The solution of differential equation $\frac{dy}{dx}+\frac{y}{x}=\sin x$ is

$x(y+\cos x)=\sin x+C$

The correct answer is Option (1) → $x(y+\cos x)=\sin x+C$ ##

Given differential equation is

$\frac{dy}{dx} + y \frac{1}{x} = \sin x$

which is linear differential equation.

On comparing it with $\frac{dy}{dx} + P \cdot y = Q$, we get

$P = \frac{1}{x} \text{ and } Q = \sin x$

$∴\text{I.F} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$

The general solution is $y \cdot \text{I.F} = \int (Q \cdot \text{I.F}) dx + C$

$y \cdot x = \int x \cdot \sin x \, dx + C \quad \dots(i)$

$xy = -x \cos x - \int -\cos x \, dx + C$

$\text{[using integration by parts]}$

$\Rightarrow xy = -x \cos x + \sin x + C$

$\Rightarrow x(y + \cos x) = \sin x + C$