Unit 7 Test - Differential Equations 11. Consider the differential equation \( \frac{d y}{d x}=\frac{e^{x}-1}{2 y} \). If \( y=4 \) when \( x=0 \) what is a value of \( y \) when \( x=1 \) ?

To solve the differential equation: \[ \frac{dy}{dx} = \frac{e^x - 1}{2y} \] with the initial condition \( y = 4 \) when \( x = 0 \), follow these steps: 1. **Separate the Variables:** Multiply both sides by \( 2y \) and \( dx \) to separate variables: \[ 2y\, dy = (e^x - 1)\, dx \] 2. **Integrate Both Sides:** Integrate both sides to find the general solution: \[ \int 2y\, dy = \int (e^x - 1)\, dx \] This gives: \[ y^2 = e^x - x + C \] 3. **Apply the Initial Condition:** Use the initial condition \( y = 4 \) when \( x = 0 \) to find the constant \( C \): \[ 4^2 = e^0 - 0 + C \quad \Rightarrow \quad 16 = 1 + C \quad \Rightarrow \quad C = 15 \] Thus, the particular solution is: \[ y^2 = e^x - x + 15 \] 4. **Find \( y \) When \( x = 1 \):** Substitute \( x = 1 \) into the equation: \[ y^2 = e^1 - 1 + 15 = e + 14 \] Therefore, \( y = \sqrt{e + 14} \). Numerically, since \( e \approx 2.71828 \): \[ y \approx \sqrt{2.71828 + 14} \approx \sqrt{16.71828} \approx 4.09 \] **Final Answer:** When \( x = 1 \), \( y = \sqrt{e + 14} \) (approximately 4.09).