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General solution of differential equation : $ (1-x^2)(1-y)dx -xy (1+y) dy=0$

$(a)\;x^2+2 \log x =(y+1)^2+6(y+1) +4 \log (y+1) +c \\ (b)\;\frac{x^2}{2}- \log x =\frac{(y-1)^2}{2}+3(y-1) +2 \log (y-1) +c \\ (c)\;\frac{x^2}{2}+\log x =\frac{(y+1)^2}{2}+3(y-1) +2 \log (y+1) +c \\ (d)\;\frac{x^2}{2}+ \log x =\frac{(y-1)^2}{2}+3(y-1) +2 \log (y-1) +c $

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$\int \large\frac{(1-x^2)dx}{x}=\int \large\frac{y(1+y)dy}{1-y}$
$\int\bigg(\large\frac{1}{x} -x\bigg)= - \int \large\frac{y(1+y)}{-1+y}$$dy$
$y-1=t$
$dy=dt$
$\log n - \large\frac{x^2}{2}$$=- \int \large\frac{(t+1)(t+2)}{t}$$dt$
$-\log n +\large\frac{x^2}{2} =+ \int\bigg( \large\frac{t^2+3t+t^2}{t}\bigg)dt$
$\large\frac{x^2}{2} $$-\log x =\int (t +3 +\frac{2}{t} )dt$
$\large\frac{x^2}{2} -\log x =\large\frac{t^2}{2} $$+3t+2 \log t +c$
$\large\frac{x^2}{2}$$- \log x =\large\frac{(y-1)^2}{2}$$+3(y-1) +2 \log (y-1) +c$
Hence b is the correct answer.
answered Feb 3, 2014 by meena.p
 

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