Video Transcript
Solve the differential equation d𝑦
by d𝑥 plus 𝑦 equals one.
A separable equation is a first
order differential equation in which the expression for d𝑦 by d𝑥 can be factored
as a function of 𝑥 times a function of 𝑦. In other words, it can be written
in the form 𝑔 of 𝑥 times 𝑓 of 𝑦. So let’s rearrange our equation d𝑦
by d𝑥 plus 𝑦 equals one so that it’s in this form. To achieve this, we’re going to
subtract 𝑦 from both sides of the equation. And we obtain d𝑦 by d𝑥 to be
equal to one minus 𝑦. Now, it may not look like it, but
we have achieved our aim. Our function in 𝑦 is one minus 𝑦,
and our function in 𝑥 is simply one.
Then, to solve this equation, we’re
going to rewrite it using differentials. Now, remember, d𝑦 by d𝑥
absolutely isn’t a fraction. But we do treat it a little like
one for the purposes of this process. We begin by dividing both sides of
our equation by one minus 𝑦. And we see that this is equivalent
to saying one over one minus 𝑦 d𝑦 equals one d𝑥. And now, we’re ready to integrate
both sides of this equation. So how do we integrate one over one
minus 𝑦 with respect to 𝑦? Well, we begin by quoting the
general result for the integral of one over 𝑥 with respect to 𝑥. It’s the natural log of the
absolute value of 𝑥 plus some constant of integration, 𝑐.
What we’re going to do is perform a
substitution for our integral. We’re going to let 𝑢 be equal to
one minus 𝑦 so that d𝑢 by d𝑦 is equal to negative one. We could write this equivalently as
negative d𝑢 equals d𝑦. And then we’re going to replace d𝑦
with negative d𝑢 and one minus 𝑦 with 𝑢. And we see that we now need to
integrate negative one over 𝑢 with respect to 𝑢. Well, that’s negative the natural
log of the absolute value of 𝑢 plus that constant of integration 𝑐. By replacing 𝑢 with one minus 𝑦,
we find the integral of one over one minus 𝑦 to be the negative natural log of the
absolute value of one minus 𝑦 plus a constant of integration which I’m going to
call 𝑎.
When we integrate one with respect
to 𝑥, it’s a little more straightforward. We get 𝑥 plus a second constant of
integration. Let’s call that 𝑏. Let’s subtract 𝑎 from both sides
of our equation and multiply through by negative one. That gives us the natural log of
the absolute value of one minus 𝑦 equals negative 𝑥 plus 𝑐 one. 𝑐 one is a new constant, and it’s
achieved by subtracting 𝑎 from 𝑏 and multiplying by negative one. And then, we notice that we can
raise both sides of this equation as a power of 𝑒. So we obtain the absolute value of
one minus 𝑦 to be equal to 𝑒 to the power of negative 𝑥 plus 𝑐 one.
By using the laws of exponents,
though, we see that we can rewrite 𝑒 to the power of negative 𝑥 plus 𝑐 one as 𝑒
to the power of negative 𝑥 times 𝑒 to the power of 𝑐 one. But, of course, 𝑒 to the power of
𝑐 one is itself a constant. Let’s call that 𝑐 two. And we see we can rewrite the
right-hand side as 𝑐 two times 𝑒 to the power of negative 𝑥. Well, of course, we’re dealing with
the absolute value of one minus 𝑦. And we can say that this means that
one minus 𝑦 could be equal to positive 𝑐 two times 𝑒 to the power of negative 𝑥
or negative 𝑐 two times 𝑒 to the power of negative 𝑥. But 𝑐 two is a constant, so we
don’t actually need to write that.
In our final step, we’re going to
add 𝑦 to both sides of the equation and then subtract 𝑐 two 𝑒 to the power of
negative 𝑥. That gives us 𝑦 equals one minus
𝑐 two times 𝑒 to the power of negative 𝑥. But of course, once again, since 𝑐
two is a constant, we can change this to plus 𝑐. And we see that 𝑦 equals one plus
𝑐 times 𝑒 to the power of negative 𝑥 is the solution to our separable
differential equation.
