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Question Number 187155 by TUN last updated on 14/Feb/23
f(x)=x.e^(2x) =>f^((n)) (x)=
$${f}\left({x}\right)={x}.{e}^{\mathrm{2}{x}} \\ $$$$=>{f}^{\left({n}\right)} \left({x}\right)= \\ $$$$ \\ $$
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!
$${sir}: \\ $$$${please}\:{post}\:{your}\:{answer}\:{as}\:“{answer}'', \\ $$$${not}\:{as}\:“{comment}''!\:{thanks}! \\ $$
Answered by mahdipoor last updated on 14/Feb/23
⇒⇒f^((0)) =a_0 xe^(2x) +b_0 =1xe^(2x) +0 f^((m−1)) =a_(m−1) xe^(2x) +b_(m−1) e^(2x) ⇒ f^((m)) =(2a_(m−1) )xe^(2x) +(a_(m−1) +2b_(m−1) )e^(2x) =a_m xe^(2x) +b_m e^(2x) ⇒⇒get f^((n)) =a_n xe^(2x) +b_n e^(2x) a_n =2a_(n−1) =4a_(n−1) =...=2^n a_0 =2^n b_n =a_(n−1) +2b_(n−1) =a_(n−1) +2a_(n−2) +4b_(n−2) =...= a_(n−1) +2a_(n−2) +4a_(n−3) +...+2^(n−1) a_0 +2^n b_0 =2^(n−1) n ⇒⇒f^((n)) =2^n xe^(2x) +2^(n−1) ne^(2x)
$$\Rightarrow\Rightarrow{f}^{\left(\mathrm{0}\right)} ={a}_{\mathrm{0}} {xe}^{\mathrm{2}{x}} +{b}_{\mathrm{0}} =\mathrm{1}{xe}^{\mathrm{2}{x}} +\mathrm{0} \\ $$$${f}^{\left({m}−\mathrm{1}\right)} ={a}_{{m}−\mathrm{1}} {xe}^{\mathrm{2}{x}} +{b}_{{m}−\mathrm{1}} {e}^{\mathrm{2}{x}} \:\:\Rightarrow \\ $$$${f}^{\left({m}\right)} =\left(\mathrm{2}{a}_{{m}−\mathrm{1}} \right){xe}^{\mathrm{2}{x}} +\left({a}_{{m}−\mathrm{1}} +\mathrm{2}{b}_{{m}−\mathrm{1}} \right){e}^{\mathrm{2}{x}} ={a}_{{m}} {xe}^{\mathrm{2}{x}} +{b}_{{m}} {e}^{\mathrm{2}{x}} \\ $$$$\Rightarrow\Rightarrow{get}\:\:{f}^{\left({n}\right)} ={a}_{{n}} {xe}^{\mathrm{2}{x}} +{b}_{{n}} {e}^{\mathrm{2}{x}} \\ $$$${a}_{{n}} =\mathrm{2}{a}_{{n}−\mathrm{1}} =\mathrm{4}{a}_{{n}−\mathrm{1}} =…=\mathrm{2}^{{n}} {a}_{\mathrm{0}} =\mathrm{2}^{{n}} \\ $$$${b}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{b}_{{n}−\mathrm{1}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{2}} +\mathrm{4}{b}_{{n}−\mathrm{2}} =…= \\ $$$${a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{2}} +\mathrm{4}{a}_{{n}−\mathrm{3}} +…+\mathrm{2}^{{n}−\mathrm{1}} {a}_{\mathrm{0}} +\mathrm{2}^{{n}} {b}_{\mathrm{0}} =\mathrm{2}^{{n}−\mathrm{1}} {n} \\ $$$$\Rightarrow\Rightarrow{f}^{\left({n}\right)} =\mathrm{2}^{{n}} {xe}^{\mathrm{2}{x}} +\mathrm{2}^{{n}−\mathrm{1}} {ne}^{\mathrm{2}{x}} \\ $$
Answered by Mathspace last updated on 15/Feb/23
f^((n)) (x)=Σ_(k=0) ^n C_n ^k (x)^((k)) (e^(2x) )^(n−k) =x(e^(2x) )^((n)) +C_n ^1 (e^(2x) )^((n−1)) =x.2^n e^(2x) +n 2^(n−1) e^(2x) =(x 2^(n ) +n2^(n−1) )e^(2x)
$${f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left({x}\right)^{\left({k}\right)} \left({e}^{\mathrm{2}{x}} \right)^{{n}−{k}} \\ $$$$={x}\left({e}^{\mathrm{2}{x}} \right)^{\left({n}\right)} +{C}_{{n}} ^{\mathrm{1}} \left({e}^{\mathrm{2}{x}} \right)^{\left({n}−\mathrm{1}\right)} \\ $$$$={x}.\mathrm{2}^{{n}} {e}^{\mathrm{2}{x}} +{n}\:\mathrm{2}^{{n}−\mathrm{1}} \:{e}^{\mathrm{2}{x}} \\ $$$$=\left({x}\:\mathrm{2}^{{n}\:} +{n}\mathrm{2}^{{n}−\mathrm{1}} \right){e}^{\mathrm{2}{x}} \\ $$

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