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 $$ x=1 $$ ?

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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).
When $$ x = 1 $$, $$ y \approx 4.09 $$.

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 \( x=1 \) ?

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