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Q)

Find the solution : $ (1+x^4) \large\frac{dy}{dx}$$-4x^3y =(x^5+x) \tan ^{-1} (x^2) \qquad y(1)=\pi$

$(a)\;y=\frac{1+x^4)}{4} \tan ^{-1} (x^2)^2+c \\ (b)\;y=\frac{1+x^4)}{4} \tan ^{-1} x^2+c \\ (c)\;y=x^4 \tan ^{-1} (x^2)^2+c\\ (d)\;y=(1+x^4)^2+c $

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A)
$\large\frac{dy}{dx} -\frac{4x^3y}{1+x^4} $$=x \tan ^{-1} (x^2)$
I.F= $e^{-\int \large\frac{4x^3}{1+x^4}dx}$
$\qquad= e^{- \log (1+x^4)}$
$\qquad= \large\frac{1}{1+x^4}$
So, $y. \large\frac{1}{1+x^4}=\int \large\frac{x \tan ^{-1} (x^2)dx}{1+x^4}$
$y. \large\frac{1}{1+x^4} =\large\frac{1}{2} \frac{(\tan ^{-1} (x^2))^2}{2}$$+c$
$y= \large\frac{(1+x^4)}{4} $$\tan^{-1}(x^2)^2+c$
Hence a is the correct answer.
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