let $f:\mathbb{R} \to \mathbb{R}$ be differentiable function let $f:\mathbb{R} \to \mathbb{R}$ be differentiable function
such that : $f(x)=f(x/2)+\dfrac{x}{2}f'(x)$
then find the $f''(20)-f''(10)+4f''(8)=?$
My work :
$$f'(x)=\frac{1}{2}f'(x/2)+\frac{1}{2}f'(x)+\frac{x}{2}f''(x) \\f'(20)=f(10)+20f''(20) \\20f''(20)-f'(10)+f(10)=0$$
I can not go on:( please help me!
 A: We have $$f(x)=f\left(\frac x2\right)+\dfrac{x}{2}f'(x)\implies f'(x)=\frac2xf(x)-\frac2xf\left(\frac x2\right)\tag{1}$$$$f'(x)=\frac12f'\left(\frac x2\right)+\frac12f'(x)+\frac x2f''(x)$$ so $$xf''(x)=f'(x)-f'\left(\frac x2\right)$$ Hence $$f''(20)=\frac1{20}f'(20)-\frac1{20}f'(10)$$ $$f''(10)=\frac1{10}f'(10)-\frac1{10}f'(5)$$ $$4f''(8)=\frac12f'(8)-\frac1{2}f'(4)$$ and from $(1)$ $$\begin{align}\small f''(20)-f''(10)+4f''(8)&=\small\frac1{20}f'(20)-\frac3{20}f'(10)+\frac12f'(8)-\frac1{10}f'(5)-\frac12f'(4)\\&=\small \frac1{200}f(20)-\frac1{200}f(10)-\frac3{100}f(10)+\frac3{100}f(5)+\frac18f(8)-\frac18f(4)\\&\small-\frac1{25}f(5)+\frac1{25}f\left(\frac52\right)-\frac14f(4)+\frac14f(2)\\\small f''(20)-f''(10)+4f''(8)&=\boxed{\small\frac1{200}f(20)-\frac7{200}f(10)+\frac18f(8)-\frac1{100}f(5)-\frac38f(4)+\frac1{25}f\left(\frac52\right)+\frac14f(2)}\end{align}$$
A: You are on the right way! Note that what you got in the end can be written $$f''(20)=\dfrac{1}{20}(f'(20)-f(10))$$
Hint: 
You can do the same job to express $f^{\prime\prime}(10)$ and then $f^{\prime\prime}(8)$ in terms of values of $f^\prime$ and $f$ only, then use the original relation given in the problem to express the values of $f^\prime$ in terms of values of $f$ only.
