हल कीजिए-
`x (dy)/(dx) +y-x+xy cot x=0, x ne 0.`

Explanation

दिया गया अवकल समीकरण है-
`x(dy)/(dx)+y-xy cot x=0`
`impliesx(dy)/(dx) +(1+ x cot x)y=x`
`implies (dy)/(dx)+((1+ cot x)/(x))y=1`
`implies(dy)/(dx)+ ((1)/(x) +cot x)y=1" "...(1)`
यहॉँ `P=1/x+cot x` और `Q=1`
`therefore I. F. =e^(int Pdx)=e^(int((1)/(x)+cotx)dx)`
`=e^(int(1)/(x)dx+intcot x dx )`
`=e ^(log|x|+ log |sin x|)`
`=e ^(log |x sinx|)`
`=x sin x`
अतः अभीष्ट हल है-
`yxx (I.F.)=int Qxx (I.F.) dx+C`
`impliesy. x sin x=int 1(x sin x)dx +C`
`implies xy sin x = int underset(I)(x ) underset(II)(sin) x dx +C`
`implies xy sin x=x int sin x dx- int 1xx (-cos x) dx +C`
`impliesxy sin x= x (- cos x)+ sin x+C`
`implies xy sin x=- x cos x+ sin x+C`
`impliesy=(-x cos x)/(x sin x) +(sin x)/(x sin x) +(C)/(x sin x)`
`impliesy= -cot x+1/x + (C)/(x sinx).`