Video: Solving a Separable First-Order Differential Equation Given in the Normal Form

Solve the differential equation d𝑦/dπ‘₯ = βˆ’5π‘₯²𝑦².

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Video Transcript

Solve the differential equation d𝑦 by dπ‘₯ equals negative five π‘₯ squared 𝑦 squared.

This is a separable differential equation. Remember, these are called separable because the function can be factored into the product of two functions of π‘₯ and 𝑦. That is, our function in π‘₯ and 𝑦 is some function of π‘₯ times some other function of 𝑦, where 𝑔 and β„Ž here must be continuous functions. Now, to begin solving this, we consider the derivative as the ratio of two differentials, d𝑦 over dπ‘₯. This means we can move dπ‘₯ to the right-hand side and then divide by our function in 𝑦 such that one over 𝑦 squared d𝑦 equals negative five π‘₯ squared dπ‘₯.

Now, of course, to be able to do this, we need to assume that 𝑦 squared is not equal to zero. But if there is a number such that 𝑦 squared is equal to zero, then this number will also be a solution of the differential equation. What happens is when we divide by this, we lose that solution. Now, in fact, if 𝑦 is equal to zero, 𝑦 squared is equal to zero. And so we need to include the solution 𝑦 equals zero in our final answer. And this is because d𝑦 by dπ‘₯ is equal to zero for a constant. And that gives zero equals zero in our differential equation.

Now, let’s write one over 𝑦 squared as 𝑦 to the power of negative two. And now, we’re going to integrate both sides with respect to their variables. When we integrate 𝑦 to the negative two, we add one to the exponent and then divide by that new value. So it’s 𝑦 to the power of negative one over negative one plus some constant of integration 𝐴. Similarly, when we integrate negative five π‘₯ squared, we get negative five π‘₯ cubed over three plus a second constant of integration 𝐡.

Now, we’re going to multiply through by negative one and combine our constants. And we get 𝑦 to the power of negative one equals five π‘₯ cubed over three plus 𝐢. But of course, we can write 𝑦 to the negative one as one over 𝑦. We really want 𝑦 to be the subject. So we’re going to multiply through by 𝑦 so that one equals 𝑦 times five π‘₯ cubed over three plus 𝐢. And then, we’re going to divide through by five π‘₯ cubed over three plus 𝐢.

Now, in fact, it’s a little bit simpler to create a common denominator of three and divide through by five π‘₯ cubed plus 𝐷 over three, where 𝐷 is a new constant. Because when we divide by a fraction, we simply multiply by the reciprocal of that fraction. So one divided by five π‘₯ cubed plus 𝐷 over three is one times three over five π‘₯ cubed plus 𝐷, which is simply three over five π‘₯ cubed plus 𝐷.

And so, we have our two solutions. It’s usual to define any constants by 𝐢. So we know the two solutions to our differential equation are 𝑦 equals three over five π‘₯ cubed plus 𝐢 or 𝑦 equals zero.

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