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

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(D) Given differential equation is: $(1+y+x^2 y) dx + (x+x^3) dy = 0$. Rearranging the terms,we get: $(x+x^3) dy = -(1+y+x^2 y) dx$. Dividing by $dx(x

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$x$

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(D) Given differential equation is: $(1+y+x^2 y) dx + (x+x^3) dy = 0$. Rearranging the terms,we get: $(x+x^3) dy = -(1+y+x^2 y) dx$. 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,$\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$. 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 $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$. Wait,let us re-evaluate the rearrangement: $(1+y+x^2 y) dx + x(1+x^2) dy = 0$. Dividing by $dx(1+x^2)$,we get $\frac{1+y(1+x^2)}{1+x^2} dx + x dy = 0$. This is not standard. Let us rewrite as $\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$. Actually,the integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. Re-checking the options,the correct 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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