Question: Solve the equation p^2+p(x+y)+xy=0. Its related to engineering math - Clairaut's equation.
Since Clairaut's equation is a differential equation, then I am assuming that p means the first differential, dy/dx, or y'.
You equation actually then is: (y')^2 + (y')*(x+y) + xy = 0.
Treating your equation as a simple quadratic equation, and using the quadratic formula to solve it,
p = {-(x+y) +/- sqrt((x+y)^2 - 4xy)}/(2*1)
p = {-(x+y) +/- sqrt((x-y)^2)}/2
p = {-(x+y) +/- (x-y)}/2
y' = {-(x+y) - (x-y)}/2, y' = {-(x+y) + (x-y)}/2,
y' = -x, y' = -y
And from these the solutions are,
y1(x) = C - (1/2)x^2, y2(x) = ke^(-x)