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Find solution : $\large\frac{dy}{dx}$$=y (x^2y^2-1)$

$(a)\;y=(x^2-2x+2+ce^{3x})^{1/3} \\ (b)\;y= (x^2-2+ce^{3x})^3 \\ (c)\;y=(x^2-2x+2+ce^{3x})^{-1/3} \\ (d)\;y=x^2-2x+2+c $

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This is Bernoulli equation
$\large\frac{dy}{dx}$$=x^2y^4-y$
$\large\frac{dy}{dx}$$+y =x^2y^4$
$\large\frac{1}{y_4} \bigg( \large\frac{dy}{dx} \bigg) +\large\frac{1}{y^3}$$=x^2$
$\large\frac{1}{y^3}$$=t$
$-3y^{-4} \large\frac{dy}{dx}=\frac{dt}{dx}$
$\large\frac{-1}{3} \frac{dt}{dx}$$+t= x^2$
$\large\frac{dt}{dx}$$-3t=-3x^2$
Integrating factor $=e^{\int -3 dx}$
$\qquad= e^{-3x}$
So $t.e^{-3x}=\int -3x^2.e^{-3x}dx$
by parts, $t.e^{-3x}=x^2 e^{-3x} -2x.e^{-3x}+2e^{-3x}+c$
$y^{-3}.e^{-3x}=x^2.e^{-3x}-2x .e^{-3x}+ 2e^{-3x}+c$
$y^{-3}=(x^2-2x+2)+ce^{3x}$
$y= (x^2-2x+2+ce^{3x})^{-\frac{1}{3}}$
Hence c is the correct answer.
answered Feb 6, 2014 by meena.p
 

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