Differential Equations – Leibniz Rule – JEE Main 24 January 2025 Shift 1

Solution:

Given the equation:
$$ 2(x+2)^{2}f(x)-3(x+2)^{2}=10\int_{0}^{x}(t+2)f(t)dt $$
Differentiate both sides with respect to $x$:
$$ \frac{d}{dx}[2(x+2)^2 f(x)] – \frac{d}{dx}[3(x+2)^2] = \frac{d}{dx}\left[10\int_{0}^{x}(t+2)f(t)dt\right] $$
Using the product rule and Leibniz integral rule:
$$ [4(x+2)f(x) + 2(x+2)^2 f'(x)] – 6(x+2) = 10(x+2)f(x) $$
Divide throughout by $2(x+2)$ (since $x \ge 0$, $x+2 \ne 0$):
$$ 2f(x) + (x+2)f'(x) – 3 = 5f(x) $$
$$ (x+2)f'(x) – 3f(x) = 3 $$
$$ (x+2)\frac{dy}{dx} – 3y = 3 $$
$$ \frac{dy}{dx} – \frac{3}{x+2}y = \frac{3}{x+2} $$

This is a linear differential equation.
Integrating Factor (I.F.):
$$ I.F. = e^{\int -\frac{3}{x+2} dx} = e^{-3 \ln(x+2)} = (x+2)^{-3} = \frac{1}{(x+2)^3} $$
Solution:
$$ y(I.F.) = \int Q(I.F.) dx + C $$
$$ \frac{y}{(x+2)^3} = \int \frac{3}{x+2} \cdot \frac{1}{(x+2)^3} dx + C $$
$$ \frac{y}{(x+2)^3} = 3 \int (x+2)^{-4} dx + C $$
$$ \frac{y}{(x+2)^3} = 3 \frac{(x+2)^{-3}}{-3} + C $$
$$ \frac{y}{(x+2)^3} = -\frac{1}{(x+2)^3} + C $$
$$ y = -1 + C(x+2)^3 $$

To find $C$, put $x=0$ in the original integral equation:
$$ 2(2)^2 f(0) – 3(2)^2 = 10(0) \Rightarrow 8f(0) – 12 = 0 \Rightarrow f(0) = \frac{12}{8} = \frac{3}{2} $$
Substitute $y(0) = 3/2$ into the solution:
$$ \frac{3}{2} = -1 + C(2)^3 \Rightarrow \frac{3}{2} + 1 = 8C \Rightarrow \frac{5}{2} = 8C \Rightarrow C = \frac{5}{16} $$
$$ f(x) = -1 + \frac{5}{16}(x+2)^3 $$

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

Ans. (19)