Question

The functions \(u = e^x \sin x\) and \(v = e^x \cos x\) satisfy the equation

• A. \(v\frac{du}{dx} - u\frac{dv}{dx} = u^2 + v^2\)
• B. \(\frac{d^2u}{dx^2} = 2v\)
• C. \(\frac{d^2v}{dx^2} = -2u\)
• D. All of these

Hint:

Functions of the form \(e^{ax}\sin bx\) and \(e^{ax}\cos bx\) have cyclic derivatives.

Answer 1

The Correct Option is D

 To solve this problem, we are given two functions \( u = e^x \sin x \) and \( v = e^x \cos x \). We need to verify the options provided and determine if all or some of them are satisfied by the given functions.

1. First, let's find the derivative of \( u = e^x \sin x \). Using the product rule, we have:
2. Similarly, calculate the derivative of \( v = e^x \cos x \):
3. We now verify each of the options one-by-one:
4. **Option 1**: Verify \( v \frac{du}{dx} - u \frac{dv}{dx} = u^2 + v^2 \).
5. The expression on the right hand side \(u^2 + v^2\) simplifies to:
6. Thus, **Option 1** is satisfied: \( v \frac{du}{dx} - u \frac{dv}{dx} = u^2 + v^2 \) is true.
7. **Option 2**: Verify \( \frac{d^2u}{dx^2} = 2v \).
8. Thus, **Option 2** is satisfied: \( \frac{d^2u}{dx^2} = 2v \) is true.
9. **Option 3**: Verify \( \frac{d^2v}{dx^2} = -2u \).
10. Thus, **Option 3** is satisfied: \( \frac{d^2v}{dx^2} = -2u \) is true.

Since all three individual verifications hold, the final answer is "All of these".