The solution of the differential equation,y,d | vedclass

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(B) Given the differential equation: $y\,dx + (x + x^2y)dy = 0$Rearranging the terms,we get: $y\,dx + x\,dy + x^2y\,dy = 0$$(y\,dx + x\,dy) = -x^2y\,d

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$-\frac{1}{xy} + \log y = c$

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(B) Given the differential equation: $y\,dx + (x + x^2y)dy = 0$Rearranging the terms,we get: $y\,dx + x\,dy + x^2y\,dy = 0$$(y\,dx + x\,dy) = -x^2y\,dy$Dividing both sides by $x^2y^2$ (assuming $x, y 
eq 0$):$\frac{y\,dx + x\,dy}{x^2y^2} = -\frac{dy}{y}$Recognizing that $d(xy) = y\,dx + x\,dy$,the equation becomes:$\frac{d(xy)}{(xy)^2} = -\frac{dy}{y}$Integrating both sides:$\int (xy)^{-2} d(xy) = -\int \frac{1}{y} dy$$-\frac{1}{xy} = -\log|y| + c$Rearranging the terms,we get:$-\frac{1}{xy} + \log|y| = c$

The solution of the differential equation,$y\,dx + (x + x^2y)dy = 0$ is

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