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Hey I need help with rearranging the following equation for y in terms of x.

$$ax+b=\frac{\sqrt c(e^{2xyc}-1)}{\sqrt y(e^{2xyc}+1)}$$

a, b and c are constants by the way

Thanks in advance.

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    $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ Commented Jan 30, 2020 at 7:50
  • $\begingroup$ What is your work on the subject ? Have you noticed for example that a $\tanh$ (hyperbolic tangent) is hidden there ? $\endgroup$
    – Jean Marie
    Commented Jan 30, 2020 at 8:12

1 Answer 1

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As said in comments $$\frac{e^{2cxy}-1}{e^{2cxy}+1}=\tanh(c x y)$$ So, you want to solve for $y$ the equation $$a x+b =\frac{\sqrt c}{\sqrt y}\tanh(c x y)$$ Let $c xy=t$ to make $$\frac{a x+b}{c \sqrt{x}}=\frac{\tanh(t)}{\sqrt t}$$ which will not show explicit solution even using special functions.

Considering, for a given $x$, the equation $$k=\frac{\tanh(t)}{\sqrt t}$$ if $k>0.7632712$, there will nor be any solution. In the other case, there are two solutions, one between $0$ and $1.088659$ (which is the maximum) and another one between $1.088659$ and $\infty$.

If you are concerned by the samllest root, you could use Taylor series to get $$k=\sqrt t\left(1-\frac{1}{3}t^2+\frac{2 }{15}t^4-\frac{17 }{315}t^6+O\left(t^8\right) \right)$$ and use series reversion to get $$t=k^2+\frac{2 }{3}k^6+\frac{43 }{45}k^{10}+\frac{1642}{945} k^{14}+O\left(k^{18}\right)$$

Trying for $k=0.5$, the above truncated expansion would give $t=\frac{404809}{1548288}\approx 0.261456$ while the exact solution, obtained using Newton method, would be $0.261472$.

However, for this case, the second solution is close to $t=4$.

Edit

When $x$ starts to be large, we could use $k\sim\frac{1}{\sqrt t}$ which would give $t \sim \frac 1 {k^2}$ which, for the worked example would exactly give $x=4$ while the exact solution is $3.99458$.

Now, remains the intermediate range that is to say around $t=1$. In such a case, wa can again use Taylor expansion $$\frac{\tanh(t)}{\sqrt t}=\alpha+(t-1) \left(-\alpha^2-\frac{\alpha}{2}+1\right)+(t-1)^2 \left(\alpha^3+\frac{\alpha^2}{2}-\frac{5 \alpha}{8}-\frac{1}{2}\right)+O\left((t-1)^3\right)$$ where $\alpha=\tanh(1)$. Now, we just face a quadratic equation in $(t-1)$. For example, using $k=0.7$ would give the roots $0.571656$ and $1.58875$ while the exact solutions are $0.620209$ and $1.84781$. The estimates are suffciently good to make Newton method converging quite fast.

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