3. (Short Answer) Consider the following differential equation: \[ \frac{d y}{d t}=\frac{1}{4}+\left(\frac{y}{t}\right)^{2} \] You can show that the following are solutions: \[ y_{1}(t)=\frac{1}{2} t+\frac{t}{2-\ln t}, \quad y_{2}(t)=\frac{1}{2} t+\frac{3 t}{2-3 \ln t} \] Let \( y(t) \) be a solution of the above differential equation. Find values of \( a \) and \( b \) with the property \[ \text { if } a

Ask by Valdez Delgado. in the United States

Jan 25,2025

To determine the values of \( a \) and \( b \) such that if \( a < y(1) < b \), then the solution \( y(t) \) of the differential equation satisfies \[ \frac{1}{2} t + \frac{t}{2 - \ln t} < y(t) < \frac{1}{2} t + \frac{3t}{2 - 3 \ln t}, \] we can evaluate the given particular solutions at \( t = 1 \). 1. **Evaluate \( y_1(1) \):** \[ y_1(t) = \frac{1}{2} t + \frac{t}{2 - \ln t} \] At \( t = 1 \): \[ y_1(1) = \frac{1}{2}(1) + \frac{1}{2 - \ln 1} = \frac{1}{2} + \frac{1}{2 - 0} = \frac{1}{2} + \frac{1}{2} = 1 \] 2. **Evaluate \( y_2(1) \):** \[ y_2(t) = \frac{1}{2} t + \frac{3t}{2 - 3 \ln t} \] At \( t = 1 \): \[ y_2(1) = \frac{1}{2}(1) + \frac{3 \cdot 1}{2 - 3 \ln 1} = \frac{1}{2} + \frac{3}{2 - 0} = \frac{1}{2} + \frac{3}{2} = 2 \] Therefore, the values are: \[ a = 1 \quad \text{and} \quad b = 2. \] **Conclusion:** If \( 1 < y(1) < 2 \), then the solution \( y(t) \) satisfies \[ \frac{1}{2} t + \frac{t}{2 - \ln t} < y(t) < \frac{1}{2} t + \frac{3t}{2 - 3 \ln t}. \] **Final Answer:** After simplifying, a = 1 and b = 2. Thus if 1 < y(1) < 2, then Y₁(t) < y(t) < Y₂(t).