Question Number 187155 by TUN last updated on 14/Feb/23
$${f}\left({x}\right)={x}.{e}^{\mathrm{2}{x}} \\ $$$$=>{f}^{\left({n}\right)} \left({x}\right)= \\ $$$$ \\ $$
Commented by 0670322918 last updated on 14/Feb/23
Commented by mr W last updated on 15/Feb/23
$${sir}: \\ $$$${please}\:{post}\:{your}\:{answer}\:{as}\:“{answer}'', \\ $$$${not}\:{as}\:“{comment}''!\:{thanks}! \\ $$
Answered by mahdipoor last updated on 14/Feb/23
$$\Rightarrow\Rightarrow{f}^{\left(\mathrm{0}\right)} ={a}_{\mathrm{0}} {xe}^{\mathrm{2}{x}} +{b}_{\mathrm{0}} =\mathrm{1}{xe}^{\mathrm{2}{x}} +\mathrm{0} \\ $$$${f}^{\left({m}−\mathrm{1}\right)} ={a}_{{m}−\mathrm{1}} {xe}^{\mathrm{2}{x}} +{b}_{{m}−\mathrm{1}} {e}^{\mathrm{2}{x}} \:\:\Rightarrow \\ $$$${f}^{\left({m}\right)} =\left(\mathrm{2}{a}_{{m}−\mathrm{1}} \right){xe}^{\mathrm{2}{x}} +\left({a}_{{m}−\mathrm{1}} +\mathrm{2}{b}_{{m}−\mathrm{1}} \right){e}^{\mathrm{2}{x}} ={a}_{{m}} {xe}^{\mathrm{2}{x}} +{b}_{{m}} {e}^{\mathrm{2}{x}} \\ $$$$\Rightarrow\Rightarrow{get}\:\:{f}^{\left({n}\right)} ={a}_{{n}} {xe}^{\mathrm{2}{x}} +{b}_{{n}} {e}^{\mathrm{2}{x}} \\ $$$${a}_{{n}} =\mathrm{2}{a}_{{n}−\mathrm{1}} =\mathrm{4}{a}_{{n}−\mathrm{1}} =…=\mathrm{2}^{{n}} {a}_{\mathrm{0}} =\mathrm{2}^{{n}} \\ $$$${b}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{b}_{{n}−\mathrm{1}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{2}} +\mathrm{4}{b}_{{n}−\mathrm{2}} =…= \\ $$$${a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{2}} +\mathrm{4}{a}_{{n}−\mathrm{3}} +…+\mathrm{2}^{{n}−\mathrm{1}} {a}_{\mathrm{0}} +\mathrm{2}^{{n}} {b}_{\mathrm{0}} =\mathrm{2}^{{n}−\mathrm{1}} {n} \\ $$$$\Rightarrow\Rightarrow{f}^{\left({n}\right)} =\mathrm{2}^{{n}} {xe}^{\mathrm{2}{x}} +\mathrm{2}^{{n}−\mathrm{1}} {ne}^{\mathrm{2}{x}} \\ $$
Answered by Mathspace last updated on 15/Feb/23
$${f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left({x}\right)^{\left({k}\right)} \left({e}^{\mathrm{2}{x}} \right)^{{n}−{k}} \\ $$$$={x}\left({e}^{\mathrm{2}{x}} \right)^{\left({n}\right)} +{C}_{{n}} ^{\mathrm{1}} \left({e}^{\mathrm{2}{x}} \right)^{\left({n}−\mathrm{1}\right)} \\ $$$$={x}.\mathrm{2}^{{n}} {e}^{\mathrm{2}{x}} +{n}\:\mathrm{2}^{{n}−\mathrm{1}} \:{e}^{\mathrm{2}{x}} \\ $$$$=\left({x}\:\mathrm{2}^{{n}\:} +{n}\mathrm{2}^{{n}−\mathrm{1}} \right){e}^{\mathrm{2}{x}} \\ $$