Let's analyze the claim **step by step**. ## Given: 1. \( e^x y'' + y' \sin(x) - (1 + x) y \geq 0 \) for \( x > 0 \) 2. \( y(0) \geq 0 \) 3. \( y'(0) > 0 \) 4. **Claim:** \( y(x) \) is an increasing function (i.e., \( y'(x) > 0 \) for all \( x > 0 \)). Let's check if this claim is true. --- ## **Step 1: Write the inequality for \( y'' \)** Given: \[ e^x y'' + y' \sin(x) - (1 + x) y \geq 0 \] \[ e^x y'' \geq (1 + x) y - y' \sin(x) \] \[ y'' \geq e^{-x} \left[(1 + x) y - y' \sin(x)\right] \] --- ## **Step 2: Analyze \( y'(x) \)** You are given \( y'(0) > 0 \). Let's see what happens for \( x > 0 \). Suppose \( y'(x) \) becomes zero at some \( x_1 > 0 \). - At \( x_1 \), \( y'(x_1) = 0 \). - Now, let's see what \( y''(x_1) \) would be: \[ y''(x_1) \geq e^{-x_1} (1 + x_1) y(x_1) \] - If \( y(x_1) > 0 \), then \( y''(x_1) > 0 \). - This means \( y'(x) \) has a **local minimum** at \( x_1 \), and since \( y'(x_1) = 0 \), \( y'(x) \) will increase for \( x > x_1 \), becoming positive again. - If \( y(x_1) = 0 \), then \( y''(x_1) \geq 0 \). - So, \( y'(x) \) will also increase for \( x > x_1 \). **So, even if \( y'(x) \) becomes zero, it immediately starts increasing and becomes positive.** --- ## **Step 3: Behavior for all \( x > 0 \)** - At \( x=0 \), \( y'(0) > 0 \). - If \( y'(x) \) were to become negative anywhere, it would have to cross zero. - But at any zero crossing, the second derivative is **non-negative** (or positive if \( y(x) > 0 \)), so \( y'(x) \) cannot remain negative. - Thus, \( y'(x) \) cannot become negative for any \( x > 0 \). --- ### **Conclusion** **Yes, your claim is true. \( y(x) \) is an increasing function for \( x > 0 \).** ### **Final Answer** \[ \boxed{\text{Yes, } y(x) \text{ is an increasing function for } x > 0.} \] **Reason:** Because if \( y'(x) \) tries to become zero or negative, the given inequality forces \( y'(x) \) to increase, so it cannot stay negative. Given \( y'(0) > 0 \), \( y'(x) > 0 \) for all \( x > 0 \).