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Solution of differential equation: $X^2= 1+\large\frac{1}{2} \bigg( \large\frac{x}{y}\bigg)^{-1} \frac{dy}{dx} + \large\frac{\bigg(\Large\frac{x}{y}\bigg)^{-2} \bigg(\Large\frac{dy}{dx}\bigg)^{2}}{4 \times 2 \times 1}+ \large\frac{\bigg(\Large\frac{x}{y}\bigg)^{-3} \bigg(\Large\frac{dy}{dx}\bigg)^{3}}{8 \times 3 \times 2 \times 1}+ \large\frac{\bigg(\Large\frac{x}{y}\bigg)^{-4} \bigg(\Large\frac{dy}{dx}\bigg)^{4}}{16 \times 4 \times 3 \times 2 \times 1}+..........$

$(a)\;y^2= 2x^2(\log x^2-1)+c \\ (b)\;y= 2x^2(\log x-1)+c \\ (c)\;y^2= 2x^2(\log x-1)+c \\ (d)\;y= 2x^2 e^{x^2}-1)+c $

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$x^2= e^{\bigg(\Large\frac{2x}{y}\bigg)^{-1} \bigg( \frac{dy}{dx}\bigg)}$
$\log x =\bigg (\large\frac{2x}{y} \bigg)^{-1} \bigg( \large\frac{dy}{dx} \bigg)$
$\log x^2= \large\frac{y}{2x} \frac{dy}{dx}$
$\int 2x \log x^2 dx=\int y dy$
$x^2= t\quad 2xdx=dt$
$\int \log t dt =\large\frac{y^2}{2} +c$
$2( t \log t -t) =y^2 +2c'$
$2( x^2 \log x -x)=y^2 +c$
Hence a is the correct answer.
answered Jan 31, 2014 by meena.p
 

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