Question

(Xy^2-e^(1/x^3))dx-x^2ydy=0

Answer 1

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• $\pink{\boxed{ \frac{1}{3}e^{\frac{1}{x^3}}- \frac{y^2}{2x^2} = C}}$

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$\to (xy^2-e^{\frac{1}{x^3}})dx - x^2ydy = 0$

$\to xy^2dx-e^{\frac{1}{x^3}}dx - x^2ydy = 0$

$\to xy^2dx - x^2ydy -e^{\frac{1}{x^3}}dx = 0$

$\to \frac{xy^2dx - x^2ydy}{x^4} -e^{\frac{1}{x^3}}\frac{dx} {x^4}= 0$

$\to \frac{1}{3}e^{\frac{1}{x^3}}\left(\frac{-3}{x^4}\right)dx -\frac{1}{2}\frac{2x^2ydy - 2xy^2dx}{{(x^2) }^{2}} = 0$

$\to \frac{1}{3}e^{\frac{1}{x^3}}\left(\frac{-3}{x^4}\right)dx -\frac{1}{2}d\left(\frac{y^2}{x^2}\right) = 0$

$\to \frac{1}{3}\int e^{\frac{1}{x^3}}\left(\frac{-3}{x^4}\right)dx -\frac{1}{2}\int d\left(\frac{y^2}{x^2}\right) = C$

Put $\frac{1}{x^3} = t$ in first integral.

$\implies \frac{-3}{x^4}dx = dt$

$\to \frac{1}{3}\int e^{t}dt -\frac{1}{2}\left(\frac{y^2}{x^2}\right) = C$

$\to \frac{1}{3}e^{t} -\frac{y^2}{2x^2} = C$

$\to \frac{1}{3}e^{\frac{1}{x^3}} -\frac{y^2}{2x^2}= C$