Video Transcript
Solve the differential equation the
derivative of 𝜃 with respect to 𝑡 is equal to 𝑡 multiplied by the sec of 𝜃
divided by 𝜃 multiplied by 𝑒 to the power of 𝑡 squared.
The first thing we can try is to
see if we’ve been given a separable differential equation. And we recall that we call a
differential equation separable if the derivative of 𝜃 with respect to 𝑡 could be
written as a function of 𝜃 multiplied by a function of 𝑡. And we see that we can separate our
differential equation. We get the first function of the
sec of 𝜃 divided by 𝜃. And then this is multiplied by 𝑡
divided by 𝑒 to the power of 𝑡 squared.
So since we have a separable
differential equation, we’ll divide both sides of our equation by 𝑓 of 𝜃. And this is the same as multiplying
by 𝜃 divided by the sec of 𝜃. This gives us 𝜃 divided by the sec
of 𝜃 multiplied by the derivative of 𝜃 with respect to 𝑡 is equal to 𝑡 divided
by 𝑒 to the power of 𝑡 squared.
It’s worth reiterating at this
point that the derivative of 𝜃 with respect to 𝑡 is not a fraction. However, when we’re solving
separable differential equations, we can treat it a little bit like a fraction. So using this gives us the
equivalent statement that 𝜃 divided by the sec of 𝜃 d𝜃 is equal to 𝑡 divided by
𝑒 to the power of 𝑡 squared d𝑡.
Finally, we integrate both sides of
this equation to get a new expression. We can simplify our integrand of 𝜃
divided by the sec of 𝜃 by recalling that the sec of 𝜃 is equal to one divided by
the cos of 𝜃. And we can simplify the integrand
in our second integral by recalling that dividing by 𝑒 to the power of 𝑡 squared
is the same as multiplying by 𝑒 to the power of negative 𝑡 squared.
This gives us two integrals to
solve. We can solve the first integral of
𝜃 multiplied by the cos of 𝜃 with respect to 𝜃 by using integration by parts. We recall that integration by parts
tells us that the integral of 𝑢 multiplied by 𝑣 prime with respect to 𝜃 is equal
to 𝑢 multiplied by 𝑣 minus the integral of 𝑢 prime multiplied by 𝑣 with respect
to 𝜃.
So to evaluate the integral of 𝜃
multiplied by the cos of 𝜃 with respect to 𝜃, we’ll set 𝑢 equal to 𝜃 and 𝑣
prime equal to the cos of 𝜃. Differentiating 𝑢 with respect to
𝜃 gives us that 𝑢 prime is equal to one. And if we have that 𝑣 prime is
equal to the cos of 𝜃, then our antiderivative 𝑣 is equal to the sin of 𝜃. So substituting this into our
integration-by-parts formula gives us the integral of 𝜃 multiplied by the cos of 𝜃
with respect to 𝜃 is equal to 𝑢 multiplied by 𝑣, which is 𝜃 sin 𝜃, minus the
integral of 𝑢 prime multiplied by 𝑣, which is one minus sin 𝜃, with respect to
𝜃.
Finally, we know that the integral
of sin 𝜃 with respect to 𝜃 is equal to negative cos of 𝜃. So we’ve shown that the integral of
𝜃 cos of 𝜃 with respect to 𝜃 is equal to 𝜃 sin 𝜃 plus cos of 𝜃 plus our
constant of integration 𝑐 one.
Now we’ve evaluated the first of
our two integrals. This gives us that 𝜃 sin 𝜃 plus
the cos of 𝜃 plus our constant of integration 𝑐 one is equal to the integral of 𝑡
multiplied by 𝑒 to the power of negative 𝑡 squared with respect to 𝑡.
Now we have to evaluate the
integral of 𝑡 multiplied by 𝑒 to the power of negative 𝑡 squared with respect to
𝑡. And there are several ways of doing
this. For example, we could use the
substitution 𝑢 is equal to 𝑡 squared. However, there’s a slightly simpler
way of doing this. If we were to try to differentiate
𝑒 to the power of negative 𝑡 squared with respect to 𝑡. Well, we recall by the chain rule
the derivative of 𝑒 to the power of 𝑓 of 𝑥 with respect to 𝑥 is equal to 𝑓
prime of 𝑥 multiplied by 𝑒 to the power of 𝑓 of 𝑥.
So to differentiate 𝑒 to the power
of negative 𝑡 squared with respect to 𝑡, we differentiate negative 𝑡 squared to
get negative two 𝑡. And then we multiply this by 𝑒 to
the power of negative 𝑡 squared. And we can now see that this is
very similar to our integrand. In fact, it’s only different by a
coefficient of negative a half. So this tells us that the
derivative with respect to 𝑡 of negative a half multiplied by 𝑒 to the power of
negative 𝑡 squared is equal to negative a half multiplied by negative two 𝑡
multiplied by 𝑒 to the power of negative 𝑡 squared.
We can then cancel the coefficient
of negative a half with the coefficient of negative two to just get 𝑡 multiplied by
𝑒 to the power of negative 𝑡 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 𝑒 to of the power of negative 𝑡 squared is our
antiderivative. So the integral of 𝑡 multiplied by
𝑒 to the power of negative 𝑡 squared with respect to 𝑡 is equal to negative a
half multiplied by 𝑒 to the power of negative 𝑡 squared plus a constant of
integration 𝑐 two. And since 𝑐 one and 𝑐 two are
both constants, we can combine both of these into a new constant, which we will call
𝑐.
Therefore, we’ve shown that the
solution to the differential equation given to us in the question is 𝜃 sin 𝜃 plus
the cos of 𝜃 is equal to negative 𝑒 to the power of negative 𝑡 squared divided by
two plus a constant of integration 𝑐.
