Find all functions such that $af(x-1)+bf(1-x)=cx$, where $a$,$b$ and $c$ are real and $a^2$ is not equal to $b^2$

Question

I tried to find $f(0)$, by taking $x=1$,

$$f(0)=\frac{c}{a+b},$$

Then i deduced that $f(x)$ is linear,

Taking $x=(u+1), af(u)+bf(-u)=(u+1)c$

Similarly, $af(-u)+bf(u)=f(1-u)c$, adding them

$2c/(a+b)=f(u)+f(-u)$

Now take u+1,

$f(u+1)+f(-u-1)=2c/(a+b)$

Equating this with the first,

$f(u+1)-f(u)=f(-u)-f(-u-1)=2c/(a+b)$

this lead me to the conclusion that f(x) is linear,

Answer 1

Let $x-1=t.$ $$af(x-1)+bf(1-x)=cx \implies af(t)+bf(-t)=c(t+1)~~~(1).$$ Let $t\rightarrow -t$; we get $$af(-t)+bf(t)=c(-t+1)~~~(2)$$ Let $f(t)=X, f(-t)=Y$, then we have $aX+bY=ct+c, aY+bX=-ct+c$ Solve for $X$ $$X=f(t)=c[\frac{1}{a+b}+\frac{t}{a-b}]\implies f(x-1)=c[\frac{1}{a+b}+\frac{(x-1)}{a-b}]$$ Finally $$f(x)=c[\frac{1}{a+b}+\frac{x}{a-b}]$$