The solution of the differential equation, `e^x(x +1)dx + (ye^y -xe^x)dy = 0` with initial condition f(0) = 0, is

Explanation

We have, `e^(x)(x+1)dx+(ye^(y)-xe^(x))dy=0`
or `e^(x)(x+1)(dx)/(dy) -xe^(x)=-ye^(y)` ………………(1)
Put `xe^(x)=t`
`rArr e^(x)(x+1)(dx)/(dy) = (dt)/(dy)`
`therefore` Equation (1), reduces to
`rArr =(dt)/(dy)-t=-ye^(y)`
`rArr I.F. e^(-intdy)=e^(-y)`
`therefore` Solution is `t xx e^(-y)=-intye^(y)e^(-y)dy`
`rArr xe^(x)e^(-y)=-y^(2)/2+C`
Since, f(0) =0, C=0
`therefore` particular solution is `2xe^(x) e^(-y)+y^(2)=0`