The integrating factor of the differential eq | vedclass

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(B) Given differential equation is $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$. Rearranging the terms,we have $(x+x^{3})dy = -(1+y(1+x^{2}))dx$. Dividing by $d

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$x$

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(B) Given differential equation is $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$. Rearranging the terms,we have $(x+x^{3})dy = -(1+y(1+x^{2}))dx$. Dividing by $dx(x+x^{3})$,we get $\frac{dy}{dx} = -\frac{1+y(1+x^{2})}{x(1+x^{2})}$. This simplifies to $\frac{dy}{dx} = -\frac{1}{x(1+x^{2})} - \frac{y(1+x^{2})}{x(1+x^{2})}$. Thus,$\frac{dy}{dx} + \left(\frac{1}{x}\right)y = -\frac{1}{x(1+x^{2})}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = -\frac{1}{x(1+x^{2})}$. The integrating factor ($I$.$F$.) is given by $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.

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

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