The integrating factor of the differential eq | vedclass

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Question

(B) Given differential equation is $(2x + 3y^2) dy = y dx$. Dividing both sides by $y dy$,we get $\frac{dx}{dy} = \frac{2x + 3y^2}{y}$. Rearranging th

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$\frac{1}{y^2}$

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(B) Given differential equation is $(2x + 3y^2) dy = y dx$. Dividing both sides by $y dy$,we get $\frac{dx}{dy} = \frac{2x + 3y^2}{y}$. Rearranging the terms,we get $\frac{dx}{dy} - \frac{2}{y}x = 3y$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{2}{y}$ and $Q(y) = 3y$. The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy}$. $IF = e^{\int -\frac{2}{y} dy} = e^{-2 \ln|y|} = e^{\ln|y^{-2}|} = y^{-2} = \frac{1}{y^2}$.

The integrating factor of the differential equation $(2x + 3y^2) dy = y dx$ $(y > 0)$ is

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