The integrating factor of the differential eq | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = -(3x^2 	an^{-1} y - x^3)(1 + y^2)$.Rearranging the terms: $\frac{dy}{dx} = x^3(1 + y^2) - 3x^2(

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$e^{x^3}$

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(B) Given the differential equation: $\frac{dy}{dx} = -(3x^2 	an^{-1} y - x^3)(1 + y^2)$.Rearranging the terms: $\frac{dy}{dx} = x^3(1 + y^2) - 3x^2(	an^{-1} y)(1 + y^2)$.Dividing both sides by $(1 + y^2)$: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} = x^3 - 3x^2 	an^{-1} y$.Rearranging to standard form: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} + 3x^2 	an^{-1} y = x^3$.Let $t = 	an^{-1} y$,then $\frac{dt}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx}$.The equation becomes: $\frac{dt}{dx} + 3x^2 t = x^3$.This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = 3x^2$.The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 3x^2 dx} = e^{x^3}$.

The integrating factor of the differential equation $\frac{dy}{dx} + (3x^2 \tan^{-1} y - x^3)(1 + y^2) = 0$ is

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