Video: Solving First-Order First-Degree Linear Differential Equations

Solve the differential equation (d𝑦/dπ‘₯) + 𝑦 = 1.

03:47

Video Transcript

Solve the differential equation d𝑦 by dπ‘₯ plus 𝑦 equals one.

A separable equation is a first order differential equation in which the expression for d𝑦 by dπ‘₯ can be factored as a function of π‘₯ times a function of 𝑦. In other words, it can be written in the form 𝑔 of π‘₯ times 𝑓 of 𝑦. So let’s rearrange our equation d𝑦 by dπ‘₯ 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 dπ‘₯ 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 π‘₯ is simply one.

Then, to solve this equation, we’re going to rewrite it using differentials. Now, remember, d𝑦 by dπ‘₯ 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 dπ‘₯. 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 π‘₯ with respect to π‘₯. It’s the natural log of the absolute value of π‘₯ plus some constant of integration, 𝑐.

What we’re going to do is perform a substitution for our integral. We’re going to let 𝑒 be equal to one minus 𝑦 so that d𝑒 by d𝑦 is equal to negative one. We could write this equivalently as negative d𝑒 equals d𝑦. And then we’re going to replace d𝑦 with negative d𝑒 and one minus 𝑦 with 𝑒. And we see that we now need to integrate negative one over 𝑒 with respect to 𝑒. Well, that’s negative the natural log of the absolute value of 𝑒 plus that constant of integration 𝑐. By replacing 𝑒 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 π‘Ž.

When we integrate one with respect to π‘₯, it’s a little more straightforward. We get π‘₯ plus a second constant of integration. Let’s call that 𝑏. Let’s subtract π‘Ž 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 π‘₯ plus 𝑐 one. 𝑐 one is a new constant, and it’s achieved by subtracting π‘Ž 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 π‘₯ plus 𝑐 one.

By using the laws of exponents, though, we see that we can rewrite 𝑒 to the power of negative π‘₯ plus 𝑐 one as 𝑒 to the power of negative π‘₯ 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 π‘₯. 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 π‘₯ or negative 𝑐 two times 𝑒 to the power of negative π‘₯. 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 π‘₯. That gives us 𝑦 equals one minus 𝑐 two times 𝑒 to the power of negative π‘₯. 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 π‘₯ is the solution to our separable differential equation.

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