The particular solution of the differential equation $\frac{dy}{dx}+\frac{3y}{x}= 0, y(1) = 1$ is

$y=\frac{1}{x^3}$

The correct answer is Option (1) → $y=\frac{1}{x^3}$

Given: $\frac{dy}{dx} + \frac{3}{x} y = 0$, with $y(1) = 1$

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = 0$

Integrating factor: $IF = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$

Multiply both sides by $x^3$: $x^3 \frac{dy}{dx} + 3x^2 y = 0$

$\Rightarrow \frac{d}{dx}(x^3 y) = 0$

$\Rightarrow x^3 y = C$

Apply $y(1) = 1$: $1^3 \cdot 1 = C \Rightarrow C = 1$