### Video Transcript

Find the function π, if π prime
of π‘ equals two sec π‘ times tan π‘ plus four sec π‘, when π‘ is greater than
negative π by two and less than π by two and π of negative π by three is
negative two.

In this question, we have a
differential equation. Remember, this is just an equation
involving derivatives. Here, thatβs π prime of π‘. And weβre also given an initial
value. That is, when π‘ is equal to
negative π by three, π of π‘ is equal to negative two. Now, weβre looking to work out what
the function π is. And so, we recall that integration
and differentiation are reverse processes. So our original function π of π‘
will be the integral of π prime of π‘ with respect to π‘. Thatβs the integral of two sec of
π‘ times tan π‘ plus four sec of π‘ with respect to π‘.

Now, this doesnβt look very
nice. Letβs first distribute the
parentheses of our integrand. When we do, we obtain our function
π of π‘ to be equal to the integral of two sec π‘ tan π‘ plus eight sec squared π‘
with respect to π‘. We can simplify this a little bit
by recalling that the integral of the sum of two functions is equal to the sum of
the integral of each of those functions. So π of π‘ is equal to the
integral of two sec π‘ tan π‘ with respect to π‘ plus the integral of eight sec
squared π‘ with respect to π‘.

Now, these might look really
nasty. But letβs recall some
derivatives. We know that the derivative of sec
of π‘ with respect to π‘ is tan π‘ sec of π‘. And we know that the derivative of
tan π‘ with respect to π‘ is sec squared π‘. This means the antiderivative of
tan of π‘ sec of π‘ must be sec of π‘. And the antiderivative of sec
squared π‘ must be tan π‘. And so, the integral of two sec of
π‘ tan π‘ must be two sec π‘. And, of course, itβs an indefinite
integral. So weβll add a constant of
integration π΄. Then the integral of eight sec
squared π‘ must be eight tan π‘. And weβll add a second constant of
integration π΅. Combining these two constants of
integration, we see that π of π‘ is equal to two sec π‘ plus eight tan π‘ plus
πΆ.

This is the general solution to our
differential equation. But we havenβt yet used the fact
that when π‘ is equal to negative π by three, our function is equal to negative
two. So letβs substitute these values
in. And when we do, we obtain negative
two to be equal to two sec of negative π by three plus eight tan of negative π by
three plus πΆ. If we recall that sec of negative
π by three is one over cos of negative π by three. Then two sec of negative π by
three must be two over cos of negative π by three, which gives us four. Then eight tan of negative π by
three is negative eight root three. So negative two is equal to four
minus eight root three plus πΆ.

Letβs subtract four from both sides
of this equation and add eight root three to both sides. And we see that πΆ is equal to
eight root three minus six. And weβve obtained the particular
solution to our differential equation to be π of π‘ equals two sec π‘ plus eight
tan π‘ plus eight root three minus six.