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General solution of differential equation: $x \large\frac{dy}{dx}$$+y=x^3 y^6$

$(a)\;1=y^3x^5 (2.5 +cx^2) \\ (b)\;1=y^5 x^3(2.5 + cx^2) \\ (c)\;1= y^3x^5(2.5 x^2+c) \\ (d)\;1=y^5x^3(2.5 x^2+c) $

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$\large\frac{x}{xy^6} \frac{y}{xy^6}$$=x^2$
$y^{-6} \large\frac{dy}{dx}$$+x^{-1}y{-5}=x^2$-----(i)
Put $y^{-5}=7$
$-5y^{-6}\large\frac{dy}{dx}=\frac{dz}{dx}$
$\therefore$ (i) become $\large\frac{-1}{5} \frac{dz}{dx}+\frac{7}{x}$$=x^2$
$\large\frac{dz}{dx}-\frac{5}{x}$$ z=-5x^2$---------(ii)
$ -\int \large\frac{5}{x} $$dx=e^{-5 \log x}=x-5$
I.f = $e^{-\int \large\frac{5}{x}dx}$
$\qquad= e^{-5 \log x}$
$\qquad= x^{-5}$
The solution of(ii) is
$z(I.F)=\int (-5x^2)$ I.f $dx+c$
$z x^{-5}= \int (-5x^2) x^{-5} dx+c$
$1= y^5x^3(2.5+cx^2)$
Hence b is the correct answer.

 

answered Feb 4, 2014 by meena.p
edited Feb 4, 2014 by meena.p
 

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