦ is equal to two dπ₯. Both sides of the equation are then
integrated. And we reach the same state as we
did before, but with less working. Always remember that this is just a
shorthand for the more rigorous steps that weβve shown here. Returning to our method, one over
π¦ squared can be expressed as π¦ to the power of negative two. You might be more comfortable
performing the integration in this form to get negative one over π¦ plus the
constant of integration π one is equal to two π₯ plus some other constant of
integration, π two.

Usually, we donβt worry about
putting a constant on both sides of this equation because we just define a new
constant, which in this case would be π two minus π one. And this could just go on one side
of the equation. Now, at this stage, we have found
the general solution to our differential equation. And itβs in implicit form, which
means itβs not in the form where π¦ is equal to some function of π₯. We found the general solution for
all lines where the gradient of the tangent at any point is equal to two times the
square of the π¦-coordinate.

Now, our next thought might be to
convert our implicit solution into an explicit solution. But as it turns out, this isnβt
necessary. This is because the general
solution is not our final destination. Rather, we are seeking a particular
solution. We want the line which passes
through the point negative eight, one. This information is an initial
value. And weβll be finding the particular
solution which matches in the following way. We know that if negative eight, one
lies on the line, when π₯ is equal to negative eight, π¦ must be equal to one. We can substitute these known
values into the equation for our general solution. We then perform some
simplification. And we find that our constant π
three is equal to 15.

We can now substitute this value
into our general solution which weβll turn it into a particular solution. Negative one over π¦ is equal to
two π₯ plus 15. Now, if we wanted, we could convert
this to an explicit solution first by multiplying by negative one and then raising
both sides to the power of negative one. To recap, we used the information
given in the question to form a separable differential equation. The particular solution to this
which passes through the point negative eight, one is π¦ equals negative one over
two π₯ plus 15. And this is the answer to our
question.