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

Solve the differential equation (dπ¦/dπ₯) + 3π₯Β² π¦ = 6π₯Β².

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### 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 doesnβt quite look like itβs in that form at the moment, so letβs 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 weβre 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; itβs 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¦ isnβt 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 πΆ. And in fact, itβs 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 πΆ. We really want to make π¦ the subject. So weβre 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 πΆ.

Now, in fact, π to the power of negative π₯ cubed plus πΆ canβt be negative. So we really donβt need these absolute value signs. And we can write π to the power of negative π₯ cubed plus πΆ as π to the power of π₯ cubed times π to the power of πΆ, which is a new constant πΆ one. By multiplying through again by negative one and then adding two to both sides, we find that π¦ is equal to two plus πΆ two π to the power of negative π₯ cubed. And thatβs because, once again, weβve done something to our constant. So itβs changed. And itβs usual to write our constant simply as πΆ. So we finish. Weβve solved our differential equation and we got π¦ equals two plus πΆ times π to the power of negative π₯ cubed.