The given equation is d/dx(e^y) = 0. Using the chain rule of differentiation, we differentiate e^y with respect to x:
$$e^y \cdot \frac{dy}{dx} = 0$$The order of a differential equation is the order of the highest derivative present in the equation. Here, the highest derivative is dy/dx, which is a first-order derivative. Thus, the order is 1.
The degree of a differential equation is the power of the highest order derivative, provided the equation is a polynomial in terms of derivatives. In the equation e^y \cdot (dy/dx) = 0, the term e^y is not a polynomial in terms of dy/dx. However, since e^y is never zero, we can divide by it to get dy/dx = 0. This is a polynomial in dy/dx with power 1. Therefore, the degree is 1.
Final Answer: 1, 1
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