Video: Solving a First-Order Separable Differential Equation

Find a relation between 𝑦 and π‘₯, given that π‘₯𝑦 𝑦′ = π‘₯Β² βˆ’ 5.

02:30

Video Transcript

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

Now, this first step isn’t entirely necessary. But it can make it a little easier to see what to do next. We write 𝑦 prime using Leibniz notation. And we see that π‘₯𝑦 d𝑦 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 d𝑦 by dπ‘₯ can be written as some function of π‘₯ times some function of 𝑦. Now, in fact, if we divide both sides of our equation by π‘₯𝑦, we see that we can achieve this. We get d𝑦 by dπ‘₯ equals π‘₯ squared minus five over π‘₯𝑦 or π‘₯ squared minus five over π‘₯ times one over 𝑦. And that’s great because 𝑔 of π‘₯ is, therefore, π‘₯ squared minus five over π‘₯. And our function of 𝑦 is one over 𝑦.

Now, in fact, we just performed this to demonstrate that we did indeed have a separable differential equation. We could’ve kept 𝑦 on the left-hand side as shown. And then, we perform this rather strange step. d𝑦 by dπ‘₯ isn’t a fraction, but we do treat it a little like one. And we say that 𝑦 d𝑦 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 𝑦 is 𝑦 squared over two plus some constant of integration π‘Ž.

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‘₯. It’s the natural log of the absolute value of π‘₯ plus 𝑐. 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 I’ve called 𝑏.

Our last step is to subtract π‘Ž from 𝑏 and then multiply the entire equation by two. When we do, we find that 𝑦 squared equals π‘₯ squared minus 10 times the natural log of the absolute value of π‘₯ plus 𝑐. This 𝑐 is a different constant achieved by subtracting π‘Ž from 𝑏 and then multiplying by two. And so, given our differential equation, we’ve found a relation between 𝑦 and π‘₯. 𝑦 squared equals π‘₯ squared minus 10 times the natural log of the absolute value of π‘₯ plus 𝑐.

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