`(1+e^(x//y))dx+e^(x//y)(1-(x)/(y))dy=0`

Explanation

`(1+e^(x//y))dx+e^(x//y)(1-(x)/(y))dy=0`
`implies (1+e^(x//y))dx=-e^(x//y)(1-(x)/(y))dy`
`implies (dx)/(dy)=(-e^(x//y)(1-(x)/(y)))/(1+e^(x//y))" ....(1)"`
यह एक समघातीय अवकल समीकरण है।
माना x = vy
`implies (dx)/(dy)=v+y.(dv)/(dy)`
समीकरण (1) में रखने पर,
`v+y.(dv)/(dy)=(-e^(v)(1-v))/(1+e^(v))`
`implies y(dv)/(dy)=(-e^(v)+ve^(v))/(1+e^(v))-v`
`implies y.(dv)/(dy)=(-e^(v)+ve^(v)-v-ve^(v))/(1+e^(v))`
`=(-(e^(v)+v))/(1+e^(v))`
`implies (1+e^(v))/(v+e^(v))dv=-(dy)/(y)`
`implies int(1+e^(v))/(v+e^(v))dv=-int(dy)/(y)`
`implies log(v+e^(v))=-logy+logc`
`implies log(v+e^(v))+logy=logc`
`implies log(vy+ye^(v))=logc`
`implies vy+ye^(v)=c`
`implies x+y.e^(x//y)=c`