Solving functional equation $f(x+y)=f(x)+f(y)+xy$ We are given $f(0)=0$. Then when $x+y=0$:
$$0=f(-y)+f(-x)+xy$$
Can I now use $x=0$ and obtain:
$$0=f(-y)?$$ Is this correct? Is there a better way to solve this equation? 
 A: Set $f(x) = g(x) +\dfrac{x^2}{2}$ then plugging in gives
$$\fbox{1}\,g(x+y)=g(x)+g(y).$$
This is Cauchy's functional equation. And under certain regularity assumptions for $g$ you get that $g(x)=ax$. But if none are given then $g$ is just a linear function over $\mathbb{Q}$ the rational numbers and so any of those would be an appropriate solution.
A: Assuming $f$ differentiable, differentiate the functional equation implicitly wrt. $x$ and $y$ to obtain two simpler equations. From these you obtain $f'(t)=t+c$ after some simple algebraic manipulations.
Assuming $f$ twice differentiable, differentiating twice wrt. $x$ gives $f''(x+y)=f''(x)$, so $f''$ does not depend on $y$. Analogously show that $f''$ does not depend on $x$. So $f''$ is constant. So...
A: if you sub $y = -x,$ you get $$0 = f(0) = f(x-x)= f(x) + f(-x) - x^2 \tag 1$$
suppose further assume that $f = ax^2 + bx.$   subbing in $(1),$ gives you $f = \frac12 x^2 + bx$ for any $b.$ 
A: I show that the only solution is $f = \frac{1}{2}x^2 + bx$ (assuming $f$ continuous).
My strategy was to transform $f(x+y) = f(x) + g(y) + xy$ into a linear equation.
Write $f(x+y) = f(x) + g(y) + xy$ as $f(x+y) - \frac{1}{2}(x+y)^2 = (f(x) - \frac{1}{2}x^2) + (g(y) - \frac{1}{2}y^2)$. Substitute $g(z) = f(z) - \frac{1}{2}z^2$. Then we have:
$g(x+y) = x+y$, which is linear, so $g(x) = bx$ (assuming continuity).
Thus $f(z) - \frac{1}{2}z^2 = bz$, so $f(z) = \frac{1}{2}z^2 + bz$.
A: From the functional equation and the initial condition $f(0)=0$ we get
$$
f(x + y) - f(x) = f(0 + y) - f(0) + xy
$$
Assuming $f$ is differentiable, dividing by $y$ and taking the limit $y\to 0$ we get
$$
f'(x) = f'(0) + x
$$
Integration gives
$$
f(x) = f'(0)\,  x + \frac{1}{2}x^2
$$
A: $F(x)=2f(x) - x^2$ satisfies $F(x+y)=F(x)+F(y)$ whose solutions are $F(x) = ax$ under the most minimal regularity assumptions (continuous is more than enough).  
As several almost identical answers are saying, after subtracting a particular solution of the equation, the inhomogeneous equation $f(x+y)=f(x)+f(y) + (\cdots)$ becomes the well understood homogeneous equation $f(x+y)=f(x)+f(y)$ so that the answer is a linear function added to the particular solution.
To remain agnostic about Axiom of Choice issues one can say that $f(x) = \frac{x^2}{2} + $ (solution of $h(x+y)=h(x)+h(y)$) instead of declaring whether nonlinear solutions to the Cauchy equation exist or not.
