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

Solve the differential equation (d𝑦/dπ‘₯) + 3π‘₯Β² 𝑦 = 6π‘₯Β².

03:18

Video Transcript

Solve the differential equation d𝑦 by dπ‘₯ plus three π‘₯ squared 𝑦 equals six π‘₯ squared.

This is a separable differential equation. This is so called because the function 𝑓 of π‘₯ and 𝑦 can be factored into the product of two functions of π‘₯ and 𝑦, that is, 𝑔 of π‘₯ times β„Ž of 𝑦, where 𝑔 of π‘₯ and β„Ž of 𝑦 are continuous functions. Now, it doesn’t quite look like it’s in that form at the moment, so let’s rearrange a little. We begin by subtracting three π‘₯ squared 𝑦 from both sides. And we get the equation d𝑦 by dπ‘₯ equals six π‘₯ squared minus three π‘₯ squared 𝑦. We want it to be of this form, some function of π‘₯ times some function of 𝑦.

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

Our next step is to integrate both sides of our equation with respect to their individual variables. So we’re going to integrate one over two minus 𝑦 with respect to 𝑦 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 𝐴.

But how do we integrate one over two minus 𝑦 with respect to 𝑦? Well, we introduce a new variable; it’s 𝑒. We let 𝑒 be equal to two minus 𝑦. Then we know that differentiating 𝑒 with respect to 𝑦 gives us negative one. d𝑒 by d𝑦 isn’t a fraction, but we treat it a little like one. And we rearrange to write negative d𝑒 equals d𝑦. And then the integral of one over two minus 𝑦 with respect to 𝑦 becomes the integral of one over 𝑒 negative d𝑒 or simply the integral of negative one over 𝑒 with respect to 𝑒.

Now, the integral of one over 𝑒 is the natural log of 𝑒. So we get the negative natural log of 𝑒 plus some constant of integration 𝐢. And in fact, it’s really the absolute value of this 𝑒. So we replace 𝑒 with two minus 𝑦. And we see that the negative natural log of the absolute value of two minus 𝑦 plus some constant of integration 𝐡 must be equal to three π‘₯ cubed over three plus 𝐴. And then that right-hand side simplifies to π‘₯ cubed plus 𝐴.

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 𝑦 is equal to negative π‘₯ cubed plus 𝐢. We really want to make 𝑦 the subject. So we’re going to raise both sides of our equation as a power of 𝑒. 𝑒 to the power of the natural log of the absolute value of two minus 𝑦 is the absolute value of two minus 𝑦. And on the right-hand side, we have 𝑒 to the power of negative π‘₯ cubed plus 𝐢.

Now, in fact, 𝑒 to the power of negative π‘₯ cubed plus 𝐢 can’t be negative. So we really don’t need these absolute value signs. And we can write 𝑒 to the power of negative π‘₯ cubed plus 𝐢 as 𝑒 to the power of π‘₯ cubed times 𝑒 to the power of 𝐢, which is a new constant 𝐢 one. By multiplying through again by negative one and then adding two to both sides, we find that 𝑦 is equal to two plus 𝐢 two 𝑒 to the power of negative π‘₯ cubed. And that’s because, once again, we’ve done something to our constant. So it’s changed. And it’s usual to write our constant simply as 𝐢. So we finish. We’ve solved our differential equation and we got 𝑦 equals two plus 𝐢 times 𝑒 to the power of negative π‘₯ cubed.

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