The given functions are \( u = e^x \sin x \) and \( v = e^x \cos x \). We need to verify which of the given equations these functions satisfy. Let's examine each option one by one.
First, let's calculate the derivatives of \( u \) and \( v \):
- \(\frac{du}{dx} = \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)\)
- \(\frac{dv}{dx} = \frac{d}{dx}(e^x \cos x) = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x)\)
Check the first option \( v\frac{du}{dx} - u\frac{dv}{dx} = u^2 + v^2 \):
Substitute the expressions:
- \(v\frac{du}{dx} = (e^x \cos x) \times e^x(\sin x + \cos x) = e^{2x}(\cos x \sin x + \cos^2 x)\)
- \(u\frac{dv}{dx} = (e^x \sin x) \times e^x(\cos x - \sin x) = e^{2x}(\sin x \cos x - \sin^2 x)\)
Check the second option \(\frac{d^2u}{dx^2} = 2v\):
- We already have \( \frac{du}{dx} = e^x(\sin x + \cos x) \).
- Thus, \(\frac{d^2u}{dx^2} = \frac{d}{dx}(e^x(\sin x + \cos x)) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x = 2v\).
Check the third option \(\frac{d^2v}{dx^2} = -2u\):
- \(\frac{dv}{dx} = e^x(\cos x - \sin x)\)
- Thus, \(\frac{d^2v}{dx^2} = \frac{d}{dx}(e^x(\cos x - \sin x)) = e^x(\cos x - \sin x) - e^x(\sin x + \cos x) = -2e^x \sin x = -2u\)
Since all three conditions are satisfied, the answer is All of these.