What is wrong? Symmetric function

Question

I need some advice here.

What is $y(\ln(4))$ if the function $y$ satisfies:

$$\frac{dy}{dx} = 1-y^2$$ and is symmetric about the point $(\ln(9),0)$.

Solving that equation I end up with:

$$y=\frac{Ce^{2x}-1}{Ce^{2x}+1}.$$

And a function is symmetric about a point if $f(−x)=−f(x)\forall x.$

Plugging it into the equation:

$$y=\frac{Ce^{2(-\ln(9))}-1}{Ce^{2(-\ln(9))}+1} = -\frac{Ce^{2\ln(9)}-1}{Ce^{2\ln(9)}+1}. $$

(The constants shouldn't really matter here should they?)

Anyways:

$$\frac{15}{17} = \frac{Ce^{2\ln(4)}-1}{Ce^{2\ln(4)}+1}.$$

Which is wrong, and I guess the problem is the constants. How should i find them?

Answer 1

In case when $\forall x f(-x)=-f(x)$ the function is symmetric about a point $(0;0)$

Answer 2

In general, a function is symmetric about $c$ if and only if $f(c+x) = f(c-x)$ (Check that this makes sense the intuitive idea of symmetry). Using the general solution for $y$, we obtain $$ \frac{Ce^{2(\log(9)-x)}-1}{Ce^{2(\log(9)-x)}+1} = \frac{Ce^{2(\log(9)+x)}-1}{Ce^{2(\log(9)+x)}+1}$$ from which we can solve for our coefficient $C$.

Answer 3

Note that the constant $C = 1$.

And note that $$\frac{Ce^{2\ln(4)}-1}{Ce^{2\ln(4)}+1} =\frac{e^{\ln(4)^2}-1}{e^{\ln(4^2)}+1} = = \frac{e^{16}-1}{e^{16}+1} = \frac{16 - 1}{16 + 1} = \frac{15}{17}$$