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

A normal is drawn at a point P(x,y) of a curve. It meets the x-axis at Q such that PQ is of constant length k . Answer the question:The differential equation describing such a curve is (A) y dy/dx=+-sqrt(k^2-x^2) (B) x dy/dx=+-sqrt(k^2-x^2) (C) y dy/dx=+-sqrt(k^2-y^2) (D) x dy/dx=+-sqrt(k^2-y^2)

An equation of the curve satisfying x dy - y dx = sqrt(x^(2)-y^(2))dx and y(1) = 0 is

Solve the differential equation x dy-y dx=sqrt(x^(2)+y^(2)) dx .

1. x(dy)/(dx)-y=x sqrt(x^(2)+y^(2))

The solution of (x d x+y dy)/(x dy-y dx)=sqrt((1-x^2-y^2)/(x^2+y^2)) is

If sqrt(1-x^2) + sqrt(1-y^2)=a(x-y) , prove that (dy)/(dx)= sqrt((1-y^2)/(1-x^2))

If sqrt(1-x^2) + sqrt(1-y^2)=a(x-y) , prove that (dy)/(dx)= sqrt((1-y^2)/(1-x^2))

Solution of the differential equation x(dy)/(dx)=y+sqrt(x^(2)+y^(2)) , is

If sqrt(1-x^(2)) + sqrt(1-y^(2))=a(x-y) , show that (dy)/(dx)= (sqrt(1-y^(2)))/(sqrt(1-x^(2)))

Solve : (dy)/(dx) sqrt(1+x+y) =x+y-1