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. \( y(0) \geq 0 \) and \( y'(0) > 0 \) 3. The claim: **\( y(x) \) is an increasing function for \( x > 0 \).** --- ## **Step 1: Simplify the Differential Inequality** The given inequality: \[ e^x y' + y' \sin(x) - (1 + x)y \geq 0 \] Factor \( y' \): \[ y' (e^x + \sin(x)) \geq (1 + x)y \] Or, \[ y' \geq \frac{(1 + x)y}{e^x + \sin(x)} \] (provided \( e^x + \sin(x) > 0 \); for all \( x > 0 \), \( e^x > 0 \), so the denominator is always positive.) --- ## **Step 2: What is Required for \( y(x) \) to be Increasing?** A function is **increasing** if \( y'(x) > 0 \) for all \( x > 0 \). From above, \[ y'(x) \geq \frac{(1 + x)y(x)}{e^x + \sin(x)} \] So, if \( y(x) > 0 \) for all \( x > 0 \), then \( y'(x) \geq 0 \) (since denominator is positive and numerator is positive). --- ## **Step 3: Is \( y(x) > 0 \) for all \( x > 0 \)?** - At \( x = 0 \), \( y(0) \geq 0 \). - \( y'(0) > 0 \). Let's analyze the **possibility of \( y(x) \) becoming negative**: Suppose there is some \( x_1 > 0 \) where \( y(x_1) = 0 \) and \( y(x) > 0 \) for \( x < x_1 \). - At \( x_1 \): \[ y'(x_1) \geq \frac{(1 + x_1) \cdot 0}{e^{x_1} + \sin(x_1)} = 0 \] Thus, \( y'(x_1) \geq 0 \). This means, at any point where \( y(x) = 0 \), the function is not decreasing (slope is non-negative). Since \( y'(0) > 0 \), the function starts by increasing from \( x=0 \). By the **uniqueness theorem** and the fact that at any zero point the slope is non-negative, once \( y(x) > 0 \) for some open interval, it cannot cross below zero because the slope cannot be negative at zero. --- ## **Step 4: Conclusion** - \( y(0) \geq 0 \), \( y'(0) > 0 \) ⇒ \( y(x) > 0 \) for \( x > 0 \) (at least for small \( x \)), and it cannot drop below zero. - For \( x > 0 \), \[ y'(x) \geq \frac{(1 + x)y(x)}{e^x + \sin(x)} > 0 \] because numerator and denominator are both positive. So, **your claim is true**: \[ \boxed{y(x) \text{ is an increasing function for } x > 0.} \] --- ### **Final Answer** **Yes, your claim is true. \( y(x) \) is an increasing function for \( x > 0 \) given the conditions.**