The solution of $(x^{2} y^{3} + x^{2}) d x + (y^{2} x^{3} + y^{2}) d y = 0$ is

(x^{3} + 1) (y^{3} + 1) = c

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(x^{3} - 1) (y^{3} - 1) = c

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(x^{3} - 1) (y^{3} + 1) = c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x^{3} + 1) (y^{3} - 1) = c

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Solution

The correct option is A $(x^{3} + 1) (y^{3} + 1) = c$
$(x^{2} y^{3} + x^{2}) d x + (y^{2} x^{3} + y^{2}) d y = 0$
$\Rightarrow x^{2} (1 + y^{3}) d x + y^{2} (1 + x^{3}) d y = 0$
$\Rightarrow y^{2} (1 + x^{3}) d y = - x^{2} (1 + y^{3}) d x$
$\Rightarrow \frac{1}{3} \int \frac{3 y^{2}}{y^{3} + 1} d y = \frac{1}{3} \int - \frac{3 x^{2}}{x^{3} + 1} d x$
$\Rightarrow \frac{1}{3} log (y^{3} + 1) = - \frac{1}{3} log (x^{3} + 1) + log c$
$\Rightarrow log (y^{3} + 1) + log (x^{3} + 1) = log c$
$\Rightarrow (x^{3} + 1) (y^{3} + 1) = c$

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The solution of $(x^{2} y^{3} + x^{2}) d x + (y^{2} x^{3} + y^{2}) d y = 0$ is