Question Video: Solving a Separable First Order Differential Equation given in the Normal Form Mathematics • Higher Education

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.