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Question Number 175490 by Linton last updated on 31/Aug/22
solve f(x)f(y)= f(x+y)+xy f:R⇒R
$${solve} \\ $$$${f}\left({x}\right){f}\left({y}\right)=\:{f}\left({x}+{y}\right)+{xy} \\ $$$${f}:\mathbb{R}\Rightarrow\mathbb{R} \\ $$
Answered by ajfour last updated on 31/Aug/22
f(x)f(0)=f(x+0)+0 ⇒ f(0)=1 f(x)f(y)=f(x+y)+xy f^( 2) (x)=f(2x)+x^2 2f(x)f′(x)=2f′(2x)+2x ⇒2{f′(x)}^2 +2f′(x)f′′(x) =4f′′(2x)+2 let x=0 2+2f′′(0)=4f′′(0)+2 ⇒ f′′(0)=0 f(x)f(h)=f(x+h)+hx ⇒ ((df(x))/dx)=((f(x+h)−f(x))/h) =((f(x)f(h)−hx−f(x))/h) =((f(x)f(h)−hx−f(x))/h) =f(x){((f(h)−f(0))/h)}−x ((df(x))/dx) =cf(x)−x ⇒ (dy/dx)−cy=−x ye^(−cx) =−∫xe^(−cx) ye^(−cx) =(1/c)xe^(−cx) +(1/c^2 )e^(−cx) c^2 y=cx+1 f(x)=y=((cx+1)/c^2 ) but f(0)=1 ⇒ c=±1 f(x)=1±x
$${f}\left({x}\right){f}\left(\mathrm{0}\right)={f}\left({x}+\mathrm{0}\right)+\mathrm{0} \\ $$$$\Rightarrow\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left({x}\right){f}\left({y}\right)={f}\left({x}+{y}\right)+{xy} \\ $$$${f}^{\:\mathrm{2}} \left({x}\right)={f}\left(\mathrm{2}{x}\right)+{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{f}\left({x}\right){f}'\left({x}\right)=\mathrm{2}{f}'\left(\mathrm{2}{x}\right)+\mathrm{2}{x} \\ $$$$\Rightarrow\mathrm{2}\left\{{f}'\left({x}\right)\right\}^{\mathrm{2}} +\mathrm{2}{f}'\left({x}\right){f}''\left({x}\right) \\ $$$$\:\:\:=\mathrm{4}{f}''\left(\mathrm{2}{x}\right)+\mathrm{2} \\ $$$${let}\:\:{x}=\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{2}{f}''\left(\mathrm{0}\right)=\mathrm{4}{f}''\left(\mathrm{0}\right)+\mathrm{2} \\ $$$$\Rightarrow\:\:\:{f}''\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right){f}\left({h}\right)={f}\left({x}+{h}\right)+{hx} \\ $$$$\Rightarrow\:\:\frac{{df}\left({x}\right)}{{dx}}=\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{f}\left({x}\right){f}\left({h}\right)−{hx}−{f}\left({x}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{f}\left({x}\right){f}\left({h}\right)−{hx}−{f}\left({x}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={f}\left({x}\right)\left\{\frac{{f}\left({h}\right)−{f}\left(\mathrm{0}\right)}{{h}}\right\}−{x} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{df}\left({x}\right)}{{dx}}\:={cf}\left({x}\right)−{x} \\ $$$$\Rightarrow\:\:\:\frac{{dy}}{{dx}}−{cy}=−{x} \\ $$$${ye}^{−{cx}} =−\int{xe}^{−{cx}} \\ $$$${ye}^{−{cx}} =\frac{\mathrm{1}}{{c}}{xe}^{−{cx}} +\frac{\mathrm{1}}{{c}^{\mathrm{2}} }{e}^{−{cx}} \\ $$$${c}^{\mathrm{2}} {y}={cx}+\mathrm{1} \\ $$$${f}\left({x}\right)={y}=\frac{{cx}+\mathrm{1}}{{c}^{\mathrm{2}} } \\ $$$${but}\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:\:\Rightarrow\:\:\:{c}=\pm\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{1}\pm{x} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 15/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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