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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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    @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.2012-06-22
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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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    $\sin^2t+\cos^2t=1$.2012-06-22
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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 $y= \frac{1+ce^t}{1 - ce^t}$ 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.