# Video: Solving a Separable First Order Differential Equation Involving Properties of Exponents

Solve the differential equation (dπ§/dπ‘) + π^(2π‘ + 2π§) = 0.

02:10

### Video Transcript

Solve the differential equation dπ§ by dπ‘ plus π to the power of two π‘ plus two π§ equals zero.

Remember, a separable differential equation is one for which the expression for dπ§ by dπ‘ can be expressed as some function of π§ times some function of π‘. So how exactly are we going to achieve that for our equation? Well, weβre going to begin by subtracting π the power of two π‘ plus two π§ from both sides of our equation. We then recall that the laws of exponents tell us that π₯ to the power of π plus π can be written as π₯ to the power of π times π₯ to the power of π. So we can write negative π to the power of two π‘ plus two π§ as negative π to the power of two π‘ times π to the power of two π§.

Now, of course, dπ§ by dπ‘ isnβt a fraction, but we treat it a little like one. And we can say that this is equivalent to one over π to the power of two π§ dπ§ equals negative π to the two π‘ dπ‘. And this is great because weβre now ready to integrate both sides. It can be simpler to express one over π to the power of two π§ as π to the power of negative two π§. And then, we quote the general result for the integral of π to the power of ππ₯ for some constant π. Itβs π to the power of ππ₯ over π plus π. So this means the integral of π to the power of negative two π§ is negative π to the power of negative two π§ over two. And the integral of negative π to the two π‘ is negative π to the two π‘ over two.

Our next step is to subtract π one from both sides of the equation and then multiply through by negative two. Remember, in solving our differential equation, ideally, we want an equation for π§ in terms of π‘, So we find that π to the power of negative two π§ is equal to π to the power of two π‘ plus π three. π three is a new constant obtained by subtracting π one from π two and then multiplying by negative two. To solve for π§, we find the natural log of both sides of this equation. But the natural log of π to the power of negative two π§ is just negative two π§. So our final step is to divide through by negative two. And weβve solved our differential equation. π§ is equal to negative one-half times the natural log of π to the power of two π‘ plus some constant; letβs call that π.