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(A) Given the equation: $2(x+2)^2 f(x) - 3(x+2)^2 = 10 \int_0^x (t+2) f(t) dt$. At $x=0$,$2(2)^2 f(0) - 3(2)^2 = 10(0) \implies 8 f(0) = 12 \implies f

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) Given the equation: $2(x+2)^2 f(x) - 3(x+2)^2 = 10 \int_0^x (t+2) f(t) dt$. At $x=0$,$2(2)^2 f(0) - 3(2)^2 = 10(0) \implies 8 f(0) = 12 \implies f(0) = \frac{3}{2}$. Differentiating both sides with respect to $x$ using the Leibniz rule: $4(x+2) f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2) f(x)$. Divide by $2(x+2)$ (assuming $x \geq 0$): $2 f(x) + (x+2) f'(x) - 3 = 5 f(x)$. $(x+2) f'(x) - 3 f(x) = 3$. This is a linear differential equation: $\frac{df}{dx} - \frac{3}{x+2} f = \frac{3}{x+2}$. Integrating factor $IF = e^{\int -\frac{3}{x+2} dx} = e^{-3 \ln(x+2)} = (x+2)^{-3}$. Multiplying by $IF$: $\frac{d}{dx} [f(x) (x+2)^{-3}] = 3(x+2)^{-4}$. Integrating both sides: $f(x) (x+2)^{-3} = \int 3(x+2)^{-4} dx = -(x+2)^{-3} + C$. $f(x) = -1 + C(x+2)^3$. Using $f(0) = \frac{3}{2}$: $\frac{3}{2} = -1 + C(2)^3 \implies \frac{5}{2} = 8C \implies C = \frac{5}{16}$. Thus,$f(x) = \frac{5}{16}(x+2)^3 - 1$. For $f(2)$: $f(2) = \frac{5}{16}(4)^3 - 1 = \frac{5}{16}(64) - 1 = 5(4) - 1 = 19$.

Let $f$ be a differentiable function such that $2(x+2)^2 f(x) - 3(x+2)^2 = 10 \int_0^x (t+2) f(t) dt$ for $x \geq 0$. Then $f(2)$ is equal to . . . . . .

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