To solve the given differential equation, observe that the equation is an infinite series similar to the Taylor expansion of the exponential function. Let's examine the equation:
\(x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{(xy)^3}{3!}\left(\frac{dy}{dx}\right)^3 + \ldots\)
This can be rewritten using the concept of exponential series:
\(x = \exp(xy\frac{dy}{dx})\)
Assuming \(xy\frac{dy}{dx} = z\), the series becomes:
\(x = e^z\)
Now, using \(z = xy\frac{dy}{dx}\), the differential equation transforms as follows:
Since \(e^z = x\), we have \(z = \log x\).
Therefore,
\(xy\frac{dy}{dx} = \log x\)
Separate the variables to solve:
\(y\frac{dy}{dx} = \frac{\log x}{x}\)
Integrate both sides:
\(\int y\frac{dy}{dx} \, dx = \int \frac{\log x}{x} \, dx\)
Solving the left side:
\(\int y \, dy = \frac{y^2}{2} + C_1\)
Solving the right side using by parts (let \(u = \log x, \, dv = \frac{1}{x} \, dx\)):
\(\int \log x \, d(\log x) = \frac{(\log x)^2}{2} + C_2\)
Setting both integrals equal gives:
\(\frac{y^2}{2} = \frac{(\log x)^2}{2} + C\)
Rearrange the equation:
\(y^2 = (\log x)^2 + 2C = (\log x)^2 + C'\).
Pursue a square root, and include the \(\pm\) sign due to the nature of square roots:
\(y = \pm e^{\frac{(\log x)^2}{2}} + C\)
Thus, the correct solution to the differential equation is:
Correct Answer: \(y = \pm e^{\frac{(\log x)^2}{2}} + C\)