The degree of the differential equation x = 1 | vedclass

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(C) The given differential equation is $x = 1 + \left(\frac{dy}{dx}\right) + \frac{1}{2!} \left(\frac{dy}{dx}\right)^2 + \frac{1}{3!} \left(\frac{dy}{

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$1$

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(C) The given differential equation is $x = 1 + \left(\frac{dy}{dx}\right) + \frac{1}{2!} \left(\frac{dy}{dx}\right)^2 + \frac{1}{3!} \left(\frac{dy}{dx}\right)^3 + \dots$ This series is the expansion of the exponential function $e^{\frac{dy}{dx}}$. Thus, the equation can be written as $x = e^{\frac{dy}{dx}}$. Taking the natural logarithm on both sides, we get $\ln(x) = \frac{dy}{dx}$. The differential equation is $\frac{dy}{dx} = \ln(x)$. In this equation, the highest order derivative is $\frac{dy}{dx}$, which is of order $1$. The power of the highest order derivative is $1$. Therefore, the degree of the differential equation is $1$.

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

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