Find the equation of the family of curves for which the slope of tangent at any point (x,y) on it, is `(x^(2)+y^(2))/(2xy)`.

Explanation

We know that the slope of the tangent at any point (x,y) of the curve is `(dy)/(dx)`.
`therefore (dy)/(dx)=(x^(2)+y^(2))/(2xy)`
On dividing the Nr and Dr of RHS of (i) by `x^(2)`, we get ,
`(dy)/(dx)={(1+(y/x)^(2))/(2(y/x))}=f(y/x)`.
So, the given differential equation is homogeneous.
Putting, `y=vx` and `(dy)/(dx)=v+x(dy)/(dx)` in (i), we get
`v+x(dv)/(dx)=(1+v^(2))/(2v)`
`rArr (2v)/(1-v^(2))={((1+v^(2))/(2v)-v}={(1-v)^(2)/(2v))`
`rArr (2v)/(1-v^(2))dv=1/xdx`
`rArr int(2v)/(v^(2)-1)dv=-int(1/x)dx` [on integrating both sides]
`rArr log|v^(2)-1|=-log|x|+log|C_(1)|`, where `C_(1)` is an arbitrary constant
`rArr log|v^(2)-1|+log|x|+log|C_(1)|`
`rArr log|(v^(2)-1)x|=log|C_(1)|`
`rArr |(v^(2)-1)x|=log|C_(1)|`
`rArr (v^(2)-1)x=+-C_(1)`
`rArr (y^(2)/x^(2)-1)x=+-C_(1)`
`rArr (y^(2)-x^(2))=+-C_(1)x`
`rArr (x^(2)-y^(2))=Cx`, where `+-C_(1)=C`.
Hence, `(x^(2)-y^(2))=Cx` is the equation of the required family of curves.