# Question Video: Solving First-Order First-Degree Linear Differential Equations Mathematics • Higher Education

Solve the differential equation (dπ¦/dπ₯) + π¦ = 1.

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### Video Transcript

Solve the differential equation dπ¦ by dπ₯ plus π¦ equals one.

A separable equation is a first order differential equation in which the expression for dπ¦ by dπ₯ can be factored as a function of π₯ times a function of π¦. In other words, it can be written in the form π of π₯ times π of π¦. So letβs rearrange our equation dπ¦ by dπ₯ plus π¦ equals one so that itβs in this form. To achieve this, weβre going to subtract π¦ from both sides of the equation. And we obtain dπ¦ by dπ₯ to be equal to one minus π¦. Now, it may not look like it, but we have achieved our aim. Our function in π¦ is one minus π¦, and our function in π₯ is simply one.

Then, to solve this equation, weβre going to rewrite it using differentials. Now, remember, dπ¦ by dπ₯ absolutely isnβ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 π¦. And we see that this is equivalent to saying one over one minus π¦ dπ¦ equals one dπ₯. And now, weβre ready to integrate both sides of this equation. So how do we integrate one over one minus π¦ with respect to π¦? Well, we begin by quoting the general result for the integral of one over π₯ with respect to π₯. Itβs the natural log of the absolute value of π₯ plus some constant of integration, π.

What weβre going to do is perform a substitution for our integral. Weβre going to let π’ be equal to one minus π¦ so that dπ’ by dπ¦ is equal to negative one. We could write this equivalently as negative dπ’ equals dπ¦. And then weβre going to replace dπ¦ with negative dπ’ and one minus π¦ with π’. And we see that we now need to integrate negative one over π’ with respect to π’. Well, thatβs negative the natural log of the absolute value of π’ plus that constant of integration π. By replacing π’ with one minus π¦, we find the integral of one over one minus π¦ to be the negative natural log of the absolute value of one minus π¦ plus a constant of integration which Iβm going to call π.

When we integrate one with respect to π₯, itβs a little more straightforward. We get π₯ plus a second constant of integration. Letβs call that π. Letβs subtract π 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 π¦ equals negative π₯ plus π one. π one is a new constant, and itβs achieved by subtracting π from π and multiplying by negative one. And then, we notice that we can raise both sides of this equation as a power of π. So we obtain the absolute value of one minus π¦ to be equal to π to the power of negative π₯ plus π one.

By using the laws of exponents, though, we see that we can rewrite π to the power of negative π₯ plus π one as π to the power of negative π₯ times π to the power of π one. But, of course, π to the power of π one is itself a constant. Letβs call that π two. And we see we can rewrite the right-hand side as π two times π to the power of negative π₯. Well, of course, weβre dealing with the absolute value of one minus π¦. And we can say that this means that one minus π¦ could be equal to positive π two times π to the power of negative π₯ or negative π two times π to the power of negative π₯. But π two is a constant, so we donβt actually need to write that.

In our final step, weβre going to add π¦ to both sides of the equation and then subtract π two π to the power of negative π₯. That gives us π¦ equals one minus π two times π to the power of negative π₯. But of course, once again, since π two is a constant, we can change this to plus π. And we see that π¦ equals one plus π times π to the power of negative π₯ is the solution to our separable differential equation.