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

Video Transcript

Solve the differential equation d𝑧 by d𝑡 plus 𝑒 to the power of two 𝑡 plus two 𝑧 equals zero.

Remember, a separable differential equation is one for which the expression for d𝑧 by d𝑡 can be expressed as some function of 𝑧 times some function of 𝑡. So how exactly are we going to achieve that for our equation? Well, we’re going to begin by subtracting 𝑒 the power of two 𝑡 plus two 𝑧 from both sides of our equation. We then recall that the laws of exponents tell us that 𝑥 to the power of 𝑎 plus 𝑏 can be written as 𝑥 to the power of 𝑎 times 𝑥 to the power of 𝑏. So we can write negative 𝑒 to the power of two 𝑡 plus two 𝑧 as negative 𝑒 to the power of two 𝑡 times 𝑒 to the power of two 𝑧.

Now, of course, d𝑧 by d𝑡 isn’t a fraction, but we treat it a little like one. And we can say that this is equivalent to one over 𝑒 to the power of two 𝑧 d𝑧 equals negative 𝑒 to the two 𝑡 d𝑡. And this is great because we’re now ready to integrate both sides. It can be simpler to express one over 𝑒 to the power of two 𝑧 as 𝑒 to the power of negative two 𝑧. And then, we quote the general result for the integral of 𝑒 to the power of 𝑘𝑥 for some constant 𝑘. It’s 𝑒 to the power of 𝑘𝑥 over 𝑘 plus 𝑐. So this means the integral of 𝑒 to the power of negative two 𝑧 is negative 𝑒 to the power of negative two 𝑧 over two. And the integral of negative 𝑒 to the two 𝑡 is negative 𝑒 to the two 𝑡 over two.

Our next step is to subtract 𝑐 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 𝑧 in terms of 𝑡, So we find that 𝑒 to the power of negative two 𝑧 is equal to 𝑒 to the power of two 𝑡 plus 𝑐 three. 𝑐 three is a new constant obtained by subtracting 𝑐 one from 𝑐 two and then multiplying by negative two. To solve for 𝑧, we find the natural log of both sides of this equation. But the natural log of 𝑒 to the power of negative two 𝑧 is just negative two 𝑧. So our final step is to divide through by negative two. And we’ve solved our differential equation. 𝑧 is equal to negative one-half times the natural log of 𝑒 to the power of two 𝑡 plus some constant; let’s call that 𝑐.