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General solution of differential equation: $\large\frac{dy}{dx}$$+x \sin 2y=x^3 \cos ^2 y$

$(a)\; \tan y=\frac{1}{2}(x^2-1)e^{x^2}+c \\ (b)\;\tan y=\frac{1}{2}(x^2+1)e^{x^2}+ce^{x^2} \\ (c)\;\tan x=\frac{1}{2}(x^2+1)e^{y^2}+ce^{x^2} \\ (d)\;\tan x=\frac{1}{2}(x^2-1)+ce^{-x^2} $
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$\large\frac{dy}{dx}$$+x \sin 2y=x^3 \cos^2 y$
dividing by $\cos^2y$
$\sec^2 y \large\frac{dy}{dx}$$+2x \tan y=x^3$
$\tan y=t$
$\sec^2 y \large\frac{dy}{dx}=\frac{dt}{dx}$
$\large\frac{dt}{dx}$$+2xt=x^3$
$I.F=e^{\int 2xdx}=e^{x^2}$
Solution of $t.e^{x^2}=\int x^3 e^{x^2}dx+c$
$x^2=k \quad 2xdx=dk$
$(\tan y)e^{x^2}=\int \large\frac{1}{2} $$k.e^{k}dk- \int e^{k} dk+c$
$2 \tan y e^{x^2}=ke^k-e^k+c$
$2 \tan y e^{x^2}=(x^2-1)e^{x^2}+c$
$\tan x=\frac{1}{2}(x^2-1)+ce^{-x^2} $
Hence d is the correct answer.
answered Feb 4, 2014 by meena.p
 

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