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General solution of differential equation ; $\large\frac{ydy}{xdx}+\frac{2(x^2+y^2)-1}{(x^2+y^2+1)}=0$

$(a)\;2x^2+2y^2+3 \log (x^2+y^2)=c \\ (b)\;x^2+2y^2+3 \log (x^2+y^2-2)=0 \\ (c)\;2x^2+2y^2+3 \log (x^2+y^2-2)=0 \\ (d)\;2x^2+y^2+3 \log (x^2+y^2-2)=c $

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Putting $x^2+y^2=t$
$2x+ 2y \large\frac{dy}{dx}=\frac{dt}{dx}$
$\large\frac{2y}{2x}=\frac{1}{2x}\frac{dt}{dx}$$-1$
$\large\frac{1}{2x} \frac{dt}{dx}$$-1+ \large\frac{2t-1}{t+1}$$=0$
$\large\frac{1}{2x} \frac{dt}{dx}$$=1 - \large\frac{2t-1}{t+1} =\frac{2-t}{t+1}$
$\large\frac{t+1}{-2+t}$$dt=-2x dx$
$\int \bigg(\large\frac{t-2}{t-2} +\frac{3}{t-2}\bigg)$$dt=-x^2$
$t+ 3\log (t-2)=-x^2+c$
$x^2+y^2+3 \log (x^2+y^2-2)=-x^2+c$
$2x^2+y^2+3 \log (x^2+y^2-2)=c $
Hence d is the correct answer.
answered Feb 4, 2014 by meena.p
 

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