Key Concepts and Formulas

• Leibniz Integral Rule: If $F(x) = \int_{a(x)}^{b(x)} g(t) \, dt$, then $F'(x) = g(b(x))b'(x) - g(a(x))a'(x)$.
• Fundamental Theorem of Calculus: $\frac{d}{dx} \int_a^x f(t) dt = f(x)$, where $a$ is a constant.
• Solving Differential Equations: Techniques for solving differential equations, including separation of variables or recognizing exact forms.

Step-by-Step Solution

Step 1: Differentiate both sides of the given equation with respect to $x$.

We have the equation $2(x+2)^2 f(x)-3(x+2)^2=10 \int_0^x(t+2) f(t) d t$ Differentiating both sides with respect to $x$ using Leibniz rule and the Fundamental Theorem of Calculus, we get: $\frac{d}{dx} \left[ 2(x+2)^2 f(x)-3(x+2)^2 \right] = \frac{d}{dx} \left[ 10 \int_0^x(t+2) f(t) d t \right]$ $2 \left[ (x+2)^2 f'(x) + 2(x+2)f(x) \right] - 6(x+2) = 10(x+2)f(x)$ $2(x+2)^2 f'(x) + 4(x+2)f(x) - 6(x+2) = 10(x+2)f(x)$

Step 2: Simplify the equation.

We can divide both sides by $2(x+2)$ (since $x \geq 0$, $x+2 \neq 0$): $(x+2)f'(x) + 2f(x) - 3 = 5f(x)$ $(x+2)f'(x) - 3f(x) = 3$

Step 3: Solve the differential equation.

We can rewrite the equation as: $f'(x) - \frac{3}{x+2} f(x) = \frac{3}{x+2}$ This is a first-order linear differential equation of the form $f'(x) + P(x)f(x) = Q(x)$, where $P(x) = -\frac{3}{x+2}$ and $Q(x) = \frac{3}{x+2}$. The integrating factor is given by: $I.F. = e^{\int P(x) dx} = e^{\int -\frac{3}{x+2} dx} = e^{-3 \ln(x+2)} = e^{\ln(x+2)^{-3}} = (x+2)^{-3}$ Multiplying the differential equation by the integrating factor, we get: $(x+2)^{-3} f'(x) - \frac{3}{(x+2)^4} f(x) = \frac{3}{(x+2)^4}$ $\frac{d}{dx} \left[ (x+2)^{-3} f(x) \right] = \frac{3}{(x+2)^4}$ Integrating both sides with respect to $x$: $\int \frac{d}{dx} \left[ (x+2)^{-3} f(x) \right] dx = \int \frac{3}{(x+2)^4} dx$ $(x+2)^{-3} f(x) = 3 \int (x+2)^{-4} dx = 3 \frac{(x+2)^{-3}}{-3} + C$ $(x+2)^{-3} f(x) = -(x+2)^{-3} + C$ Multiplying by $(x+2)^3$: $f(x) = -1 + C(x+2)^3$

Step 4: Determine the value of $C$ using the initial condition.

From the original equation, when $x = 0$: $2(0+2)^2 f(0) - 3(0+2)^2 = 10 \int_0^0 (t+2)f(t) dt$ $2(4) f(0) - 3(4) = 0$ $8 f(0) = 12$ $f(0) = \frac{12}{8} = \frac{3}{2}$ Substituting $x = 0$ in the general solution: $f(0) = -1 + C(0+2)^3$ $\frac{3}{2} = -1 + 8C$ $\frac{5}{2} = 8C$ $C = \frac{5}{16}$ Thus, $f(x) = -1 + \frac{5}{16}(x+2)^3$

Step 5: Calculate $f(2)$.

