### Video Transcript

Solve the differential equation dπ¦ by dπ₯ equals negative five π₯ squared π¦ 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 π¦. That is, our function in π₯ and π¦ is some function of π₯ times some other function of π¦, where π and β here must be continuous functions. Now, to begin solving this, we consider the derivative as the ratio of two differentials, dπ¦ over dπ₯. This means we can move dπ₯ to the right-hand side and then divide by our function in π¦ such that one over π¦ squared dπ¦ equals negative five π₯ squared dπ₯.

Now, of course, to be able to do this, we need to assume that π¦ squared is not equal to zero. But if there is a number such that π¦ 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 π¦ is equal to zero, π¦ squared is equal to zero. And so we need to include the solution π¦ equals zero in our final answer. And this is because dπ¦ by dπ₯ is equal to zero for a constant. And that gives zero equals zero in our differential equation.

Now, letβs write one over π¦ squared as π¦ to the power of negative two. And now, weβre going to integrate both sides with respect to their variables. When we integrate π¦ to the negative two, we add one to the exponent and then divide by that new value. So itβs π¦ to the power of negative one over negative one plus some constant of integration π΄. Similarly, when we integrate negative five π₯ squared, we get negative five π₯ cubed over three plus a second constant of integration π΅.

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

Now, in fact, itβs a little bit simpler to create a common denominator of three and divide through by five π₯ cubed plus π· over three, where π· 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 π· over three is one times three over five π₯ cubed plus π·, which is simply three over five π₯ cubed plus π·.

And so, we have our two solutions. Itβs usual to define any constants by πΆ. So we know the two solutions to our differential equation are π¦ equals three over five π₯ cubed plus πΆ or π¦ equals zero.