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In this video, we’ll learn the definition of a separable differential equation and how we can rewrite these to obtain equality between two integrals that we can then evaluate.
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To this end, it’s important you’re confident in evaluating integrals of a variety of functions — such as polynomials, trigonometric, and exponential functions — before accessing this video.
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A separable equation is a first order differential equation in which the expression for d𝑦 by d𝑥 can be factored as a function of 𝑥 times a function of 𝑦.
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In other words, it can be written in the form d𝑦 by d𝑥 equals 𝑔 of 𝑥 times 𝑓 of 𝑦.
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The name separable comes from the fact that the expression on the right-hand side can be separated into a function of 𝑥 and a function of 𝑦.
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Equivalently, if 𝑓 of 𝑦 is not equal to zero, we can write our equation as d𝑦 by d𝑥 equals 𝑔 of 𝑥 over ℎ of 𝑦, where ℎ of 𝑦 is one over 𝑓 of 𝑦.
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To solve this equation, we write it in the differential form ℎ of 𝑦 d𝑦 equals 𝑔 of 𝑥 d𝑥 Essentially, we needed to get all 𝑦s on one side of the equation and all 𝑥s on the other.
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And then we can integrate both sides of the equation.
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Now, it’s important to realize that d𝑦 by d𝑥 isn’t a fraction.
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But for the purposes of solving separable differential equations, we do treat it a little like one.
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And what we’ve done here is defined 𝑦 implicitly as a function in 𝑥.
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And in many cases, we’ll be able to solve for 𝑦 in terms of 𝑥.
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Let’s have a look at an example of how this might work.
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Solve the differential equation d𝑦 by d𝑥 plus 𝑦 equals one.
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A separable equation is a first order differential equation in which the expression for d𝑦 by d𝑥 can be factored as a function of 𝑥 times a function of 𝑦.
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In other words, it can be written in the form 𝑔 of 𝑥 times 𝑓 of 𝑦.
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So let’s rearrange our equation d𝑦 by d𝑥 plus 𝑦 equals one so that it’s in this form.
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To achieve this, we’re going to subtract 𝑦 from both sides of the equation.
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And we obtain d𝑦 by d𝑥 to be equal to one minus 𝑦.
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Now, it may not look like it, but we have achieved our aim.
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Our function in 𝑦 is one minus 𝑦, and our function in 𝑥 is simply one.
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Then, to solve this equation, we’re going to rewrite it using differentials.
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Now, remember, d𝑦 by d𝑥 absolutely isn’t a fraction.
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But we do treat it a little like one for the purposes of this process.
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We begin by dividing both sides of our equation by one minus 𝑦.
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And we see that this is equivalent to saying one over one minus 𝑦 d𝑦 equals one d𝑥.
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And now, we’re ready to integrate both sides of this equation.
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So how do we integrate one over one minus 𝑦 with respect to 𝑦?
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Well, we begin by quoting the general result for the integral of one over 𝑥 with respect to 𝑥.
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It’s the natural log of the absolute value of 𝑥 plus some constant of integration, 𝑐.
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What we’re going to do is perform a substitution for our integral.
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We’re going to let 𝑢 be equal to one minus 𝑦 so that d𝑢 by d𝑦 is equal to negative one.
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We could write this equivalently as negative d𝑢 equals d𝑦.
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And then we’re going to replace d𝑦 with negative d𝑢 and one minus 𝑦 with 𝑢.
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And we see that we now need to integrate negative one over 𝑢 with respect to 𝑢.
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Well, that’s negative the natural log of the absolute value of 𝑢 plus that constant of integration 𝑐.
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By replacing 𝑢 with one minus 𝑦, we find the integral of one over one minus 𝑦 to be the negative natural log of the absolute value of one minus 𝑦 plus a constant of integration which I’m going to call 𝑎.
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When we integrate one with respect to 𝑥, it’s a little more straightforward.
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We get 𝑥 plus a second constant of integration.
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Let’s call that 𝑏.
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Let’s subtract 𝑎 from both sides of our equation and multiply through by negative one.
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That gives us the natural log of the absolute value of one minus 𝑦 equals negative 𝑥 plus 𝑐 one.
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𝑐 one is a new constant, and it’s achieved by subtracting 𝑎 from 𝑏 and multiplying by negative one.
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And then, we notice that we can raise both sides of this equation as a power of 𝑒.
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So we obtain the absolute value of one minus 𝑦 to be equal to 𝑒 to the power of negative 𝑥 plus 𝑐 one.
