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

Solve the differential equation dπ¦/dπ₯ = β5π₯β(π¦).

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### Video Transcript

Solve the differential equation dπ¦ by dπ₯ equals negative five π₯ times the square root of π¦.

This equation is whatβs known as a separable differential equation. Given an equation dπ¦ by dπ₯ equals some function of π₯ and π¦, itβs called separable if this function of π₯ and π¦ could be factored into the product of two functions of π₯ and π¦, where π of π₯ and β of π¦ are continuous functions. Weβre then able to consider the derivative dπ¦ 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 π¦, in this case, divide by the square root of π¦.

Now, of course, this only follows if the square root of π¦ is not equal to zero. If there is some number π¦ such that the square root of π¦ 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 π¦, we lose that solution. Now, in this case, when π¦ is equal to zero, the square root of π¦ 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 some constant. And that gives zero equals zero in the differential equation.

Now, going back to what we were doing before, letβs rewrite one over the square root of π¦ as π¦ to the power of negative one-half. And then, weβre going to integrate both sides of this equation with respect to their individual variables. Now, to integrate π¦ to the power of negative one-half, we add one to the exponent and then divide by that new value. So, we get π¦ to the power of one-half divided by one-half plus some constant of integration π΄. 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 π΅.

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

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