If `(dy)/(dx)=y+3 and y(0)=2`, then y(ln 2) is equal to

Options

A. 7
B. 5
C. 13
D. `-2`

Explanation

To solve the differential equation \(\frac{dy}{dx} = y + 3\) with the initial condition \(y(0) = 2\) and find \(y(\ln 2)\), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ \frac{dy}{dx} = y + 3 \]
We can separate the variables by rewriting it as:
\[ \frac{dy}{y + 3} = dx \]

**Hint**: To separate variables, we want all \(y\) terms on one side and \(x\) terms on the other.

### Step 2: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{dy}{y + 3} = \int dx \]
The left side integrates to:
\[ \ln |y + 3| = x + C \]
where \(C\) is the constant of integration.

**Hint**: Remember that the integral of \(\frac{1}{u}\) is \(\ln |u|\).

### Step 3: Solve for the Constant of Integration
We use the initial condition \(y(0) = 2\) to find \(C\). Substituting \(x = 0\) and \(y = 2\) into the equation gives:
\[ \ln |2 + 3| = 0 + C \implies \ln 5 = C \]

**Hint**: Plug in the initial condition to find the constant.

### Step 4: Substitute Back the Constant
Now we substitute \(C\) back into our equation:
\[ \ln |y + 3| = x + \ln 5 \]

**Hint**: Make sure to keep track of the logarithmic properties.

### Step 5: Exponentiate Both Sides
To eliminate the logarithm, we exponentiate both sides:
\[ |y + 3| = e^{x + \ln 5} = 5e^x \]
Since \(y + 3\) is positive (as we will see), we can drop the absolute value:
\[ y + 3 = 5e^x \]

**Hint**: Remember that \(e^{\ln a} = a\).

### Step 6: Solve for \(y\)
Now we solve for \(y\):
\[ y = 5e^x - 3 \]

**Hint**: Isolate \(y\) to express it in terms of \(x\).

### Step 7: Find \(y(\ln 2)\)
Now we need to find \(y(\ln 2)\):
\[ y(\ln 2) = 5e^{\ln 2} - 3 = 5 \cdot 2 - 3 = 10 - 3 = 7 \]

**Hint**: Use the property \(e^{\ln a} = a\) to simplify.

### Final Answer
Thus, the value of \(y(\ln 2)\) is:
\[ \boxed{7} \]