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By using the laws of exponents, though, we see that we can rewrite 𝑒 to the power of negative 𝑥 plus 𝑐 one as 𝑒 to the power of negative 𝑥 times 𝑒 to the power of 𝑐 one.
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But, of course, 𝑒 to the power of 𝑐 one is itself a constant.
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Let’s call that 𝑐 two.
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And we see we can rewrite the right-hand side as 𝑐 two times 𝑒 to the power of negative 𝑥.
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Well, of course, we’re dealing with the absolute value of one minus 𝑦.
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And we can say that this means that one minus 𝑦 could be equal to positive 𝑐 two times 𝑒 to the power of negative 𝑥 or negative 𝑐 two times 𝑒 to the power of negative 𝑥.
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But 𝑐 two is a constant, so we don’t actually need to write that.
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In our final step, we’re going to add 𝑦 to both sides of the equation and then subtract 𝑐 two 𝑒 to the power of negative 𝑥.
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That gives us 𝑦 equals one minus 𝑐 two times 𝑒 to the power of negative 𝑥.
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But of course, once again, since 𝑐 two is a constant, we can change this to plus 𝑐.
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And we see that 𝑦 equals one plus 𝑐 times 𝑒 to the power of negative 𝑥 is the solution to our separable differential equation.
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We’ll now have a look at an example which requires some further techniques for integration.
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Find a relation between 𝑦 and 𝑥 given that 𝑥𝑦 times 𝑦 prime is equal to 𝑥 squared minus five.
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Now, this first step isn’t entirely necessary.
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But it can make it a little easier to see what to do next.
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We write 𝑦 prime using Leibniz notation.
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And we see that 𝑥𝑦 d𝑦 by d𝑥 is equal to 𝑥 squared minus five.
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And then we recall that a separable differential equation is one in which the expression for d𝑦 by d𝑥 can be written as some function of 𝑥 times some function of 𝑦.
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Now, in fact, if we divide both sides of our equation by 𝑥𝑦, we see that we can achieve this.
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We get d𝑦 by d𝑥 equals 𝑥 squared minus five over 𝑥𝑦 or 𝑥 squared minus five over 𝑥 times one over 𝑦.
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And that’s great because 𝑔 of 𝑥 is, therefore, 𝑥 squared minus five over 𝑥.
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And our function of 𝑦 is one over 𝑦.
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Now, in fact, we just performed this to demonstrate that we did indeed have a separable differential equation.
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We could’ve kept 𝑦 on the left-hand side as shown.
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And then, we perform this rather strange step. d𝑦 by d𝑥 isn’t a fraction, but we do treat it a little like one.
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And we say that 𝑦 d𝑦 equals 𝑥 squared minus five over 𝑥 d𝑥.
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Our next step is to integrate both sides of this equation.
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Remember, to integrate a polynomial term whose exponent is not equal to negative one, we add one to the exponent and divide by that new number.
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So the integral of 𝑦 is 𝑦 squared over two plus some constant of integration 𝑎.
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Then, it might look like we need to perform some sort of substitution to evaluate this integral.
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But actually, if we separate our fraction into 𝑥 squared over 𝑥 minus five over 𝑥, we find that we need to integrate 𝑥 minus five over 𝑥 with respect to 𝑥.
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Then, we recall the general result for the integral of one over 𝑥.
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It’s the natural log of the absolute value of 𝑥 plus 𝑐.
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And so, when we integrate the right-hand side of our equation, we get 𝑥 squared over two minus five times the natural log of the absolute value of 𝑥 plus some second constant of integration which I’ve called 𝑏.
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Our last step is to subtract 𝑎 from 𝑏 and then multiply the entire equation by two.
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When we do, we find that 𝑦 squared equals 𝑥 squared minus 10 times the natural log of the absolute value of 𝑥 plus 𝑐.
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This 𝑐 is a different constant achieved by subtracting 𝑎 from 𝑏 and then multiplying by two.
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And so, given our differential equation, we’ve found a relation between 𝑦 and 𝑥.
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𝑦 squared equals 𝑥 squared minus 10 times the natural log of the absolute value of 𝑥 plus 𝑐.
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In this next example, we’ll see how some clever factoring can create a much simpler expression to separate.
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Find a relation between 𝑢 and 𝑡 given that d𝑢 by d𝑡 equals one plus 𝑡 to the fourth power over 𝑢𝑡 squared plus 𝑢 to the fourth power times 𝑡 squared.
