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Question Number 187155 by TUN last updated on 14/Feb/23

f(x)=x.e^(2x)   =>f^((n)) (x)=

f(x)=x.e2x =>f(n)(x)=

Commented by0670322918 last updated on 14/Feb/23

Commented bymr W last updated on 15/Feb/23

sir:  please post your answer as ♮answerε,  not as ♮commentε! thanks!

sir: pleasepostyouranswerasanswerε, notascommentε!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)

⇒⇒f(0)=a0xe2x+b0=1xe2x+0 f(m1)=am1xe2x+bm1e2x f(m)=(2am1)xe2x+(am1+2bm1)e2x=amxe2x+bme2x ⇒⇒getf(n)=anxe2x+bne2x an=2an1=4an1=...=2na0=2n bn=an1+2bn1=an1+2an2+4bn2=...= an1+2an2+4an3+...+2n1a0+2nb0=2n1n ⇒⇒f(n)=2nxe2x+2n1ne2x

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(n)(x)=k=0nCnk(x)(k)(e2x)nk =x(e2x)(n)+Cn1(e2x)(n1) =x.2ne2x+n2n1e2x =(x2n+n2n1)e2x

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