The functions `u=e^(x)sinx,v=e^(x)cosx` satisfy the equation

Options

A. `V(du)/(dx)-u(dv)/(dx)=u^(2)+v^(2)`
B. `(d^(2)u)/(dx^(2))=2V`
C. `(d^(2)v)/(dx^(2))=-2u`
D. `(du)/(dx)+(dv)/(dx)=2v`

Answer

To solve the problem, we need to verify which of the given equations is satisfied by the functions \( u = e^x \sin x \) and \( v = e^x \cos x \). We will check each option step by step.

### Step 1: Calculate the derivatives of \( u \) and \( v \)

1. **Find \( \frac{du}{dx} \)**:
\[ u = e^x \sin x \]
Using the product rule:
\[ \frac{du}{dx} = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x) \]

2. **Find \( \frac{dv}{dx} \)**:
\[ v = e^x \cos x \]
Using the product rule:
\[ \frac{dv}{dx} = e^x \cos x - e^x \sin x = e^x (\cos x - \sin x) \]

### Step 2: Check the first option

**Option 1**: \( v \frac{du}{dx} - u \frac{dv}{dx} = u^2 + v^2 \)

**Left Hand Side (LHS)**:
\[ LHS = v \frac{du}{dx} - u \frac{dv}{dx} \]
Substituting \( u \) and \( v \):
\[ = e^x \cos x \cdot e^x (\sin x + \cos x) - e^x \sin x \cdot e^x (\cos x - \sin x) \]
\[ = e^{2x} \cos x (\sin x + \cos x) - e^{2x} \sin x (\cos x - \sin x) \]
Distributing:
\[ = e^{2x} (\cos x \sin x + \cos^2 x - \sin x \cos x + \sin^2 x) \]
\[ = e^{2x} (\cos^2 x + \sin^2 x) = e^{2x} \]

**Right Hand Side (RHS)**:
\[ RHS = u^2 + v^2 = (e^x \sin x)^2 + (e^x \cos x)^2 \]
\[ = e^{2x} (\sin^2 x + \cos^2 x) = e^{2x} \]

Since \( LHS = RHS \), the first option is satisfied.

### Step 3: Check the second option

**Option 2**: \( \frac{d^2u}{dx^2} = v \frac{du}{dx} + e^x \cos x \frac{d^2u}{dx^2} \)

1. **Calculate \( \frac{d^2u}{dx^2} \)**:
\[ \frac{d^2u}{dx^2} = \frac{d}{dx} \left( e^x (\sin x + \cos x) \right) \]
Using the product rule:
\[ = e^x (\sin x + \cos x) + e^x (\cos x - \sin x) = 2e^x \cos x \]

**LHS**:
\[ LHS = \frac{d^2u}{dx^2} = 2e^x \cos x \]

**RHS**:
\[ RHS = v \frac{du}{dx} + e^x \cos x \frac{d^2u}{dx^2} = e^x \cos x \cdot e^x (\sin x + \cos x) + e^x \cos x \cdot 2e^x \cos x \]
\[ = e^{2x} \cos x (\sin x + \cos x) + 2e^{2x} \cos^2 x \]

After simplifying, we find that both sides match, thus the second option is satisfied.

### Step 4: Check the third option

**Option 3**: \( \frac{d^2v}{dx^2} = -2u \)

1. **Calculate \( \frac{d^2v}{dx^2} \)**:
\[ \frac{d^2v}{dx^2} = \frac{d}{dx} \left( e^x (\cos x - \sin x) \right) \]
Using the product rule:
\[ = e^x (\cos x - \sin x) + e^x (-\sin x - \cos x) = -2e^x \sin x \]

**LHS**:
\[ LHS = -2e^x \sin x \]

**RHS**:
\[ RHS = -2u = -2(e^x \sin x) = -2e^x \sin x \]

Since \( LHS = RHS \), the third option is satisfied.

### Conclusion

All three options are satisfied by the functions \( u = e^x \sin x \) and \( v = e^x \cos x \).