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Find solution : $(x-2y+1) dy -(3x-6y+2)dx=0$

$(a)\;5(x-2y)+10 \log (x -2y+\frac{3}{5})+x=c \\ (b)\;\frac{1}{5}(x-2y)+\frac{2}{25} \log (x -2y+\frac{3}{5})+x=c \\ (c)\;x-2y+\frac{2}{5}\log (x -2y+\frac{3}{5})+x=c \\ (d)\;5(x-2y)+10 \log (x +2y+\frac{3}{5})+x=c $
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$\large\frac{dy}{dx}=\frac{3(x-2y)+2}{(x^2-y)+1}$
$x-2y=v$
$1-2y'=v'$
$v'= 1-\large\frac{2(3v+2)}{v+1}$
$\quad= -\large\frac{5v+3}{v+1}$
Separating variables
$\large\frac{(v+1)dv}{5v+3}$$=-dx$
or $\large\frac{1}{5}\bigg[1+ \Large\frac{2/5}{v+(3/5)}\bigg]$$dv=dx$
Integrate $\large\frac{1}{5}$$v+\frac{2}{25} \log (v+\frac{3}{5})=-x+c$
$ \large\frac{1}{5}$$(x-2y)+\large\frac{2}{25}$$ \log (x -2y+\frac{3}{5})+x=c$
Hence b is the correct answer.
answered Feb 7, 2014 by meena.p
 

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