Differential Equations – Linear Differential Equations – JEE Main 02 April 2025 Shift 2

Solution:

$$10\int_{1}^{x}f(t)dt = 5xf(x) – x^{5} – 9$$

Differentiating both sides with respect to $x$:
$$10f(x) = 5f(x) + 5xf'(x) – 5x^{4}$$
$$5f(x) + 5x^{4} = 5xf'(x)$$
$$f(x) + x^{4} = xf'(x)$$

$$\frac{dy}{dx} – \frac{1}{x}y = x^{3}$$

$$\text{IF} = e^{\int \left(-\frac{1}{x}\right) dx} = e^{-\ln|x|} = \frac{1}{x} \quad (\text{for } x \ge 1)$$

$$y \cdot \text{IF} = \int Q \cdot \text{IF} dx + C$$
$$y \left(\frac{1}{x}\right) = \int x^{3} \left(\frac{1}{x}\right) dx + C$$
$$\frac{y}{x} = \int x^{2} dx + C$$
$$\frac{y}{x} = \frac{x^{3}}{3} + C$$

Substitute $x=1$ in the originally given integral equation:
$$10\int_{1}^{1}f(t)dt = 5(1)f(1) – (1)^{5} – 9$$
$$0 = 5f(1) – 10 \Rightarrow f(1) = 2$$

Substitute $x=1$ and $y=2$:
$$\frac{2}{1} = \frac{1}{3} + C \Rightarrow C = \frac{5}{3}$$

$$\frac{f(x)}{x} = \frac{x^{3}}{3} + \frac{5}{3}$$

Substitute $x=3$:
$$\frac{f(3)}{3} = \frac{3^{3}}{3} + \frac{5}{3} = \frac{27}{3} + \frac{5}{3}$$
$$\frac{f(3)}{3} = \frac{32}{3}$$

$$f(3) = 32$$

Ans. (2)