Show that the family of curves for which the slope of the tangent at any point (x, y) on it is `(x^2+y^2)/(2x y)` , is given by `x^2" -"y^2=" "c x` .

Answer

`y^2-x^2=cx`

Explanation

We know that the slop of the tangent at any point on a curve is `(dy)/(dx)` Therefore,


`(dy)/(dx)=(x^2+y^2)(2 x y)`


` (d y)/(d x)=(1+(y^2)/(x^2))((2 y)/(x))---(1)`


Clearly, equation `(1)` is a homogeneous differential equation. To solve it we make substitution


`Let (y)=(vx)`


`(d y)/(d x)=v+x (d v)/(d x)`


Putting the value of `y` and `(d y)/(d x)` in equation `(1)`, we get


`v+x (d v)/(d x)=(1+v^2)/(2 v)`


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