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Find a relation between π¦ and π₯ given that π₯π¦ times π¦ prime is equal to π₯ squared minus five.
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Now, this first step isnβt 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 π¦ prime using Leibniz notation.
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And we see that π₯π¦ dπ¦ by dπ₯ is equal to π₯ squared minus five.
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And then we recall that a separable differential equation is one in which the expression for dπ¦ by dπ₯ can be written as some function of π₯ times some function of π¦.
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Now, in fact, if we divide both sides of our equation by π₯π¦, we see that we can achieve this.
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We get dπ¦ by dπ₯ equals π₯ squared minus five over π₯π¦ or π₯ squared minus five over π₯ times one over π¦.
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And thatβs great because π of π₯ is, therefore, π₯ squared minus five over π₯.
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And our function of π¦ is one over π¦.
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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 couldβve kept π¦ on the left-hand side as shown.
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And then, we perform this rather strange step. dπ¦ by dπ₯ isnβt a fraction, but we do treat it a little like one.
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And we say that π¦ dπ¦ equals π₯ squared minus five over π₯ dπ₯.
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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 π¦ is π¦ squared over two plus some constant of integration π.
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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 π₯ squared over π₯ minus five over π₯, we find that we need to integrate π₯ minus five over π₯ with respect to π₯.
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Then, we recall the general result for the integral of one over π₯.
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Itβs the natural log of the absolute value of π₯ plus π.
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And so, when we integrate the right-hand side of our equation, we get π₯ squared over two minus five times the natural log of the absolute value of π₯ plus some second constant of integration which Iβve called π.
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Our last step is to subtract π from π and then multiply the entire equation by two.
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When we do, we find that π¦ squared equals π₯ squared minus 10 times the natural log of the absolute value of π₯ plus π.
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This π is a different constant achieved by subtracting π from π and then multiplying by two.
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And so, given our differential equation, weβve found a relation between π¦ and π₯.
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π¦ squared equals π₯ squared minus 10 times the natural log of the absolute value of π₯ plus π.