Functional equation $2f\left(\frac{x+y}{2}\right)-f(y)=f''(x)$

Question

Let $f:\Bbb R\to \Bbb R$ be a differentiable function satisfying $2f\left(\frac{x+y}{2}\right)-f(y)=f''(x)$ for $x,y\in \Bbb R$.

If $f(0)=5$ and $f'(5)=-1$, then $f(x)$ is?

My Work:

Replacing y with x: $f(x)=f''(x)$

$\int f'(x)f(x)=\int f'(x)f''(x)$ which gives

$\frac{f^2(x)}{2}+c=\frac {(f'(x))^2}{2}$ after this I get stuck because I can't find a way to calculate the value of c.

Any help would be appreciated.

Answer 1

Let $y=x$ in the equation, then $f''(x)=f(x)$ for all $x$.

Let $y=0$ in the equation, then $2f(\frac x2)=f(0)+f''(x)=f(0)+f(x)$ for all $x$.

Differentiate twice: $\frac12f''(\frac x2)=f''(x)$ for all $x$.

Since $f''(x)=f(x)$, you have $\frac12f(\frac x2)=f(x)$ for all $x$.

Hence $f(0)=0$.

There is thus no solution with $f(0)\ne0$.

Actually, from $\frac12f(\frac x2)=f(x)$ and using the fact that $f$ is continuous, you get more.

For all integer $n\ge0$, you have $f(x)=\frac{1}{2^n}f(\frac{x}{2^n})\to0$ as $n\to \infty$, hence $f(x)=0$ for all $x$.

Answer 2

Hint: With $y=x$, we arrive at $$f(x)=f''(x). $$