Key Concepts and Formulas

• Separation of Variables: A technique to solve first-order differential equations of the form $\frac{dy}{dx} = f(x)g(y)$ by separating the variables and integrating.
• Integration: The process of finding the integral of a function. Key integrals include $\int \frac{1}{u} du = \ln|u| + C$ and $\int dx = x + C$.
• Initial Value Problem: Finding a particular solution to a differential equation that satisfies a given initial condition $y(x_0) = y_0$.

Step-by-Step Solution

Step 1: Separate the Variables

• Explanation: We want to rewrite the given differential equation $\frac{dy}{dx} = y + 3$ so that all terms involving $y$ are on one side with $dy$, and all terms involving $x$ are on the other side with $dx$. This allows us to integrate both sides independently.
• Action: Divide both sides by $(y+3)$ and multiply both sides by $dx$.
• Constraint Consideration: The problem states $y+3 > 0$. This is important because it ensures that we are not dividing by zero and that the argument of any logarithm we encounter will be positive.

$\frac{dy}{y+3} = dx$

Step 2: Integrate Both Sides

• Explanation: Now that the variables are separated, we integrate both sides with respect to their respective variables to find the general solution for $y(x)$.
• Action: Integrate the left side with respect to $y$ and the right side with respect to $x$. Remember to add a constant of integration, $C$, on one side (typically the side with $x$).

$\int \frac{dy}{y+3} = \int dx$

• Integration:

  • The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u|$. So, $\int \frac{dy}{y+3} = \ln|y+3|$.
  • The integral of $1$ with respect to $x$ is $x$. So, $\int dx = x$.

  This gives us: $\ln|y+3| = x + C$

• Applying the Constraint: Since the problem states $y+3 > 0$, we can remove the absolute value sign. $\ln(y+3) = x + C$

Step 3: Apply the Initial Condition to Find the Constant $C$

• Explanation: The solution we found in Step 2 is a general solution because it contains an arbitrary constant $C$. To find the particular solution that satisfies the given conditions, we use the initial condition $y(0)=2$. This means when $x=0$, $y=2$.
• Action: Substitute $x=0$ and $y=2$ into our general solution.

$\ln(2+3) = 0 + C$ $\ln 5 = C$

Step 4: Formulate the Particular Solution

• Explanation: Now that we have the value of $C$, we substitute it back into the general solution to get the unique function $y(x)$ that satisfies both the differential equation and the initial condition.
• Action: Substitute $C = \ln 5$ into $\ln(y+3) = x + C$.

$\ln(y+3) = x + \ln 5$

Step 5: Solve for y(x)

• Explanation: To evaluate $y\left( {\ln 2} \right)$, we need to find an explicit expression for $y(x)$.
• Action: Exponentiate both sides of the equation using the exponential function $e^x$. $e^{\ln(y+3)} = e^{x + \ln 5}$ $y+3 = e^x e^{\ln 5}$ $y+3 = 5e^x$ $y = 5e^x - 3$

Step 6: Evaluate y(ln 2)

• Explanation: Now that we have the explicit function $y(x)$, we can evaluate it at $x = \ln 2$.
• Action: Substitute $x = \ln 2$ into the expression for $y(x)$. $y(\ln 2) = 5e^{\ln 2} - 3$ $y(\ln 2) = 5(2) - 3$ $y(\ln 2) = 10 - 3$ $y(\ln 2) = 7$
• Note: This value does not match the given correct answer. We need to review.

Step 5 (Revised): Solve for y(x) - Alternative Approach

• Explanation: An alternative to isolating $y$ directly is to use properties of logarithms to rewrite the equation before exponentiating. This may reduce errors.
• Action: From $\ln(y+3) = x + \ln 5$, exponentiate both sides to obtain $y+3 = e^{x + \ln 5} = e^x e^{\ln 5} = 5e^x$. Then $y = 5e^x - 3$. This is the same as before.

Step 6 (Revised): Evaluate y(ln 2)

• Explanation: We still have the correct expression $y = 5e^x - 3$. Let's re-evaluate at $x = \ln 2$.
• Action: Substitute $x = \ln 2$ into the expression for $y(x)$. $y(\ln 2) = 5e^{\ln 2} - 3$ $y(\ln 2) = 5(2) - 3$ $y(\ln 2) = 10 - 3$ $y(\ln 2) = 7$

It appears there is an error in the provided answer.

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always remember to add a constant of integration when performing indefinite integrals.
• Incorrectly Applying Logarithm/Exponential Properties: Be careful when manipulating logarithmic and exponential functions. Double-check your application of the properties.
• Not Checking the Constraint: The condition $y+3 > 0$ is important for justifying the removal of absolute value signs.

Summary

We solved the given first-order differential equation using separation of variables. We integrated both sides, applied the initial condition to find the constant of integration, and obtained an explicit expression for $y(x)$. Finally, we evaluated $y(\ln 2)$. The result is 7. However, since this is a multiple-choice question and 7 is option (D) but the given answer is (A) 5, there is likely an error in the problem statement or provided answer. It appears the correct answer should be 7. I will proceed as if the answer is 7.

Final Answer The final answer is \boxed{7}, which corresponds to option (D).