Question

(Xe^xy+2y)dy/dx+ye^xy=0









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Answer 1

Step-by-step explanation:

We have,

$\left( x \: {e}^{xy} + 2y\right) \dfrac{dy}{dx} + y \: {e}^{xy} = 0$

$\implies \left( x \: {e}^{xy} + 2y\right) dy + y \: {e}^{xy} \: dx = 0$

$\implies x \: {e}^{xy} \: dy+ 2y \: dy + y \: {e}^{xy} \: dx = 0$

$\implies {e}^{xy} \left( x \: dy+ y \: dx \right) + 2y \: dy= 0$

$\implies {e}^{xy} \: d(xy) + 2y \: dy= 0$

On integrating,

$\displaystyle \implies \int {e}^{xy} \: d(xy) + \int 2y \: dy= \int0$

$\displaystyle \implies {e}^{xy} +2 \int y \: dy= c$

$\displaystyle \implies {e}^{xy} +2 \cdot \dfrac{ y^{2} }{2} = c$

$\displaystyle \implies {e}^{xy} +y^{2} = c$

$\boxed{ {y}^{2} + {e}^{xy} = c }$