Question

x dy/dx + y - x + xy cot x = 0

Answer 1

$y = \frac{\sin x - x \cos x + C}{x \sin x}$

Answer 2

The given differential equation is: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ Rearranging to the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$: $\frac{dy}{dx} + y \left(\frac{1}{x} + \cot x\right) = 1$ Here, $P(x) = \frac{1}{x} + \cot x$ and $Q(x) = 1$.

The integrating factor (IF) is calculated as: $IF = e^{\int P(x) dx} = e^{\int \left(\frac{1}{x} + \cot x\right) dx}$ $IF = e^{\ln|x| + \ln|\sin x|} = e^{\ln|x \sin x|} = |x \sin x|$ We use $IF = x \sin x$.

The general solution of a linear first-order differential equation is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$. $y (x \sin x) = \int 1 \cdot (x \sin x) dx + C$ $y x \sin x = \int x \sin x dx + C$ Using integration by parts for $\int x \sin x dx$ (with $u=x$ and $dv=\sin x dx$), we get: $\int x \sin x dx = -x \cos x + \sin x$ Substituting this result back into the solution equation: $y x \sin x = -x \cos x + \sin x + C$ Solving for $y$, we get the general solution: $y = \frac{-x \cos x + \sin x + C}{x \sin x}$ This can also be written as: $y = \frac{\sin x - x \cos x + C}{x \sin x}$