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General solution of differential equation: $\large\frac{xdy}{x^2+y^2} $$+\bigg(1- \large\frac{y}{x^2+y^2} \bigg)$$dx=0$ is

$(a)\;x^2+\tan^{-1}(\frac{y}{h})=c \\ (b)\;x+\tan^{-1} (\frac{x}{y})=c \\ (c)\;x^2-\tan^{-1}(\frac{y}{x})=c \\ (d)\;x-\tan^{-1} (\frac{x}{y})=c $

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$\large\frac{xdy-ydx}{x^2+y^2}$$+dx=0$
$dx= \large\frac{y dx-xdy}{x^2+y^2}$
$dx= \large\frac{ydx-xdy}{y^2\bigg(\Large\frac{x^2}{y^2}+1\bigg)}$
$dx=\int \large\frac{d(x/y)}{1+(x/y)^2}$
$dx=d( \tan^{-1} (\frac{x}{y}))$
$x=\tan^{-1} (\frac{x}{y})+c$
$x-\tan^{-1} \bigg(\frac{x}{y}\bigg)=c$
Hence d is the correct answer.
answered Feb 3, 2014 by meena.p
 

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