Question

For the differential equation, find the general solution:

${\displaystyle x\frac{dy}{dx}+y-x+xy\cot x=0(x\ne 0)}$

Answer 1

Given differential equation

${\displaystyle x \frac{dy}{dx}+y-x+xy \cot x=0}$

⇒ ${\displaystyle x \frac{dy}{dx}+y(1+x \cot x)=x}$

or ${\displaystyle \frac{dy}{dx}+(\frac{1}{x}+\cot x)y=1}$            ...(i)

Comparing with ${\displaystyle \frac{dy}{dx}+Py=Q}$

${\displaystyle P=\frac{1}{x}+\cot x}$ and  Q = 1

∴ ${\displaystyle I.F.=e^{\int Pdx}=e^{\int (\frac{1}{x}+\cot x)dx}}$

${\displaystyle =e^{\log x}+\log \sin x}$

${\displaystyle \Rightarrow e^{\log \left(x\sin x\right)}=x\sin x}$

Hence the required solution

∴ ${\displaystyle y\times I.F.=\int I.F.\times Q dx+C}$

${\displaystyle \Rightarrow y\times x\sin x=\int 1\cdot x\sin xdx+C}$

${\displaystyle \Rightarrow xy\sin x=-x\cos x+\int 1\cos xdx+C}$

${\displaystyle \Rightarrow xy\sin x=-x\cos x+\sin x+C}$

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

⇒ ${\displaystyle y=\frac{1}{x}-\cot x+\frac{C}{x\sin x}}$