हल कीजिए-
`(dy)/(dx) =-(x+y cos x)/(1+ sin x).`

Explanation

दिया गया अवकल समीकरण है-
`(dy)/(dx) =(x+y cos x)/(1+sin x)`
`implies(dy)/(dx) =-(x)/(1+sin x) -(y cos x)/(1+sin x)`
`=(dy)/(dx) +((cos x)/(1+sin x)) y= -(x)/(1+sin x)" "...(1)`
यहाँ `P =(cos x)/(1+sin x)` और `Q=-(x)/(1+sin x)`
`therefore I. F. =e ^(int Pdx) =e ^(int(cos x)/(1+sin )dx)`
`=e ^(int(dt)/(dt)),`
`["माना"1+sin x=timplies cos sdx = dt]`
`=e^(log |t|) =e^(log |sin x|)=1+ sin x`
अतः अभीष्ट हल है-
`y(1+ sin x) =int ((-x)/(1+sin x)xx 1+sin x) dx+C`
`impliesy (1+ sin x) =-int xdx +C`
`implies y(1+ sin x) =-(x^(2))/(2)+C`
`implies y (1+sin x) =(2C-x^(2))/(2)`
`implies y=(2C-x^(2))/(2(1+ sin x))`