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(1 + e^(x)/(y)) dx + e^(x)/(y)(1 - (x)/(...

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

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`(1+e^(x/y))dx+e^(y/x)(1-x/y)dy=0`
or `(dx)/(dy) = (-e^(x/y)(1-x/y))/(1+e^(x/y))`
The given differential equation is a homogenous equation.
Put x=vy
`rArr (dx)/(dy) = v+y(dv)/(dy)`
Substituing the values of x and `(dx)(dy)` in equation (1), we get
`v+y(dy)/(dx)=(-e^(v)(1-v))/(1+e^(v))`
or `y(dv)/(dy) = (-e^(v)+ve^(v))/(1+e^(v))-v`
`=(dv)/(dy) = -[(v+e^(Y))/(1+e^(y))]`
`[(1+e^(y))/(v+e^(y))]dv=-(dv)/(y)`
Integrating both sides, we get,
`log(v+e^(v))=-logy+logC=log(C/y)`
`[x/y+e^(x/y)]=C/y`
or `x+ye^(x/y)+C`
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