The degree of the differential equation y(x) | vedclass

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(C) Given the differential equation: $y = 1 + \frac{dy}{dx} + \frac{1}{2!} \left( \frac{dy}{dx} \right)^2 + \frac{1}{3!} \left( \frac{dy}{dx} \right)^

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$1$

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(C) Given the differential equation: $y = 1 + \frac{dy}{dx} + \frac{1}{2!} \left( \frac{dy}{dx} \right)^2 + \frac{1}{3!} \left( \frac{dy}{dx} \right)^3 + \dots$ Let $t = \frac{dy}{dx}$. The series expansion is $y = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \dots$ This is the Taylor series expansion for $e^t$,so $y = e^t$. Substituting $t = \frac{dy}{dx}$ back,we get $y = e^{\frac{dy}{dx}}$. Taking the natural logarithm on both sides: $\ln(y) = \frac{dy}{dx}$. The differential equation is now in the form $\frac{dy}{dx} = \ln(y)$. The highest order derivative is $\frac{dy}{dx}$,and its power is $1$. Therefore,the degree of the differential equation is $1$.

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

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