Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are continuous functions of $x$.
• Integrating Factor (I.F.): The integrating factor is given by I.F. $= e^{\int P(x) dx}$.
• General Solution: The general solution of a first-order linear differential equation is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Transform the given differential equation into the standard linear form.

The given differential equation is: $(x + 1)dy = ((x + 1)^2 + y – 3)dx$ Our goal is to rewrite this in the form $\frac{dy}{dx} + P(x)y = Q(x)$.

1. Isolate $\frac{dy}{dx}$: Divide both sides by $(x+1)dx$: $\frac{dy}{dx} = \frac{(x + 1)^2 + y – 3}{x + 1}$

2. Separate terms and move the term containing $y$ to the left side: $\frac{dy}{dx} = \frac{(x + 1)^2}{x + 1} + \frac{y}{x + 1} - \frac{3}{x + 1}$ $\frac{dy}{dx} = (x + 1) + \frac{y}{x + 1} - \frac{3}{x + 1}$ $\frac{dy}{dx} - \frac{1}{x + 1}y = (x + 1) - \frac{3}{x + 1}$ This is now in the standard linear form.

Step 2: Identify $P(x)$ and $Q(x)$.

Comparing the equation $\frac{dy}{dx} - \frac{1}{x + 1}y = (x + 1) - \frac{3}{x + 1}$ with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we can identify: $P(x) = -\frac{1}{x + 1}$ $Q(x) = (x + 1) - \frac{3}{x + 1}$

Step 3: Calculate the Integrating Factor (I.F.).

The integrating factor is given by $\text{I.F.} = e^{\int P(x) dx}$. So, $\text{I.F.} = e^{\int -\frac{1}{x + 1} dx}$ $\text{I.F.} = e^{-\int \frac{1}{x + 1} dx}$ $\text{I.F.} = e^{-\ln|x + 1|}$ Since $x \ge 0$, $x+1 > 0$, so $|x+1| = x+1$. $\text{I.F.} = e^{-\ln(x + 1)}$ $\text{I.F.} = e^{\ln(x + 1)^{-1}}$ $\text{I.F.} = (x + 1)^{-1} = \frac{1}{x + 1}$

Step 4: Apply the General Solution Formula.

The general solution is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$. Substituting the values of I.F. and $Q(x)$, we get: $y \left( \frac{1}{x + 1} \right) = \int \left( (x + 1) - \frac{3}{x + 1} \right) \left( \frac{1}{x + 1} \right) dx + C$

Step 5: Perform the Integration.

Simplify the integrand: $\frac{y}{x + 1} = \int \left( \frac{x + 1}{x + 1} - \frac{3}{(x + 1)^2} \right) dx + C$ $\frac{y}{x + 1} = \int \left( 1 - \frac{3}{(x + 1)^2} \right) dx + C$ Integrate term by term: $\frac{y}{x + 1} = \int 1 \, dx - 3 \int (x + 1)^{-2} \, dx + C$ $\frac{y}{x + 1} = x - 3 \frac{(x + 1)^{-1}}{-1} + C$ $\frac{y}{x + 1} = x + \frac{3}{x + 1} + C$

Step 6: Use the Initial Condition to Find $C$.

We are given $y(2) = 0$. Substitute $x = 2$ and $y = 0$ into the general solution: $\frac{0}{2 + 1} = 2 + \frac{3}{2 + 1} + C$ $0 = 2 + \frac{3}{3} + C$ $0 = 2 + 1 + C$ $C = -3$

Step 7: Write the Particular Solution.

Substitute $C = -3$ back into the general solution: $\frac{y}{x + 1} = x + \frac{3}{x + 1} - 3$

Step 8: Evaluate $y(3)$.

We want to find $y(3)$, so substitute $x = 3$: $\frac{y}{3 + 1} = 3 + \frac{3}{3 + 1} - 3$ $\frac{y}{4} = 3 + \frac{3}{4} - 3$ $\frac{y}{4} = \frac{3}{4}$ $y = 3$ Thus, $y(3) = 3$.

Common Mistakes & Tips:

• Always rewrite the differential equation into the standard form before identifying $P(x)$ and $Q(x)$.
• Pay close attention to signs, especially when calculating the integrating factor and integrating.
• Do not forget to add the constant of integration, $C$, after performing the indefinite integration.

Summary:

We transformed the given differential equation into the standard linear form, calculated the integrating factor, found the general solution, and used the initial condition to determine the constant of integration. Finally, we evaluated $y(3)$ to find the answer.

The final answer is \boxed{3}.