Show that x^3 3xy^2=1 is an solution of differential equation 2xydy/dx+x2+y2=0 on interval (0,1)
If x^3 + 3xy^2 = 1 is a solution of the given differential equation, then differentiating the given solution will result in the given DE.
Starting with,
x^3 + 3xy^2 = 1
Differentiating both sides wrt x,
3x^2 + 3y^2 + 6xy.y' = 0 (where y' = dy/dx)
Taking out the common factor of 3,
2xy.y' + x^2 + y^2 = 0
The above is the given DE
Therefore the eqn, x^3 + 2xy^2 = 1, is a solution of the DE