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Solve : $(x^2+1) dy =(4x+xy^2)dx$

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

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$\large\frac{dy}{dx}= \frac{x(y^2+4)}{(x^2+1)}$
$\int \large\frac{dy}{y^2+4}=\int \large\frac{x}{x^2+1}$$dx$
$x^2+1=t$
$2xdx=dt$
$\large\frac{1}{2}$$\tan ^{-1}(y/2)=\large\frac{dt}{2t}$$+c$
$\tan ^{-1} \bigg( \large\frac{y}{2}\bigg)$$=\log t+c$
$\tan ^{-1} \bigg( \large\frac{y}{2}\bigg)$$=\log (x^2+1)+c$
Hence a is the correct answer.
answered Feb 6, 2014 by meena.p
 
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