Question Video: Solving a Separable First-Order Differential Equation Given in the Normal Form Mathematics • Higher Education

Video Transcript

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

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 d𝑥 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 𝑥 squared, we get negative five 𝑥 cubed over three plus a second constant of integration 𝐵.

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 𝑥 cubed over three plus 𝐶. 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 𝑥 cubed over three plus 𝐶. And then, we’re going to divide through by five 𝑥 cubed over three plus 𝐶.

Now, in fact, it’s a little bit simpler to create a common denominator of three and divide through by five 𝑥 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 𝑥 cubed plus 𝐷 over three is one times three over five 𝑥 cubed plus 𝐷, which is simply three over five 𝑥 cubed plus 𝐷.

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