Compute $sumlimits_{n=0}^infty a_nx^n$ if $a_0=3$, $a_1=5$, and $na_n=frac23a_{n-1}-(n-1)a_{n-1}$ for every...












1












$begingroup$



Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.




I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$



$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$



I want to know how to continue it.



Edit: (after reading ideas by @JV.Stalker)



I made the following supplement



$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$






sequences-and-series power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $a_{n-1}$ twice and two conditions ?
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 11:05












  • $begingroup$
    @ClaudeLeibovici I don't know why it's written like this, but it's correct
    $endgroup$
    – Jack throne
    Dec 10 '18 at 11:11










  • $begingroup$
    Did you check the answer below before accepting it?
    $endgroup$
    – Did
    Dec 14 '18 at 11:06










  • $begingroup$
    Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
    $endgroup$
    – Jack throne
    Dec 14 '18 at 14:22
















1












$begingroup$



Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.




I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$



$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$



I want to know how to continue it.



Edit: (after reading ideas by @JV.Stalker)



I made the following supplement



$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$






sequences-and-series power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $a_{n-1}$ twice and two conditions ?
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 11:05












  • $begingroup$
    @ClaudeLeibovici I don't know why it's written like this, but it's correct
    $endgroup$
    – Jack throne
    Dec 10 '18 at 11:11










  • $begingroup$
    Did you check the answer below before accepting it?
    $endgroup$
    – Did
    Dec 14 '18 at 11:06










  • $begingroup$
    Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
    $endgroup$
    – Jack throne
    Dec 14 '18 at 14:22














1












1








1


1



$begingroup$



Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.




I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$



$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$



I want to know how to continue it.



Edit: (after reading ideas by @JV.Stalker)



I made the following supplement



$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$






sequences-and-series power-series







share|cite|improve this question











$endgroup$





Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.




I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$



$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$



I want to know how to continue it.



Edit: (after reading ideas by @JV.Stalker)



I made the following supplement



$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$






sequences-and-series power-series




sequences-and-series power-series






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share|cite|improve this question













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share|cite|improve this question








edited Dec 14 '18 at 11:14









Did

247k23223460




247k23223460










asked Dec 10 '18 at 10:59









Jack throneJack throne

184




184












  • $begingroup$
    $a_{n-1}$ twice and two conditions ?
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 11:05












  • $begingroup$
    @ClaudeLeibovici I don't know why it's written like this, but it's correct
    $endgroup$
    – Jack throne
    Dec 10 '18 at 11:11










  • $begingroup$
    Did you check the answer below before accepting it?
    $endgroup$
    – Did
    Dec 14 '18 at 11:06










  • $begingroup$
    Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
    $endgroup$
    – Jack throne
    Dec 14 '18 at 14:22


















  • $begingroup$
    $a_{n-1}$ twice and two conditions ?
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 11:05












  • $begingroup$
    @ClaudeLeibovici I don't know why it's written like this, but it's correct
    $endgroup$
    – Jack throne
    Dec 10 '18 at 11:11










  • $begingroup$
    Did you check the answer below before accepting it?
    $endgroup$
    – Did
    Dec 14 '18 at 11:06










  • $begingroup$
    Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
    $endgroup$
    – Jack throne
    Dec 14 '18 at 14:22
















$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05






$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05














$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11




$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11












$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06




$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06












$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22




$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22










1 Answer
1






active

oldest

votes


















-1












$begingroup$

$na_n=frac{5}{3}a_{n-1}-na_{n-1}$



Multiply by $x^n$ both sides and sum from $n=1$ to $infty$



$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$



Reindex of the RHS:



$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$



After sorting the eqution:



$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$



We have the following differential eqution:



$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$



Finally $f(x)=(x+1)^frac{2}{3}+c$



$sumlimits_{n=1}^infty a_n x^n$ is convergent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
    $endgroup$
    – JV.Stalker
    Dec 10 '18 at 14:30












  • $begingroup$
    I made the following supplement after your method, and Thanks again! @JV.Stalker
    $endgroup$
    – Jack throne
    Dec 13 '18 at 13:46












  • $begingroup$
    @Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
    $endgroup$
    – JV.Stalker
    Dec 13 '18 at 15:30










  • $begingroup$
    "Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
    $endgroup$
    – Did
    Dec 14 '18 at 11:05










  • $begingroup$
    Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
    $endgroup$
    – Did
    Dec 14 '18 at 11:08











