Video: Solving a First-Order Separable Differential Equation

Find a relation between π¦ and π₯, given that π₯π¦ π¦β² = π₯Β² β 5.

02:30

Video Transcript

Find a relation between π¦ and π₯ given that π₯π¦ times π¦ prime is equal to π₯ squared minus five.

Now, this first step isnβt entirely necessary. But it can make it a little easier to see what to do next. We write π¦ prime using Leibniz notation. And we see that π₯π¦ dπ¦ by dπ₯ is equal to π₯ squared minus five. And then we recall that a separable differential equation is one in which the expression for dπ¦ by dπ₯ can be written as some function of π₯ times some function of π¦. Now, in fact, if we divide both sides of our equation by π₯π¦, we see that we can achieve this. We get dπ¦ by dπ₯ equals π₯ squared minus five over π₯π¦ or π₯ squared minus five over π₯ times one over π¦. And thatβs great because π of π₯ is, therefore, π₯ squared minus five over π₯. And our function of π¦ is one over π¦.

Now, in fact, we just performed this to demonstrate that we did indeed have a separable differential equation. We couldβve kept π¦ on the left-hand side as shown. And then, we perform this rather strange step. dπ¦ by dπ₯ isnβt a fraction, but we do treat it a little like one. And we say that π¦ dπ¦ equals π₯ squared minus five over π₯ dπ₯. Our next step is to integrate both sides of this equation. Remember, to integrate a polynomial term whose exponent is not equal to negative one, we add one to the exponent and divide by that new number. So the integral of π¦ is π¦ squared over two plus some constant of integration π.

Then, it might look like we need to perform some sort of substitution to evaluate this integral. But actually, if we separate our fraction into π₯ squared over π₯ minus five over π₯, we find that we need to integrate π₯ minus five over π₯ with respect to π₯. Then, we recall the general result for the integral of one over π₯. Itβs the natural log of the absolute value of π₯ plus π. And so, when we integrate the right-hand side of our equation, we get π₯ squared over two minus five times the natural log of the absolute value of π₯ plus some second constant of integration which Iβve called π.

Our last step is to subtract π from π and then multiply the entire equation by two. When we do, we find that π¦ squared equals π₯ squared minus 10 times the natural log of the absolute value of π₯ plus π. This π is a different constant achieved by subtracting π from π and then multiplying by two. And so, given our differential equation, weβve found a relation between π¦ and π₯. π¦ squared equals π₯ squared minus 10 times the natural log of the absolute value of π₯ plus π.