The integrating factor of the differential eq | vedclass

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(D) Given the differential equation: $y dx - (x + 2y^2) dy = 0$.Rearranging the terms,we get: $y dx - x dy = 2y^2 dy$.Dividing both sides by $y^2$ (wh

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$\frac{1}{y}$

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(D) Given the differential equation: $y dx - (x + 2y^2) dy = 0$.Rearranging the terms,we get: $y dx - x dy = 2y^2 dy$.Dividing both sides by $y^2$ (where $y 
eq 0$): $\frac{y dx - x dy}{y^2} = 2 dy$.Recognizing the derivative of the quotient: $d\left(\frac{x}{y}ight) = 2 dy$.Integrating both sides: $\int d\left(\frac{x}{y}ight) = \int 2 dy$.This implies $\frac{x}{y} = 2y + C$.Alternatively,rewriting the equation as $\frac{dx}{dy} - \frac{x}{y} = 2y$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y$.The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y|^{-1}} = \frac{1}{y}$.Thus,the correct option is $D$.

The integrating factor of the differential equation $y dx - (x + 2y^2) dy = 0$ is . . . . . . .

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