Video: Solving a Separable First Order Differential Equation given in the Normal Form

Find a relation between 𝑒 and 𝑑 given that d𝑒/d𝑑 = (1 + 𝑑⁴)/(𝑒𝑑² + 𝑒⁴ 𝑑²).

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Video Transcript

Find a relation between 𝑒 and 𝑑 given that d𝑒 by d𝑑 equals one plus 𝑑 to the fourth power over 𝑒𝑑 squared plus 𝑒 to the fourth power times 𝑑 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 d𝑒 by d𝑑 can be written as some function of 𝑒 times some function of 𝑑. So how are we going to achieve that? Well, we begin by factoring the denominator by 𝑑 squared. And we find that d𝑒 by d𝑑 is equal to one plus 𝑑 to the fourth power over 𝑑 squared times 𝑒 plus 𝑒 to the fourth power. We can now write this as some function of 𝑑 times some function of 𝑒. It’s one plus 𝑑 to the fourth power over 𝑑 squared times one over 𝑒 plus 𝑒 to the fourth power.

Let’s begin by multiplying both sides of this equation by 𝑒 plus 𝑒 to the fourth power. Then, we recall that whilst d𝑒 by d𝑑 is not a fraction, we treat it a little like one. And we can say that 𝑒 plus 𝑒 to the fourth power d𝑒 is equal to one plus 𝑑 to the fourth power over 𝑑 squared d𝑑. 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 𝑒 is 𝑒 squared over two. And the integral of 𝑒 to the fourth power is 𝑒 to the fifth power over five. Don’t forget, since this is an indefinite integral, we need that constant of integration π‘Ž.

On the right-hand side, we’re going to rewrite our integrand as one over 𝑑 squared plus 𝑑 to the fourth power over 𝑑 squared. And that simplifies to one over 𝑑 squared plus 𝑑 squared. But, of course, one over 𝑑 squared is the same as 𝑑 to the power of negative two. And we can now integrate as normal. When we integrate 𝑑 to the power of negative two, we add one to the exponent to get negative one and then we divide by that number. 𝑑 squared becomes 𝑑 cubed over three. And we have a second constant of integration; I’ve called that 𝑏. Now, of course, we can rewrite 𝑑 to the power of negative one over negative one as negative one over 𝑑.

Our final step is to subtract our constant π‘Ž from both sides of the equation. That gives us a new constant 𝑐. That’s just the difference between 𝑏 and π‘Ž. So we’ve found a relation between 𝑒 and 𝑑. It’s 𝑒 to the fifth power over five plus 𝑒 squared over two equals negative one over 𝑑 plus two cubed over three plus 𝑐.

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