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

$na_n=frac{5}{3}a_{n-1}-na_{n-1}$



Multiply by $x^n$ both sides and sum from $n=1$ to $infty$



$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$



Reindex of the RHS:



$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$



After sorting the eqution:



$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$



We have the following differential eqution:



$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$



Finally $f(x)=(x+1)^frac{2}{3}+c$



$sumlimits_{n=1}^infty a_n x^n$ is convergent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
    $endgroup$
    – JV.Stalker
    Dec 10 '18 at 14:30












  • $begingroup$
    I made the following supplement after your method, and Thanks again! @JV.Stalker
    $endgroup$
    – Jack throne
    Dec 13 '18 at 13:46












  • $begingroup$
    @Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
    $endgroup$
    – JV.Stalker
    Dec 13 '18 at 15:30










  • $begingroup$
    "Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
    $endgroup$
    – Did
    Dec 14 '18 at 11:05










  • $begingroup$
    Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
    $endgroup$
    – Did
    Dec 14 '18 at 11:08
















-1












$begingroup$

$na_n=frac{5}{3}a_{n-1}-na_{n-1}$



Multiply by $x^n$ both sides and sum from $n=1$ to $infty$



$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$



Reindex of the RHS:



$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$



After sorting the eqution:



$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$



We have the following differential eqution:



$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$



Finally $f(x)=(x+1)^frac{2}{3}+c$



$sumlimits_{n=1}^infty a_n x^n$ is convergent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
    $endgroup$
    – JV.Stalker
    Dec 10 '18 at 14:30












  • $begingroup$
    I made the following supplement after your method, and Thanks again! @JV.Stalker
    $endgroup$
    – Jack throne
    Dec 13 '18 at 13:46












  • $begingroup$
    @Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
    $endgroup$
    – JV.Stalker
    Dec 13 '18 at 15:30










  • $begingroup$
    "Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
    $endgroup$
    – Did
    Dec 14 '18 at 11:05










  • $begingroup$
    Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
    $endgroup$
    – Did
    Dec 14 '18 at 11:08














-1












-1








-1





$begingroup$

$na_n=frac{5}{3}a_{n-1}-na_{n-1}$



Multiply by $x^n$ both sides and sum from $n=1$ to $infty$



$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$



Reindex of the RHS:



$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$



After sorting the eqution:



$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$



We have the following differential eqution:



$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$



Finally $f(x)=(x+1)^frac{2}{3}+c$



$sumlimits_{n=1}^infty a_n x^n$ is convergent.






share|cite|improve this answer









$endgroup$



$na_n=frac{5}{3}a_{n-1}-na_{n-1}$



Multiply by $x^n$ both sides and sum from $n=1$ to $infty$



$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$



Reindex of the RHS:



$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$



After sorting the eqution:



$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$



Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$



We have the following differential eqution:



$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$



Finally $f(x)=(x+1)^frac{2}{3}+c$



$sumlimits_{n=1}^infty a_n x^n$ is convergent.







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answered Dec 10 '18 at 14:24









JV.StalkerJV.Stalker

64839




64839












  • $begingroup$
    I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
    $endgroup$
    – JV.Stalker
    Dec 10 '18 at 14:30












  • $begingroup$
    I made the following supplement after your method, and Thanks again! @JV.Stalker
    $endgroup$
    – Jack throne
    Dec 13 '18 at 13:46












  • $begingroup$
    @Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
    $endgroup$
    – JV.Stalker
    Dec 13 '18 at 15:30










  • $begingroup$
    "Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
    $endgroup$
    – Did
    Dec 14 '18 at 11:05










  • $begingroup$
    Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
    $endgroup$
    – Did
    Dec 14 '18 at 11:08


















  • $begingroup$
    I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
    $endgroup$
    – JV.Stalker
    Dec 10 '18 at 14:30












  • $begingroup$
    I made the following supplement after your method, and Thanks again! @JV.Stalker
    $endgroup$
    – Jack throne
    Dec 13 '18 at 13:46












  • $begingroup$
    @Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
    $endgroup$
    – JV.Stalker
    Dec 13 '18 at 15:30










  • $begingroup$
    "Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
    $endgroup$
    – Did
    Dec 14 '18 at 11:05










  • $begingroup$
    Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
    $endgroup$
    – Did
    Dec 14 '18 at 11:08
















$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30






$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30














$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46






$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46














$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30




$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30












$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05




$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05












$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08




$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08


















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