The integrating factor of the differential eq | vedclass

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(D) The given differential equation is $(1 - y^2)\frac{dx}{dy} + yx = ay$.Dividing by $(1 - y^2)$,we get $\frac{dx}{dy} + \frac{y}{1 - y^2}x = \frac{a

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

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(D) The given differential equation is $(1 - y^2)\frac{dx}{dy} + yx = ay$.Dividing by $(1 - y^2)$,we get $\frac{dx}{dy} + \frac{y}{1 - y^2}x = \frac{ay}{1 - y^2}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{y}{1 - y^2}$ and $Q(y) = \frac{ay}{1 - y^2}$.The integrating factor $(I.F.)$ is given by $e^{\int P(y) dy}$.$I.F. = e^{\int \frac{y}{1 - y^2} dy}$.Let $1 - y^2 = t$,then $-2y dy = dt$,or $y dy = -\frac{1}{2} dt$.$I.F. = e^{-\frac{1}{2} \int \frac{1}{t} dt} = e^{-\frac{1}{2} \ln|t|} = e^{\ln|t|^{-1/2}} = |t|^{-1/2} = \frac{1}{\sqrt{1 - y^2}}$.Thus,the correct option is $D$.

The integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay$ for $(-1 < y < 1)$ is

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