हल कीजिए-
`(1+y^(2))+ x-e ^(tan^(-1y))(dy)/(dx)=0.`

Explanation

दिया गया अवकल समीकरण है-
`(1+y^(2))+(x-e^(tan^(-1)y))(dy)/(dx) =0`
`implies(1+ y^(2))=-(x-e^(tan ^(-1)y))(dy)/(dx)`
`implies (dx)/(dy)=(-(x-e^(tan^(-1)y)))/(1+y^(2))`
`implies (dx)/(dy) =-(x)/(1+y^(2))+(e^(tan^(-1)y))/(1+y^(2))`
`implies(dx)/(dy)+ (1)/(1+y^(2)).x = (e ^(tan^(-1)y))/(1+y^(2))" "...(1)`
जो कि x में रैखिक अवकल समीकरण है .
समी (1 ) कि तुलना `(dx)/(dy)+Px =Q` से करने पर,
`P =(1)/(1+y^(2))` और `Q=(e^(tan^(-1)y))/(1+y^(2))`
`thereforeI. F. =e^(intPdy )=e ^(int(1)/(1+y^(2))dy)=e ^(tan ^(-1)y)`
अतः अभीष्ट हल है-
`x xx(I. F. ) =int Q xx (I. F. )dy +C`
`impliesxe ^(tan^(-1)y)=int ((e^(tan^(-1)y))/(1+y^(2)), e^(tan ^(-1)y))dy+C`
`impliesxe ^(tan ^(-1)y)=int e ^(t). e^(t) dt +C,`
`["माना" tan ^(-1)y=timplies (1)/(1+y^(2))dy = dt]`
`impliesxe^(tan ^(-1)y)=int e^(2t) dt+C`
`impliesxe^(tan ^(-1)y)= (e ^(2t))/(2)+C`
`impliesx.e^(tan ^(-1)y)=1/2 e ^(2 tan^(-1)y)+C.`