Kamran011 wrote: Find all functions $u:R\rightarrow{R}$ for which there exists a strictly monotonic function $f:R\rightarrow{R}$ such that $f(x+y)=f(x)u(y)+f(y)$ for all $x,y\in{\mathbb{R}}$ Let $P(x,y)$ be the assertion $f(x+y)=f(x)u(y)+f(y)$ $P(0,x)$ $\implies$ $f(0)u(x)=0$ and so $f(0)=0$ (else we get $f(x+y)=f(y)$ and so $f(x)$ is not strictly monotonic. $P(x,0)$ $\implies$ $f(x)(u(0)-1)=0$ and so $u(0)=1$ (else $f(x)$ is not strictly monotonic If $u(t)=0$ for some $t$, then $P(x,t)$ implies $f(x+t)=f(t)$, impossible since $f(x)$ strictly monotonic. So $u(x)\ne 0$ $\forall x$ Subtracting $P(x,y)$ from $P(y,x)$, we get New assertion $Q(x,y)$ : $f(x)(u(y)-1)=f(y)(u(x)-1)$ If $u(t)=1$ for some $t\ne 0$ (so that $f(t)\ne 0$), then $Q(x,t)$ $\implies$ $u(x)=1$ $\forall x$ which indeed is a solution (choose $f(x)\equiv x$ for example). If $u(x)=1$ $\iff$ $x=0$ : $Q(x,1)$ $\implies$ $f(x)=c(u(x)-1)$ for some $c=\frac{f(1)}{u(1)-1}\ne 0$ Plugging this in original equation, we get $u(x+y)=u(x)u(y)$ Since we previously got $u(x)\ne 0$ $\forall x$, this easily implies $u(x)>0$ $\forall x$ (set $x=y$ And so $\ln u(x)$ is additive. So $u(x)=e^{g(x)}$ where $g(x)$ is some additive function. If $g(x)$ is not linear, graph of $g(x)$ is dense in $\mathbb R^2$ and so graph of $f(x)$ is dense in $\mathbb R\times (-c,+\infty)$ (or $(-\infty,-c)$ if $c<0$, in contradiction with $f(x)$ stricly monotonic. So $g(x)$ is linear and we got $\boxed{u(x)=e^{ax}\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$, choosing for example $f(x)=e^{ax}-1$ if $a\ne 0$ or $f(x)=x$ if $a=0$

Same solution as post #2, posting anyways Let $P(x,y)$ be the assertion of f.e as usual. Let the strictly monotonic property be shown as $(\clubsuit)$. Clearly $u(x) \neq 0$ and $f$ is injective. $$\begin{align*} &P(0,x):f(0)u(x)=0 \stackrel{(\clubsuit)}{\Rightarrow} f(0)=0 \\ &P(x,0): f(x)=f(x)u(0) \stackrel{(\clubsuit)}{\Rightarrow} u(0)=1 \\ &P(-x,x): f(-x)u(x)+f(x)=0 \Rightarrow u(x) = \frac{ -f(x)}{f(-x)} \qquad &(\spadesuit) \\ &P(x,y)-P(y,x)= f(x)\big(u(y)-1 \big)= f(y)\big( u(x)-1 \big) \Rightarrow \frac{f(x)}{f(y)}= \frac{u(x)-1}{u(y)-1} &\qquad (1) \end{align*}$$ Assume $u(a)=1$ for some $a \neq 0$. By $(\clubsuit)$ if $x \neq 0$ in $(\spadesuit)$ we have $f(x) \neq f(-x)$. if $-f(x)=f(-x)$, or that is, $f$ is odd function, analyzing $P(-x,x)$ once more yields $$f(x) \big( -u(x)+1\big)=0$$ Since $f$ is obviously not constant, we must have $u(x)=1$ which is a clear solution. Hence, for sake of finding another solutions, assume $u(a)=1$ only if $a=0$. Observe $(1)$ now. We have $f(x)= au(x)-a$ for some $a$ such that $a \neq 0$. Put into the original equation and we have $$u(x+y)= u(x)u(y)$$ Which is a well known problem and the only solutions are $u(x)=e^{ax}$ for some parameter $a$ which indeed works, hence we are done.