### Video Transcript

Solve the differential equation dπ¦ by dπ₯ plus three π₯ squared π¦ equals six π₯ squared.

This is a separable differential equation. This is so called because the function π of π₯ and π¦ can be factored into the product of two functions of π₯ and π¦, that is, π of π₯ times β of π¦, where π of π₯ and β of π¦ are continuous functions. Now, it doesnβt quite look like itβs in that form at the moment, so letβs rearrange a little. We begin by subtracting three π₯ squared π¦ from both sides. And we get the equation dπ¦ by dπ₯ equals six π₯ squared minus three π₯ squared π¦. We want it to be of this form, some function of π₯ times some function of π¦.

So we factor the right-hand side. The highest common factor of each term is three π₯ squared. And therefore, we get the equation dπ¦ by dπ₯ equals three π₯ squared times two minus π¦. And then this next step is a little bit strange. We consider our derivative as the ratio of two differentials, dπ¦ and dπ₯. We then move the dπ₯ to the right-hand side and divide both sides of our equation by two minus π¦. And we get one over two minus π¦ dπ¦ equals three π₯ squared dπ₯.

Our next step is to integrate both sides of our equation with respect to their individual variables. So weβre going to integrate one over two minus π¦ with respect to π¦ and three π₯ squared with respect to π₯. Now, the integral of three π₯ squared is fairly straightforward. We add one to the exponent and then divide by that new value. So we get three π₯ cubed over three plus some constant of integration π΄.

But how do we integrate one over two minus π¦ with respect to π¦? Well, we introduce a new variable; itβs π’. We let π’ be equal to two minus π¦. Then we know that differentiating π’ with respect to π¦ gives us negative one. dπ’ by dπ¦ isnβt a fraction, but we treat it a little like one. And we rearrange to write negative dπ’ equals dπ¦. And then the integral of one over two minus π¦ with respect to π¦ becomes the integral of one over π’ negative dπ’ or simply the integral of negative one over π’ with respect to π’.

Now, the integral of one over π’ is the natural log of π’. So we get the negative natural log of π’ plus some constant of integration πΆ. And in fact, itβs really the absolute value of this π’. So we replace π’ with two minus π¦. And we see that the negative natural log of the absolute value of two minus π¦ plus some constant of integration π΅ must be equal to three π₯ cubed over three plus π΄. And then that right-hand side simplifies to π₯ cubed plus π΄.

Next, we multiply through by negative one and combine our constants. And we find that the natural log of the absolute value of two minus π¦ is equal to negative π₯ cubed plus πΆ. We really want to make π¦ the subject. So weβre going to raise both sides of our equation as a power of π. π to the power of the natural log of the absolute value of two minus π¦ is the absolute value of two minus π¦. And on the right-hand side, we have π to the power of negative π₯ cubed plus πΆ.

Now, in fact, π to the power of negative π₯ cubed plus πΆ canβt be negative. So we really donβt need these absolute value signs. And we can write π to the power of negative π₯ cubed plus πΆ as π to the power of π₯ cubed times π to the power of πΆ, which is a new constant πΆ one. By multiplying through again by negative one and then adding two to both sides, we find that π¦ is equal to two plus πΆ two π to the power of negative π₯ cubed. And thatβs because, once again, weβve done something to our constant. So itβs changed. And itβs usual to write our constant simply as πΆ. So we finish. Weβve solved our differential equation and we got π¦ equals two plus πΆ times π to the power of negative π₯ cubed.