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Solve $xy (1+ x y^2) \frac{dy}{dx} = 1$
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Solution:

$\therefore \frac{d x}{d y}=x y+x^{2} y^{3}$

$\therefore \frac{1}{x^{2}} \frac{d x}{d y}-\frac{1}{x} y=y^{3} \quad$ Now, put $-\frac{1}{x}=v$

$\therefore \frac{d v}{d y}+v y=y^{3} \quad \ldots \ldots \ldots \ldots\left(\frac{1}{x^{2}} \frac{d x}{d y}=\frac{d v}{d y}\right)$

This is linear differential eqn.

Integrating Factor=e $^{\int y d y}=e^{\frac{y^{2}}{2}}$

The solution of linear diff. eqn is given by,

$v .($ I.F. $)=\int(I . F .)(R . H . S)+c$

$=v e^{\frac{y^{2}}{2}}=\int e^{\frac{y^{2}}{2}} \cdot y^{3} d y+c$

$=-\frac{1}{x} e^{\frac{y^{2}}{2}}=e^{\frac{y^{2}}{2}}\left(y^{2}-2\right)+c$

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