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Now, this differential equation might look really nasty.
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But it is, in fact, a separable differential equation.
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In this case, that’s one where an expression for d𝑢 by d𝑡 can be written as some function of 𝑢 times some function of 𝑡.
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So how are we going to achieve that?
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Well, we begin by factoring the denominator by 𝑡 squared.
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And we find that d𝑢 by d𝑡 is equal to one plus 𝑡 to the fourth power over 𝑡 squared times 𝑢 plus 𝑢 to the fourth power.
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We can now write this as some function of 𝑡 times some function of 𝑢.
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It’s one plus 𝑡 to the fourth power over 𝑡 squared times one over 𝑢 plus 𝑢 to the fourth power.
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Let’s begin by multiplying both sides of this equation by 𝑢 plus 𝑢 to the fourth power.
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Then, we recall that whilst d𝑢 by d𝑡 is not a fraction, we treat it a little like one.
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And we can say that 𝑢 plus 𝑢 to the fourth power d𝑢 is equal to one plus 𝑡 to the fourth power over 𝑡 squared d𝑡.
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And that’s great because we’re now ready to integrate both sides of our equation.
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The left-hand side is quite straightforward to integrate.
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The integral of 𝑢 is 𝑢 squared over two.
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And the integral of 𝑢 to the fourth power is 𝑢 to the fifth power over five.
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Don’t forget, since this is an indefinite integral, we need that constant of integration 𝑎.
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On the right-hand side, we’re going to rewrite our integrand as one over 𝑡 squared plus 𝑡 to the fourth power over 𝑡 squared.
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And that simplifies to one over 𝑡 squared plus 𝑡 squared.
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But, of course, one over 𝑡 squared is the same as 𝑡 to the power of negative two.
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And we can now integrate as normal.
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When we integrate 𝑡 to the power of negative two, we add one to the exponent to get negative one and then we divide by that number.
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𝑡 squared becomes 𝑡 cubed over three.
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And we have a second constant of integration; I’ve called that 𝑏.
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Now, of course, we can rewrite 𝑡 to the power of negative one over negative one as negative one over 𝑡.
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Our final step is to subtract our constant 𝑎 from both sides of the equation.
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That gives us a new constant 𝑐.
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That’s just the difference between 𝑏 and 𝑎.
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So we’ve found a relation between 𝑢 and 𝑡.
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It’s 𝑢 to the fifth power over five plus 𝑢 squared over two equals negative one over 𝑡 plus two cubed over three plus 𝑐.
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In our final example, we’ll see how this process works for exponential functions.
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Solve the differential equation d𝑧 by d𝑡 plus 𝑒 to the power of two 𝑡 plus two 𝑧 equals zero.
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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.
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So how exactly are we going to achieve that for our equation?
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Well, we’re going to begin by subtracting 𝑒 the power of two 𝑡 plus two 𝑧 from both sides of our equation.
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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.
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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.
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Now, of course, d𝑧 by d𝑡 isn’t a fraction, but we treat it a little like one.
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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.
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And this is great because we’re now ready to integrate both sides.
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It can be simpler to express one over 𝑒 to the power of two 𝑧 as 𝑒 to the power of negative two 𝑧.
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And then, we quote the general result for the integral of 𝑒 to the power of 𝑘𝑥 for some constant 𝑘.
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It’s 𝑒 to the power of 𝑘𝑥 over 𝑘 plus 𝑐.
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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.
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And the integral of negative 𝑒 to the two 𝑡 is negative 𝑒 to the two 𝑡 over two.
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Our next step is to subtract 𝑐 one from both sides of the equation and then multiply through by negative two.
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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.
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𝑐 three is a new constant obtained by subtracting 𝑐 one from 𝑐 two and then multiplying by negative two.
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To solve for 𝑧, we find the natural log of both sides of this equation.
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But the natural log of 𝑒 to the power of negative two 𝑧 is just negative two 𝑧.
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So our final step is to divide through by negative two.
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And we’ve solved our differential equation.
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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.
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In this video, we’ve seen that a separable equation is a first order differential equation of the form d𝑦 by d𝑥 equals some function of 𝑥 times some function of 𝑦.
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We saw that to solve these kinds of equations, we separate all our 𝑥s onto one side and all our 𝑦s onto the other, and then we integrate.
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Finally, we emphasized that whilst we do treat d𝑦 by d𝑥 a little like a fraction in this method, d𝑦 by d𝑥 is absolutely not a fraction.