### Video Transcript

Find the equation of the curve that
passes through the point zero, negative one given dπ¦ by dπ₯ is equal to negative
six π₯ minus four divided by four π¦ plus 13.

For this question, we first noticed
that we have been given a separable differential equation. This is an equation that can be
written in the form dπ¦ by dπ₯ is equal to π of π₯ multiplied by π of π¦. Solving this separable differential
equation will give us a general solution, which will include a constant π. Now, the question has also given us
an initial value, which is the fact that the curve should pass through the point
zero, negative one. Weβll be able to use this
information to narrow down the general solution that we find into a specific
solution. For now, letβs just work on the
general solution.

For now, weβll put aside our
initial value and just work towards the general solution. To find this, weβll first express
our differential equation in the following form, where we have some function of π¦
dπ¦ by dπ₯ is equal to π of π₯. For our equation, we can do this by
multiplying both sides by four π¦ plus 13. This gives us that four π¦ plus 13
dπ¦ by dπ₯ is equal to negative six π₯ minus four. The next thing we can do is to
treat dπ¦ by dπ₯ somewhat like a fraction. If we pretend weβre multiplying
both sides by dπ₯, we obtain an equivalent statement four π¦ plus 13 dπ¦ is equal to
negative six π₯ minus four dπ₯.

The next step we take is to then
integrate both sides of our equation, giving us an integral with respect to π¦ on
the left-hand side and an integral with respect to π₯ on the right-hand side. Now, you should always remember
that what we have done here is essentially a trick. dπ¦ by dπ₯ is not actually a
fraction and cannot always be treated like one. However, weβve done so here to save
us from working since it gives us an equivalent answer. Going back to our method, we now
perform our integrations. Doing so gives us the
following.

Now, itβs worth noting that both of
our integrals would actually have given us a constant. However, we can simply combine
these two constants into one term. Instead of π one and π two, we
simply have a π, which is equal to π one minus π two. Again, this leads to less
unnecessary work for us. After some small simplifications,
we have arrived at the general solution to our differential equation. This means that we have found an
equation which describes all the lines for which dπ¦ by dπ₯ is equal to negative six
π₯ minus four divided by four π¦ plus 13.

However, we need to find which one
of these lines passes through the point that weβve been given: zero, negative
one. This initial value that we have
been given will allow us to find that particular solution. We know that if the line passes
through the point zero, negative one that when π₯ is equal to zero, π¦ must be equal
to negative one. We can, therefore, take our general
solution, substitute in π¦ equals negative one and π₯ equals zero, and solve this to
find the value of our constant π. With a little bit of working, we
find that the value of π here is negative 11. We can now substitute this value of
π back into our general solution to find the specific solution. This is the specific solution to
the differential equation which passes through the point zero, negative one, which
is our initial value.

Now, you might notice that the
solution that we have been given is implicit, which means itβs not in the form π¦
equals some function of only π₯. Usually, itβs a good idea to try
and express your solution in explicit form. However, in this case, you should
not worry. It is not actually possible to give
the answer to this problem in the form of a single explicit equation since we have a
multiple powers of π¦ on the left-hand side. Again, this is no cause for
concern, since the implicit form is a perfectly valid solution. And we have indeed answered our
question.