Question

Find the particular solution of the differential equation \[ \frac{dy}{dx} + y \cot x = 2x \quad (x \neq 0), \quad \text{when } y = 0 \text{ if } x = \frac{\pi}{2}. \]

Hint:

For first-order linear differential equations, always find the integrating factor and use it to simplify the equation.

Answer 1

Thegivendifferentialequationis \[ \frac{dy}{dx}+y\cotx=2x. \] Theintegratingfactor(IF)iscalculatedas: \[ IF=e^{\int\cotx\,dx}=\sinx. \] Multiplyingthroughby\(\sinx\),theequationbecomes: \[ \sinx\frac{dy}{dx}+y\sinx\cotx=2x\sinx\quad\Rightarrow\quad\frac{d}{dx}(y\sinx)=2x\sinx. \] Integratingbothsides: \[ y\sinx=\int2x\sinx\,dx. \] Usingintegrationbyparts: \[ \int2x\sinx\,dx=-2x\cosx+2\int\cosx\,dx=-2x\cosx+2\sinx. \] Thus: \[ y\sinx=-2x\cosx+2\sinx+C. \] Substitute\(x=\frac{\pi}{2},y=0\)tofind\(C\): \[ 0\cdot\sin\left(\frac{\pi}{2}\right)=-2\cdot\frac{\pi}{2}\cdot\cos\left(\frac{\pi}{2}\right)+2\cdot\sin\left(\frac{\pi}{2}\right)+C\quad\Rightarrow\quadC=-2. \] Theparticularsolutionis: \[ y=\frac{-2x\cosx+2\sinx-2}{\sinx}. \]