سوال ۲۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۳

We are given the integral equation

$f(x) = 3 + 2\int_{1}^{x} t\, f(t)\, dt, \qquad x \ge 1.$

Step 1 — Initial value. Setting $x = 1$ collapses the integral (lower and upper limits coincide), so

$f(1) = 3 + 2\int_{1}^{1} t\, f(t)\, dt = 3 + 0 = 3.$

Step 2 — Differentiate (Fundamental Theorem of Calculus). The integrand depends on $x$ only through the upper limit, so by the FTC / Leibniz rule

$f'(x) = 2\cdot \big[x\, f(x)\big] = 2x\, f(x).$

Step 3 — Solve the separable ODE. Separating variables,

$\frac{f'(x)}{f(x)} = 2x \;\Longrightarrow\; \ln|f(x)| = x^{2} + C \;\Longrightarrow\; f(x) = A\, e^{x^{2}}.$

Step 4 — Fix the constant. Using $f(1) = 3$:

$3 = A\, e^{1} \;\Longrightarrow\; A = 3e^{-1},$

so

$f(x) = 3e^{-1} e^{x^{2}} = 3\, e^{\,x^{2} - 1}.$

Step 5 — Evaluate at $x = 3$.

$f(3) = 3\, e^{\,3^{2} - 1} = 3\, e^{\,9 - 1} = 3e^{8}.$

This matches option 4.

Exam trick: Convert a Volterra-type integral equation $f(x)=c+\int_a^x g(t,f(t))\,dt$ into an ODE by differentiating, and read off the initial condition $f(a)=c$ from the empty-interval integral. Don't forget the $e^{-1}$ factor: the solution is $3e^{x^2-1}$, not $3e^{x^2}$.