Question

If $x^2=2x-y$ and $|x|<1$, then
$y+\frac{y^3}{3}+\frac{y^5}{5}+...\infty =c(x+\frac{x^3}{3}+\frac{x^5}{5}+...\infty )$Find 2c

Answer 1

Given $x^{2}y=2x-y$
$y(1+x^2)=2x$
$\therefore$ $y=\frac{2x}{1+x^2}$ ...(A)
Now
$y+\frac{y^3}{3}+\frac{y^5}{5}+...\infty$
=$\frac{1}{2}{\log }_e\left(\frac{1+y}{1-y}\right)$
=$\frac{1}{2}{\log }_e\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)$
=$\frac{1}{2}{\log }_e\left(\frac{1+x^2+2x}{1+x^2-2x}\right)$
=$\frac{1}{2}{\log }_e\left(\frac{(1+x{)}^2}{(1-x{)}^2}\right)$
=$\frac{1}{2}{\log }_e{\left(\frac{1+x}{1-x}\right)}^2$
=$\frac{1}{2}.2\log \left(\frac{1+x}{1-x}\right)$
=${\log }_e\left(\frac{1+x}{1-x}\right)$ given |x| < 1
=$2(x+\frac{x^3}{3}+\frac{x^5}{5}+...\infty )$

$\Rightarrow 2c=2\times 2=4$