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(C) The given equation is $x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \dots$ This is the Taylor series expansion of the

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$y = \pm \sqrt{(\log_e x)^2 + 2C}$

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(C) The given equation is $x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \dots$ This is the Taylor series expansion of the exponential function $e^u$,where $u = xy\frac{dy}{dx}$. Thus,the equation can be written as $x = e^{xy\frac{dy}{dx}}$. Taking the natural logarithm on both sides,we get $\log_e x = xy\frac{dy}{dx}$. Rearranging the terms to separate the variables,we have $y \, dy = \frac{\log_e x}{x} \, dx$. Integrating both sides,$\int y \, dy = \int \frac{\log_e x}{x} \, dx$. Let $t = \log_e x$,then $dt = \frac{1}{x} \, dx$. So,$\frac{y^2}{2} = \frac{(\log_e x)^2}{2} + C$. Multiplying by $2$,we get $y^2 = (\log_e x)^2 + 2C$. Therefore,$y = \pm \sqrt{(\log_e x)^2 + 2C}$.

The solution of the differential equation $x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{(xy)^3}{3!}\left(\frac{dy}{dx}\right)^3 + \dots$ is

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