Verify the given function (or relation) defines a solution to the given differential equations and sketch some of the solution curves. If an initial condition is given, label the solution curve corresponding to the resulting unique solution. (in these problems, c denotes an arbitrary constant). (x-c)^2 +y^2 = c^2, (dy/dx) = (y^2 - x^2)/2xy , y(2)=2 Solution differntiate this equation (x-c)^2+y^2=c^2 with respect to x.. we get 2(x- c)+2y(dy/dx)=0 (x-c)=-y(dy/dx)< c=x+y(dy/dx) substitute the value of c in the above written equation we have (x-x+y(dy/dx))^2+y^2=(x+y(dy/dx)^2 then we have (y(dy/dx))^2+y^2=x^2+(y(dy/dx))^2+2xy(dy/dx) the term (y(dy/dx))^2 cancels on both sides then we have y^2=x^2+2xy(dy/dx) y^2-x^2=2xy(dy/dx) (dy/dx)=(y^2-x^2)/2xy hence we can say that the given equation is a solution to the differential equation.

1. Verify the given function (or relation) defines a solution to the given differential equations and
sketch some of the solution curves. If an initial condition is given, label the solution curve
corresponding to the resulting unique solution. (in these problems, c denotes an arbitrary
constant).
(x-c)^2 +y^2 = c^2, (dy/dx) = (y^2 - x^2)/2xy , y(2)=2
Solution
differntiate this equation (x-c)^2+y^2=c^2 with respect to x.. we get 2(x-
c)+2y(dy/dx)=0 (x-c)=-y(dy/dx)< c=x+y(dy/dx) substitute the value of c in the above written
equation we have (x-x+y(dy/dx))^2+y^2=(x+y(dy/dx)^2 then we have
(y(dy/dx))^2+y^2=x^2+(y(dy/dx))^2+2xy(dy/dx) the term (y(dy/dx))^2 cancels on both sides
then we have y^2=x^2+2xy(dy/dx) y^2-x^2=2xy(dy/dx) (dy/dx)=(y^2-x^2)/2xy hence we can
say that the given equation is a solution to the differential equation