### Video Transcript

Solve the differential equation the
derivative of π with respect to π‘ is equal to π‘ multiplied by the sec of π
divided by π multiplied by π to the power of π‘ squared.

The first thing we can try is to
see if weβve been given a separable differential equation. And we recall that we call a
differential equation separable if the derivative of π with respect to π‘ could be
written as a function of π multiplied by a function of π‘. And we see that we can separate our
differential equation. We get the first function of the
sec of π divided by π. And then this is multiplied by π‘
divided by π to the power of π‘ squared.

So since we have a separable
differential equation, weβll divide both sides of our equation by π of π. And this is the same as multiplying
by π divided by the sec of π. This gives us π divided by the sec
of π multiplied by the derivative of π with respect to π‘ is equal to π‘ divided
by π to the power of π‘ squared.

Itβs worth reiterating at this
point that the derivative of π with respect to π‘ is not a fraction. However, when weβre solving
separable differential equations, we can treat it a little bit like a fraction. So using this gives us the
equivalent statement that π divided by the sec of π dπ is equal to π‘ divided by
π to the power of π‘ squared dπ‘.

Finally, we integrate both sides of
this equation to get a new expression. We can simplify our integrand of π
divided by the sec of π by recalling that the sec of π is equal to one divided by
the cos of π. And we can simplify the integrand
in our second integral by recalling that dividing by π to the power of π‘ squared
is the same as multiplying by π to the power of negative π‘ squared.

This gives us two integrals to
solve. We can solve the first integral of
π multiplied by the cos of π with respect to π by using integration by parts. We recall that integration by parts
tells us that the integral of π’ multiplied by π£ prime with respect to π is equal
to π’ multiplied by π£ minus the integral of π’ prime multiplied by π£ with respect
to π.

So to evaluate the integral of π
multiplied by the cos of π with respect to π, weβll set π’ equal to π and π£
prime equal to the cos of π. Differentiating π’ with respect to
π gives us that π’ prime is equal to one. And if we have that π£ prime is
equal to the cos of π, then our antiderivative π£ is equal to the sin of π. So substituting this into our
integration-by-parts formula gives us the integral of π multiplied by the cos of π
with respect to π is equal to π’ multiplied by π£, which is π sin π, minus the
integral of π’ prime multiplied by π£, which is one minus sin π, with respect to
π.

Finally, we know that the integral
of sin π with respect to π is equal to negative cos of π. So weβve shown that the integral of
π cos of π with respect to π is equal to π sin π plus cos of π plus our
constant of integration π one.

Now weβve evaluated the first of
our two integrals. This gives us that π sin π plus
the cos of π plus our constant of integration π one is equal to the integral of π‘
multiplied by π to the power of negative π‘ squared with respect to π‘.

Now we have to evaluate the
integral of π‘ multiplied by π to the power of negative π‘ squared with respect to
π‘. And there are several ways of doing
this. For example, we could use the
substitution π’ is equal to π‘ squared. However, thereβs a slightly simpler
way of doing this. If we were to try to differentiate
π to the power of negative π‘ squared with respect to π‘. Well, we recall by the chain rule
the derivative of π to the power of π of π₯ with respect to π₯ is equal to π
prime of π₯ multiplied by π to the power of π of π₯.

So to differentiate π to the power
of negative π‘ squared with respect to π‘, we differentiate negative π‘ squared to
get negative two π‘. And then we multiply this by π to
the power of negative π‘ squared. And we can now see that this is
very similar to our integrand. In fact, itβs only different by a
coefficient of negative a half. So this tells us that the
derivative with respect to π‘ of negative a half multiplied by π to the power of
negative π‘ squared is equal to negative a half multiplied by negative two π‘
multiplied by π to the power of negative π‘ squared.

We can then cancel the coefficient
of negative a half with the coefficient of negative two to just get π‘ multiplied by
π to the power of negative π‘ squared. And we can see that this is exactly
equal to our integrand.

So what we have shown is that
negative a half multiplied by π to of the power of negative π‘ squared is our
antiderivative. So the integral of π‘ multiplied by
π to the power of negative π‘ squared with respect to π‘ is equal to negative a
half multiplied by π to the power of negative π‘ squared plus a constant of
integration π two. And since π one and π two are
both constants, we can combine both of these into a new constant, which we will call
π.

Therefore, weβve shown that the
solution to the differential equation given to us in the question is π sin π plus
the cos of π is equal to negative π to the power of negative π‘ squared divided by
two plus a constant of integration π.