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Find solution of differential equation $(y-x+1) dy -(y+x+2)dx=0$

$(a)\;\bigg[(\frac{2y+3}{2x+1}-1)^2-2 \bigg] c^2 \bigg(\frac{2x+1}{2}\bigg)^2=1 \\ (b)\;\bigg[(\frac{2x+1}{2y+3}-1)^2-2 \bigg] c^2 \bigg(\frac{2x+1}{2}\bigg)^2=1 \\ (c)\;\bigg[(\frac{x+1}{y+3}-1)^2-2 \bigg] c^2 \bigg(\frac{2x+1}{2}\bigg)^2=1 \\ (d)\;\bigg[(\frac{y+3}{x+1}-1)^2-2 \bigg] c^2 \bigg(\frac{2x+1}{2}\bigg)^2=1 $
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$x= x+h$
$y=y+k$
$\large\frac{dY}{dX} =\frac{Y+X+(K+h_2)}{Y- X +(K-h+1)}$
Choose h.K such that $K+2=0$
$K-h+1=0$
So $ h =-\large\frac{-1}{2}$$,K= \large\frac{-3}{2}$
$Y= vX$
$Xv'=+v =\large\frac{v+1}{v-1}$
$Xv'=\large\frac{v+1}{v-1}$$-v$
$\qquad= \large\frac{2v+1-v^2}{v-1}$
$\large\frac{(v-1)dv}{(v-1)^2-2}=\large\frac{-dx}{x}$
$\large\frac{1}{2}$$ \log ((v-1)^2 -2) =-\log x-\log c$
$\qquad= -\log cx$
$[(v-1)^2-2]c^2x^2=1$
Substitute $V= \large\frac{Y}{X}=\large\frac{y-K}{x-h}$
$\qquad= \large\frac{y+3/2}{x+1/2}$
$\bigg[(\frac{2y+3}{2x+1}-1)^2-2 \bigg] c^2 \bigg(\frac{2x+1}{2}\bigg)^2=1$
Hence a is the correct answer.
answered Feb 7, 2014 by meena.p
 

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