Differential Equations
PYP_JEE_ADV_2023_P2
Grade None
Question:
Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function such that $f(1) = \frac{1}{3}$ and $$3 \int_1^x f(t) dt = x f(x) - \frac{x^3}{3}, x \in [1, \infty).$$ Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is
$\frac{e^2 + 4}{3}$
$\frac{\log_e 4 + e}{3}$
$\frac{4e^2}{3}$
$\frac{e^2 - 4}{3}$
Step-by-Step Solution
Key Concept: Using an integrating factor to solve a first-order linear ordinary differential equation, and applying boundary conditions.
**Step 1: Differentiate both sides**
Differentiating $3 \int_1^x f(t) dt = x f(x) - \frac{x^3}{3}$ with respect to $x$ using Leibniz's rule gives $3f(x) = f(x) + x f'(x) - x^2$. This simplifies to $x f'(x) - 2f(x) = x^2$.
**Step 2: Solve the linear differential equation**
Rewrite as $f'(x) - \frac{2}{x} f(x) = x$. The integrating factor is $e^{\int -2/x \,dx} = x^{-2} = \frac{1}{x^2}$. Multiplying by the integrating factor gives $\frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = \frac{1}{x}$. Integrating both sides yields $\frac{f(x)}{x^2} = \ln x + C$ for $x \ge 1$.
**Step 3: Apply initial condition**
Since $f(1) = 1/3$, we have $1/3 = \ln(1) + C$, so $C = 1/3$. The function is $f(x) = x^2 \left(\ln x + \frac{1}{3}\right)$.
**Step 4: Calculate f(e)**
Substitute $x = e$ to get $f(e) = e^2 \left(1 + \frac{1}{3}\right) = \frac{4e^2}{3}$.
Correct Answer: 3