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

Explanation

`x dy-y dx=sqrt(x^(2)+y^(2))dx`
`implies x dy=(y+sqrt(x^(2)+y^(2)))dx`
`implies (dy)/(dx)=(y+sqrt(x^(2)+y^(2)))/(x)" ....(1)"`
माना y = vx
`implies (dy)/(dx)=v+x(dv)/(dx)`
समीकरण (1) से,
`v+x(dv)/(dx)=(vx+sqrt(x^(2)+v^(2)x^(2)))/(x)`
`=v+sqrt(v^(2)+1)`
`implies x(dv)/(dx)=sqrt(v^(2)+1)`
`implies (dv)/(sqrt(v^(2)+1))=(dx)/(x)`
`implies int(1)/(sqrt(v^(2)+1))dv=int(dx)/(x)`
`implies log(v+sqrt(v^(2)+1))=logx+logc=log(cx)`
`implies v+sqrt(v^(2)+1)=cx`
`implies (y)/(x)+sqrt((y^(2))/(x^(2))+1)=cx`
`implies y+sqrt(y^(2)+x^(2))=cx^(2)`