Question

If $y+\frac{y^3}{3}+\frac{y^5}{5}+...\infty =2[x+\frac{x^3}{3}+\frac{x^5}{5}+...\infty ]$, then the relation between $x$ and $y$ is ...$x^{2}y=ax-y$ Find $2a+1$

Answer 1

$y+\frac{y^3}{3}+\frac{y^5}{5}+...\infty =2[x+\frac{x^3}{3}+\frac{x^5}{5}+...\infty ]$
$\Rightarrow \frac{\log (1+y)-\log (1-y)}{2}=2\left[\frac{\log (1+x)-\log (1-x)}{2}\right]$
$\Rightarrow \log \left(\frac{1+y}{1-y}\right)=\log {\left(\frac{1+x}{1-x}\right)}^2$
$\Rightarrow \frac{1+y}{1-y}=\frac{1+x^2+2x}{1+x^2-2x}$
$\Rightarrow \frac{1}{y}=\frac{1+x^2}{2x}$ ....[By componendo-dividendo]
$\Rightarrow y{x}^2=2x-y$
Therefore, $a=2$ and $2a+1=5$
Ans: 5