Solve: y = (1+p)x + ap^2?

Question

Hint: Use Clairaut's Equational form
Comment: Please don't think me as a student.. I'm a Post graduate in Management (Graduate in terms of US educational system) with Economics and Mathematics as my background subjects.. I give these problems for refreshing my brain in the weekends after a long week of monotonous office job as well as to know many things from you also... Exchange of ideas what I mean.
:)

Answer 1

y = (1+p)x + ap^2
differentiating

p = 1+p+(x+2ap)p'

Thus,

dx/dp+x = -2ap
Thus,
x = 2a(1-p)+cexp(-p)

Substitute x in original eqn.

Answer 2

based on the hint i differentiate both sides to get

p = (1+p)+x +a 2pp'
or 1+x+ 2app' =0
or 2app' = - (1+x)
lhs is differerntial of ap^2

integrate both sides
ap^2 = - ln(1+x)
p^2 = (-ln(1+x))/a

p = sqrt(-ln(1+x)/a)
dy/dx = sqrt(-ln(1+x)/a)

by integrating both sides you get y

Answer 3

Y(x) = x(dy/dx) + f(dy/dx) → dy/dx = dy/dx + x(d²y/dx²) + f́(dy/dx)(d²y/dx²) → 0 = x(d²y/dx²) + f́(dy/dx)(d²y/dx²) = [x + f́(dy/dx)](d²y/dx²)
So, either (d²y/dx²) = 0 or [x + f́(dy/dx)] = 0

In your question, (dy/dx) = (1+p) and f(dy/dx) = ap² in other word, f(1+p) = ap² = a(1+p)² - 2a(1+p) + a → f(x) = ax² - 2ax + a = a(x-1)² → f́(x) = 2a(x-1)

In the case of (d²y/dx²) = 0, (dy/dx) = C →
y(x) = Cx + f(C) = Cx + a(C-1)²


On the other case, [x + f́(dy/dx)] = 0, x + 2a[(dy/dx) - 1) = 0 → (dy/dx) = (-x +2a)/(2a) → y(x) = -x²/(4a) + x + C