`x dy - y dx = sqrt(x^(2) + y^(2)) dx`

Explanation

`xdy-ydx=sqrt(x^(2)+y^(2))`dx
or `xdy-[y+sqrt(x^(2)+y^(2))]`dx
or `xdy= [y+sqrt(x^(2)+y^(2))]dx`
or `(dy)/(dx) = y+sqrt(x^(2)+y^(2))/(x)`…………..(1)
Let `F(x,y) = y+sqrt(x^(2)+y^(2))/(x)`
`therefore F(lambdax,lambday)=lambdax+sqrt((lambdax)^(2)+(lambday)^(2))/(lambdax) = y+sqrt(x^(2)+y^(2))/(x) = lambda^(0)F(x,y)`
Therefore, the given differential equation is a homogenous equation,
To, solve it, we make the substitution as
`y=vx`
or `(dy)/(dx)=v+x(dv)/(dx)`
Substituting the values of v and `(dy)/(dx)` in equation (1), we get
`v+x(dv)/(dx) = vx+sqrt(x^(2)+(vx)^(2))/(x)`
or `v+x(dv)/(dx) = v+sqrt(1+v^(2))`
or `(dv)/sqrt(1+v^(2))=(dx)/(x)`
Integrating both sides, we get
`log|v+sqrt(1+v^(2))|=log|x|+logC`
or `log|y/x+sqrt(1+y^(2)/x^(2))|+log|Cx|`
or `y+sqrt(x^(2)+y^(2))=Cx^(2)`