An integrating factor of the equation (1+y+x^ | vedclass

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(D) Given, the differential equation is: $(1+y+x^2 y) dx + (x+x^3) dy = 0$ Rearranging the terms: $(x+x^3) dy = -(1+y+x^2 y) dx$ $\frac{dy}{dx} = -\fr

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$x$

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(D) Given, the differential equation is: $(1+y+x^2 y) dx + (x+x^3) dy = 0$ Rearranging the terms: $(x+x^3) dy = -(1+y+x^2 y) dx$ $\frac{dy}{dx} = -\frac{1+y(1+x^2)}{x(1+x^2)}$ $\frac{dy}{dx} = -\frac{1}{x(1+x^2)} - \frac{y(1+x^2)}{x(1+x^2)}$ $\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$ 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 $(IF)$ is given by: $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$. Wait, let us re-evaluate the rearrangement. $(1+x^2)y dx + x(1+x^2) dy + dx = 0$ Dividing by $(1+x^2)$: $y dx + x dy + \frac{dx}{1+x^2} = 0$ $d(xy) + \frac{dx}{1+x^2} = 0$ This is an exact differential equation. However, looking at the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, the equation is $\frac{dy}{dx} + \frac{1}{x}y = -\frac{1}{x(1+x^2)}$. $IF = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. Thus, the integrating factor is $x$.

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

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