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Find solution : $\large\frac{dy}{dx} $$= 1+ \sqrt {y-x+3}$

$(a)\;2 \sqrt {y-x+3}=x+c \\ (b)\;y-x+3=\sqrt x +c \\ (c)\;\sqrt {y-x+3}=x+c \\ (d)\;y-x+3=\sqrt y+c $

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$y-x=v$
$\large\frac{dv}{dx}=\frac{dy}{dx}$$-r$
So, $\large\frac{dv}{dx}=\frac{dy}{dx}$$-1$
$\large\frac{dv}{\sqrt {v+3}}$$=dx$
$2 \sqrt {v+3}=x+c$
$2 \sqrt {y-x+3}=x+c$
Hence a is the correct answer.
answered Feb 6, 2014 by meena.p
 

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