If 4left[ {{x^2} + frac{{{x^6}}}{3} + frac{{{ | vedclass

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Question

(A) The given equation is $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3}

Vedclass · Accepted Answer

${x^2}y = 2x - y$

Vedclass · Answer

(A) The given equation is $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3} + \dots$Using the logarithmic series expansions $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$ and $\ln(1-t) = -t - \frac{t^2}{2} - \frac{t^3}{3} - \dots$,we have:$2 \ln\left( \frac{1+x^2}{1-x^2} \right) = -\ln(1-y^2)$This simplifies to $\ln\left( \frac{1+x^2}{1-x^2} \right)^2 = \ln\left( \frac{1}{1-y^2} \right)$Taking the exponential of both sides:$\left( \frac{1+x^2}{1-x^2} \right)^2 = \frac{1}{1-y^2}$Expanding and rearranging terms leads to the relation ${x^2}y = 2x - y$.

If $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3} + \dots$,then

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