The following question is taken from the practice set of JEE exam.
Let $f : \mathbb R^+ → \mathbb R$ satisfies the equation $f(xy) = e^{xy – x – y} (e^y f(x) + e^x f(y)) , x , y \in \mathbb R^+$. If $f ’ (1) = e$ , then find $f(x)$.
My attempt:
$$f(xy)=e^{xy}\left(\dfrac{f(x)}{e^x}+\dfrac{f(y)}{e^y}\right)$$
Putting $x=1=y$, I get $$f(1)=2f(1)\implies f(1)=0$$
Putting $y=\dfrac1x$, I get $$f(1)=e\left(\frac{f(x)}{e^x}+\frac{f(\frac1x)}{e^{\frac1x}}\right)\\\implies \frac{f(x)}{e^x}=-\frac{f(\frac1x)}{e^{\frac1x}}$$
Taking derivative on both sides, I get$$\frac{e^xf'(x)-f(x)e^x}{e^{2x}}=-\frac{e^{\frac1x}f'(\frac1x)\frac{-1}{x^2}-f(\frac1x)e^{\frac1x}\frac{-1}{x^2}}{e^{\frac2x}}\\\implies\frac{f'(x)-f(x)}{e^x}=\frac{f'(\frac1x)-f(\frac1x)}{x^2e^{\frac1x}}$$
Putting $x=1$ is not giving anything.
First, I use a non-rigorous method to solve the problem, then I will fill in the gaps. If $f$ is differentiable, then I can use $y=1+\varepsilon$ in the limit $\varepsilon\rightarrow 0$ to find the differential equation $$f^\prime(x)=f(x) + \frac{e^x}{x}$$ which (using $f(1)=0$) solves as $$\boxed{f(x)=\log (x) e^x}$$
Now, using the insight, I define $u(x)\equiv f(x)e^x$ and easily see that $u(xy)=u(x)+u(y)$ hence the rigorous proof.