Video Transcript
Find a relation between 𝑢 and 𝑡
given that d𝑢 by d𝑡 equals one plus 𝑡 to the fourth power over 𝑢𝑡 squared plus
𝑢 to the fourth power times 𝑡 squared.
Now, this differential equation
might look really nasty. But it is, in fact, a separable
differential equation. In this case, that’s one where an
expression for d𝑢 by d𝑡 can be written as some function of 𝑢 times some function
of 𝑡. So how are we going to achieve
that? Well, we begin by factoring the
denominator by 𝑡 squared. And we find that d𝑢 by d𝑡 is
equal to one plus 𝑡 to the fourth power over 𝑡 squared times 𝑢 plus 𝑢 to the
fourth power. We can now write this as some
function of 𝑡 times some function of 𝑢. It’s one plus 𝑡 to the fourth
power over 𝑡 squared times one over 𝑢 plus 𝑢 to the fourth power.
Let’s begin by multiplying both
sides of this equation by 𝑢 plus 𝑢 to the fourth power. Then, we recall that whilst d𝑢 by
d𝑡 is not a fraction, we treat it a little like one. And we can say that 𝑢 plus 𝑢 to
the fourth power d𝑢 is equal to one plus 𝑡 to the fourth power over 𝑡 squared
d𝑡. And that’s great because we’re now
ready to integrate both sides of our equation. The left-hand side is quite
straightforward to integrate. The integral of 𝑢 is 𝑢 squared
over two. And the integral of 𝑢 to the
fourth power is 𝑢 to the fifth power over five. Don’t forget, since this is an
indefinite integral, we need that constant of integration 𝑎.
On the right-hand side, we’re going
to rewrite our integrand as one over 𝑡 squared plus 𝑡 to the fourth power over 𝑡
squared. And that simplifies to one over 𝑡
squared plus 𝑡 squared. But, of course, one over 𝑡 squared
is the same as 𝑡 to the power of negative two. And we can now integrate as
normal. When we integrate 𝑡 to the power
of negative two, we add one to the exponent to get negative one and then we divide
by that number. 𝑡 squared becomes 𝑡 cubed over
three. And we have a second constant of
integration; I’ve called that 𝑏. Now, of course, we can rewrite 𝑡
to the power of negative one over negative one as negative one over 𝑡.
Our final step is to subtract our
constant 𝑎 from both sides of the equation. That gives us a new constant
𝑐. That’s just the difference between
𝑏 and 𝑎. So we’ve found a relation between
𝑢 and 𝑡. It’s 𝑢 to the fifth power over
five plus 𝑢 squared over two equals negative one over 𝑡 plus two cubed over three
plus 𝑐.
