Differential equations and family of function solutions

I can't follow what Stewart is doing in his book. I can easily follow his work but his conclusion doesn't make any sense to me.

"Show that ever member of the family of functions

$$y = \frac{1+ce^t}{1 - ce^t}$$ is a solution of the differential equation $$y' = \frac{1}{2} (y^2 - 1)$$

He starts by differentiating the right side of the first term and then just setting that equal to the $y'$ from the question.

I do not see how these are equal or what this means or what is going on at all really, and the book doens't feel the need to explain this anyways, so maybe it isn't important and maybe I just need to memorize that a solution is just the differential. But I don't see hwy this is important or how this helps anything.

For his final answer he gets $$y' = \frac{2ce^t}{(1-ce^t)^2}$$

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Plug in the value of $y$ into $\frac{1}{2}(y^2-1)$. Verify that you get $\frac{2ce^t}{(1-ce^t)^2}$.

In other words, what you are doing is plugging in the proffered solutions into the equation and verifying that you get an equality.

It's exactly as if you had been asked to verify that $x=4$ is a solution to the equation $$x^2 - 5x + 6 = 2x^2 - 8x + 2.$$ All you need to do is plug in $x=4$ into the left hand side and compute, $4^2-5(4) + 6 = 16-20+6 = 2$; then plug it into the right hand side and compute, $2(4)^2 - 8(4) + 2 = 32 - 32 + 2 = 2$. And then say: "yes, it's a solution, because it satisfies the equality." You don't have to solve anything, just plug and verify.

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Maybe I do not understand this correctly, but this is not the method that the book used is it? Also how is an equality verified? – user138246 Jun 22 '12 at 20:16
@Jordan: An equality is verified by evaluating each side of the equality and checking that the answer is the same. Exactly like I did in the example I gave you. As to how exactly the book does it, since as usual you did not bother to say which of the many books Stewart has written you are looking at, or where in that 700+ page book you are looking at, I can only try to guess (which is rather annoying). In my copy of the 4th edition of "Calculus, Early Transcendentals", this is Example 1 in Chapter 9, Section 1, page 584. (cont) – Arturo Magidin Jun 22 '12 at 20:27
@Jordan: What he does is exactly the method I say: first, he computes $y'$. Then he plugs in the value of $y$ into the right hand side, and simplifies. After he is done simplifying, he simply observes that what he obtained on the right hand side of $$y' = \frac{1}{2}(y^2-1)$$ by plugging in this value of $y$ is exactly the same thing he obtained by plugging $y$ into the left hand side of that equation (namely, computing $y'$). Since both sides give the same thing, then the given $y$ satisfies the equation. Exactly like I said. – Arturo Magidin Jun 22 '12 at 20:29
I have 7e, its on pg 583 Example 1. I did not specify that because I typed up what he did basically. I am just confused because you never used the quotient rule like he did. – user138246 Jun 22 '12 at 20:34
@Jordan. In a differential equation, the unknown is a function. So what you plug into the equation is a function. "Evaluating $y'$ at $y=\frac{1+ce^t}{1-ce^t}$", or "plugging in $y=\frac{1+ce^t}{1-ce^t}$ into $y'$" means evaluating $$\left(\frac{1+ce^t}{1-ce^t}\right)^'.$$ That is, finding the derivative of $y$ (the first thing Stewart did). "Plugging in $y=\frac{1+ce^t}{1-ce^t}$ into $\frac{1}{2}(y^2-1)$" means evaluating $$\frac{1}{2}\left(\left(\frac{1+ce^t}{1-ce^t}\right)^2-1\right)$$which is the second thing he did. – Arturo Magidin Jun 22 '12 at 20:41

From $y=\frac{1+ce^t}{1-ce^t}$, we get : $e^t=\frac{y-1}{c(y+1)}$ and derive this , you get : $e^t=\frac{2y'}{c(y+1)^2}$ then : $$e^t=\frac{y-1}{c(y+1)}=\frac{2y'}{c(y+1)^2}$$ who gives desired result

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To show that a function is a solution to a differential equation, one must "plug it in" to the given equation. In this case, we have $$y=\frac{1+ce^t}{1-ce^t},$$ and must confirm that $$y'=\frac{1}{2}(y^2-1).$$

Indeed, on the one hand,

$\begin{eqnarray*} \frac{1}{2}(y^2-1) & = & \frac{1}{2}\left[\frac{1+2ce^t+c^2e^{2t}}{1-2ce^t+c^2e^{2t}}-1\right]\\ & = & \frac{1}{2}\left[\frac{1+2ce^t+c^2e^{2t}}{1-2ce^t+c^2e^{2t}}-\frac{1-2yce^t+c^2e^{2t}}{1-2ce^t+c^2e^{2t}}\right]\\ & = & \frac{2ce^t}{1-2ce^t+c^2e^{2t}}\\ & = & \frac{2ce^t}{(1-ce^t)^2}, \end{eqnarray*}$

and on the other, we have by quotient rule that

$\begin{eqnarray*} y' & = & \frac{dy}{dt}\\ & = & \frac{ce^t(1-ce^t)-(-ce^t)(1+ce^t)}{(1-ce^t)^2}\\ & = & \frac{ce^t-c^2e^{2t}+ce^t+c^2e^{2t}}{(1-ce^t)^2}\\ & = & \frac{2ce^t}{(1-ce^t)^2}. \end{eqnarray*}$

For contrast, suppose you'd been asked to verify whether $y=\sin t$ was a solution to $y'=\frac{1}{2}(y^2-1).$ Well, $y'=\cos t$, but $$\frac{1}{2}(y^2-1)=-\frac{1}{2}\cos^2t$$ by Pythagorean identity. But when $t=0$, we see that $y'=\cos 0=1$ and $\frac{1}{2}(y^2-1)= -\frac{1}{2}\cos^20=-\frac{1}{2}$, so if $y=\sin t$, then $y'\neq\frac{1}{2}(y^2-1)$. Thus, $y=\sin t$ is not a solution to the differential equation given.

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How is that final quotient that you have equal to the differential equation? – user138246 Jun 22 '12 at 20:20
I'm not sure I understand what you mean. If you want to show that $a=b$, one way to do so is to show that $a=c$ and $b=c$. That's what I've done here. Or do you mean how did I get to the final quotient? I will add a step in case that's what you meant. Also, I'll add a further remark. – Cameron Buie Jun 22 '12 at 20:33
I understand how you got what you did, I just thought it was suppose to be an a = b sort of situation. – user138246 Jun 22 '12 at 20:37
Yes, it is. Let me know if my expanded answer clarifies things at all. – Cameron Buie Jun 22 '12 at 20:48
$\sin^2t+\cos^2t=1$. – Cameron Buie Jun 22 '12 at 20:52

To show that $y= \frac{1+ce^t}{1 - ce^t}$ is a solution of $y' = \frac{1}{2}(y^2-1)$, you just have to check that if you replace $y= \frac{1+ce^t}{1 - ce^t}$ into $y' = \frac{1}{2}(y^2-1)$ you get a valid equality.

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