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

Solve the differential equation d𝑦/dπ‘₯ = βˆ’5π‘₯√(𝑦).

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Video Transcript

Solve the differential equation d𝑦 by dπ‘₯ equals negative five π‘₯ times the square root of 𝑦.

This equation is what’s known as a separable differential equation. Given an equation d𝑦 by dπ‘₯ equals some function of π‘₯ and 𝑦, it’s called separable if this function of π‘₯ and 𝑦 could be factored into the product of two functions of π‘₯ and 𝑦, where 𝑔 of π‘₯ and β„Ž of 𝑦 are continuous functions. We’re then able to consider the derivative d𝑦 by dπ‘₯ as the ratio of two differentials. And we can move dπ‘₯ to the right-hand side and then divide by the function in 𝑦, in this case, divide by the square root of 𝑦.

Now, of course, this only follows if the square root of 𝑦 is not equal to zero. If there is some number 𝑦 such that the square root of 𝑦 is equal to zero, then this number will also be a solution of the differential equation. And what happens is when we divide by the square root of 𝑦, we lose that solution. Now, in this case, when 𝑦 is equal to zero, the square root of 𝑦 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 some constant. And that gives zero equals zero in the differential equation.

Now, going back to what we were doing before, let’s rewrite one over the square root of 𝑦 as 𝑦 to the power of negative one-half. And then, we’re going to integrate both sides of this equation with respect to their individual variables. Now, to integrate 𝑦 to the power of negative one-half, we add one to the exponent and then divide by that new value. So, we get 𝑦 to the power of one-half divided by one-half plus some constant of integration 𝐴. Similarly, to integrate negative five π‘₯, we add one to the exponent. So that becomes two. And then, we divide by two. So, the integral of negative five π‘₯ is negative five π‘₯ squared over two plus the second constant of integration 𝐡.

Now, of course, dividing by one-half is the same as multiplying by two. So, this becomes two 𝑦 to the power of one-half plus 𝐴 on the left-hand side. Next, we’re going to divide the entire equation by two and combine our constants of integration. And when we do, we get 𝑦 to the one-half equals negative five π‘₯ squared over four plus 𝐢. And of course, 𝑦 to the power of one-half is the square root of 𝑦.

And so, we have the two solutions to our differential equation. One is the square root of 𝑦 equals negative five π‘₯ squared over four plus 𝐢 and the other is 𝑦 is equal to zero.

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