Solve the integral equation $\displaystyle\int_0^{-x}f(t)dt= f(x) + x$.
Like it says, I'm playing around with even and odd functions and require a function such that $$\int_0^{-x}f(t)dt= f(x) + x\,.$$ I can't think how to go about it, any help appreciated.
If $F(x)=\int_0^{x}f(t)dt$, then $$F(-x)=F'(x)+x$$ with $F(0)=0$. Then by differentiating both sides, we get $$-F'(-x)=F''(x)+1.$$ How to solve this differential equation?
From $$\displaystyle\int_0^{-x}\,f(t)\,\text{d}t=f(x)+x\,,\tag{#}$$ we see that $f$ is differentiable, and by taking the derivative of this integral equation with respect to $x$, we have $$-f(-x)=f'(x)+1\,.\tag{*}$$ That is, $f'$ is also differentiable, and by taking derivative of (*) with respect to $x$, we get $$f'(-x)=f''(x)\,.$$ From (*), we get $f'(-x)=-f(x)-1$. Therefore, $$f''(x)=f'(-x)=-f(x)-1\,.$$ This means $$f(x)=a\cos(x)+b\sin(x)-1$$ for some constants $a$ and $b$. Now, $$x+a\cos(x)+b\sin(x)-1=f(x)+x=\int_0^{-x}\,f(t)\,\text{d}t=x-a\sin(x)-b\cos(x)+b\,.$$ This shows that $b=-1$ and $a=-b=1$. That is, $$f(x)=\cos(x)-\sin(x)-1\,.$$
Alternatively, we can continue from the results $$F(-x)=F'(x)+x\text{ and }-F''(-x)=F'(x)+1\,,$$ by noting that $$F(x)=F'(-x)+(-x)\text{ and }-F''(x)=F'(-x)+1\,.$$ Therefore, $$F'(-x)=F(x)+x\text{ so }-F''(x)=\big(F(x)+x\big)+1\,.$$ Hence, $$F''(x)+F(x)=-x-1\,.$$ This shows that $$F(x)=A\cos(x)+B\sin(x)-x-1\,.$$ Because $F(0)=0$, we get $A=1$, so $$F(x)=\cos(x)+B\sin(x)-x-1\,.$$ That is, $$F'(x)+x=-\sin(x)+B\cos(x)+x-1$$ and $$F(-x)=\cos(x)-B\sin(x)+x-1\,.$$ This means $B=1$, and so $F(x)=\cos(x)+\sin(x)-x-1$, and we get $$f(x)=F'(x)=F(-x)-x=\cos(x)-\sin(x)-1\,.$$
P.S. Since there have been many questions regarding differentiability of $f$, I shall explain. First, we at least need the assumption that $f$ is integrable. Therefore, $\int_0^{-x}\,f(t)\,\text{d}t$ is a continuous function in $x$. Then, (#) implies that $f$ is continuous. Now, as $f$ is continuous, $\int_0^{-x}\,f(t)\,\text{d}t$ is a differentiable function in $x$, whence (#) tells us again that $f$ is differentiable. By doing this cyclically infinitely (but countably) many times, we can conclude that $f$ is in fact a smooth function.