Find the general solution of the following differential equation : `(1y2)+((x-e^"tan"^((-1_"y")))dy)/(dx)=0`

Explanation

Given, differential equation is
`" "(1+y^(2))=(x-e^(tan^(-1)y)(dy)/(dx)`
`rArr" "(1+y^(2))=-(x-e^(tan^(-1)y))(dy)/(dx)`
`" "(1+y^(2))(dx)/(dy)=-x+e^(tan^(-1))y`
`rArr" "(1+y^(2))(dx)/(dy)+x=e^(tan^(-1))y`
`rArr" "(dx)/(dy)+(x)/(1+y^(2))=(e^(tan^(-1))y)/(1+y^(2))" "` [dividing throughout by `(1+y^(2))`]
which is a linear differential equation.
On comparing it with `(dx)/(dy)+Px=Q`, we get
`" "P=(1)/(1+y^(2)),Q=(e^(tan^(-1)y))/(1+y^(2))`
`" "IF=e^(intPdy)=e^(int(1)/(1+y^(2)))=e^(tan^(-1)y)`
The general solution is `x*e^(tan^(-1)y)=int(e^(tan^(-1)y))/(1+y^(2))*e^(tan^(-1)y)dy+C`
`rArr" "x*e^(tan^(-1)y)=int((e^(tan^(-1)y)))/(1+y^(2))*dy+C`
Put `tan^(-1)y=trArr " "(1)/(1+y^(2))dy=dt`
`therefore" "x*e^(tan^(-1)y)=inte^(2t)dt+C`
`rArr" "x*e^(tan^(-1)y)=(1)/(2)e^(2tan^(-1)y)+C`
`rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1)y)+2C`
`rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1)y)+K" "[becauseK=2C]`