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Question Number 175490 by Linton last updated on 31/Aug/22

$$solve$$$$f\left(x\right)f\left(y\right)=f(x+y)+xy$$$$f:\mathbb{R}\Rightarrow \mathbb{R}$$

Answered by ajfour last updated on 31/Aug/22

$$f\left(x\right)f\left(0\right)=f(x+0)+0$$$$\Rightarrow f\left(0\right)=1$$$$f\left(x\right)f\left(y\right)=f(x+y)+xy$$$$f^2\left(x\right)=f\left(2x\right)+x^2$$$$2f\left(x\right)f^{\prime }\left(x\right)=2f^{\prime }\left(2x\right)+2x$$$$\Rightarrow 2{\left\{f^{\prime }\left(x\right)\right\}}^2+2f^{\prime }\left(x\right)f^{″}\left(x\right)$$$$=4f^{″}\left(2x\right)+2$$$$letx=0$$$$2+2f^{″}\left(0\right)=4f^{″}\left(0\right)+2$$$$\Rightarrow f^{″}\left(0\right)=0$$$$f\left(x\right)f\left(h\right)=f(x+h)+hx$$$$\Rightarrow \frac{df\left(x\right)}{dx}=\frac{f(x+h)-f\left(x\right)}{h}$$$$=\frac{f\left(x\right)f\left(h\right)-hx-f\left(x\right)}{h}$$$$=\frac{f\left(x\right)f\left(h\right)-hx-f\left(x\right)}{h}$$$$=f\left(x\right)\left\{\frac{f\left(h\right)-f\left(0\right)}{h}\right\}-x$$$$\frac{df\left(x\right)}{dx}=cf\left(x\right)-x$$$$\Rightarrow \frac{dy}{dx}-cy=-x$$$${ye}^{-cx}=-\int {xe}^{-cx}$$$${ye}^{-cx}=\frac{1}{c}{xe}^{-cx}+\frac{1}{c^2}{e}^{-cx}$$$$c^{2}y=cx+1$$$$f\left(x\right)=y=\frac{cx+1}{c^2}$$$$butf\left(0\right)=1\Rightarrow c=\pm 1$$$$f\left(x\right)=1\pm x$$$$$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

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