$f(2) = -1 + \frac{5}{16}(2+2)^3 = -1 + \frac{5}{16}(4)^3 = -1 + \frac{5}{16}(64) = -1 + 5(4) = -1 + 20 = 19$

Step 6: Re-examine the initial condition and calculations

Let's go back to the original equation: $2(x+2)^2 f(x) - 3(x+2)^2 = 10 \int_0^x (t+2)f(t) dt$ When we differentiated, we got $2(x+2)^2 f'(x) + 4(x+2)f(x) - 6(x+2) = 10(x+2)f(x)$ $(x+2)f'(x) + 2f(x) - 3 = 5f(x)$ $(x+2)f'(x) - 3f(x) = 3$ $f'(x) - \frac{3}{x+2}f(x) = \frac{3}{x+2}$ Integrating factor is $(x+2)^{-3}$ $(x+2)^{-3}f'(x) - 3(x+2)^{-4}f(x) = 3(x+2)^{-4}$ $\frac{d}{dx} [(x+2)^{-3} f(x)] = 3(x+2)^{-4}$ $(x+2)^{-3} f(x) = \int 3(x+2)^{-4} dx = 3 \frac{(x+2)^{-3}}{-3} + C = -(x+2)^{-3} + C$ $f(x) = -1 + C(x+2)^3$ At $x=0$, $f(0) = \frac{3}{2}$. $\frac{3}{2} = -1 + C(2)^3 = -1 + 8C$, so $8C = \frac{5}{2}$ and $C = \frac{5}{16}$. Then $f(x) = -1 + \frac{5}{16} (x+2)^3$. $f(2) = -1 + \frac{5}{16} (4)^3 = -1 + \frac{5}{16} (64) = -1 + 20 = 19$

There is an error in the problem statement or the given answer. Let's check if $f(2) = 4$ is correct: If $f(2) = 4$, then $4 = -1 + C(4)^3 = -1 + 64C$, so $64C = 5$, and $C = \frac{5}{64}$. Then $f(x) = -1 + \frac{5}{64} (x+2)^3$. Then $f(0) = -1 + \frac{5}{64} (8) = -1 + \frac{5}{8} = -\frac{3}{8}$. However, $f(0) = \frac{3}{2}$. Thus $f(2) = 4$ is not correct.

Let's recalculate $f(2)$ based on our derived $f(x) = -1 + \frac{5}{16}(x+2)^3$. $f(2) = -1 + \frac{5}{16} (2+2)^3 = -1 + \frac{5}{16} (4^3) = -1 + \frac{5}{16} (64) = -1 + 5(4) = -1 + 20 = 19$.

Step 7: We need to reconcile the answer. Let $f(x) = a$. Then $2(x+2)^2 a - 3(x+2)^2 = 10\int_0^x (t+2)a \, dt = 10a \int_0^x (t+2) dt = 10a [\frac{t^2}{2} + 2t]_0^x = 10a (\frac{x^2}{2} + 2x)$. Then $2(x+2)^2 a - 3(x+2)^2 = 5ax^2 + 20ax$. $2(x^2 + 4x + 4)a - 3(x^2 + 4x + 4) = 5ax^2 + 20ax$. $2ax^2 + 8ax + 8a - 3x^2 - 12x - 12 = 5ax^2 + 20ax$. Comparing coefficients: $x^2: 2a - 3 = 5a$, so $3a = -3$ and $a = -1$. $x: 8a - 12 = 20a$, so $12a = -12$ and $a = -1$. constant: $8a - 12 = 0$, so $8a = 12$ and $a = 3/2$. This is a contradiction.

There seems to be an error in the question or the answer provided. The derivation leads to $f(2)=19$.

Common Mistakes & Tips

• Careful Differentiation: Ensure correct application of the Leibniz rule and chain rule during differentiation.
• Integrating Factor: Double-check the calculation of the integrating factor and its application.
• Initial Conditions: Correctly use the initial condition derived from the original equation to find the constant of integration.

Summary

We differentiated the given integral equation and obtained a first-order linear differential equation. We solved this differential equation using the integrating factor method. Applying the initial condition $f(0)=\frac{3}{2}$, we obtained the particular solution $f(x) = -1 + \frac{5}{16}(x+2)^3$. Evaluating this at $x=2$, we found $f(2) = 19$. However, the provided answer is 4. There seems to be an error with the problem or the given answer. Based on our calculations, $f(2)=19$.

Final Answer

The final answer is \boxed{19}. There appears to be an error with the options provided.