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I have given:
$$f''(x) - 2f'(x) + f(x) = 0$$ $$f(0) = 0$$ $$f'(0) = 1$$
$$find::---- f(x)=?$$.

I wanted to try it by assuming that the series $$f\left(x\right)\:=\:\sum _{n=0}^{\infty }\:a_n\cdot x^n$$ is a solution. But I got stuck here. I thought maybe using Taylor expansion and find something, but I dont know if it can help me...

Is my idea correct of what I am trying? I got $f(x)$, I need to derivative it, I will get $f'(x)$ and then again: $f''(x)$.
But the problem I will get this if I use given values: $$\:\sum _{n=0}^{\infty }\:a_n\cdot x^n\:-\:2\sum _{n=1}^{\infty }\:a_n\cdot n\cdot x^{n-1}+\sum _{n=2}^{\infty }\:a_n\cdot n\left(n-1\right)\cdot x^{n-2}$$ But it wont help me... Also using the other given values wont help me.. How should I start answer this question? What am I supposed to think when I see this question?

EDIT: Please help someone, I reached that point: $$a_n=2a_{n+1}\left(n+1\right)-a_{n+2}\left(n+1\right)\left(n+2\right)$$ With $a_0 = 0$ and $a_1 = 1$ But I am stuck here.. Someone here tried to help me, but I cant reach that function, we have not learned ODE yet.

Math_begineer
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  • Equate the coefficients of $x^n$ on both sides. Be careful with boundary conditions. – copper.hat Jun 11 '22 at 20:46
  • It's not clear what you are trying to find in this problem. – Matt E. Jun 11 '22 at 20:46
  • @copper.hat what are you talking about? I dont understand :\ please explain. – Math_begineer Jun 11 '22 at 20:46
  • @MattE. I am trying to find the function f(x), sorry for not explaining it very good. – Math_begineer Jun 11 '22 at 20:47
  • You need to figure out $a_0,a_1,...$. For example if $a x^2 + b x = x^2 + 3x$ for all $x$ then I know that $a=1, b=3$. – copper.hat Jun 11 '22 at 20:57
  • @copper.hat yea I thought I had to find the recursion formula for $a_n$, but the question is, How to do it... since I need to use some information for it, but I cant eq power series. If I put $x=0$ on all series, I will just receive 0. which is problematic. – Math_begineer Jun 11 '22 at 20:58
  • @copper.hat Please help now, I updated the question, how do I continue from here? – Math_begineer Jun 12 '22 at 21:16
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    I have not checked your arithmetic, but that it what I would have expected, You would need to solve the difference equation to get an expression for $a_n$ which, base on the accepted answer, would be $a_n = {1 \over (n-1)!}$ for $n >1$. – copper.hat Jun 12 '22 at 21:22
  • I have to use ODE for it? since we really have not learned it :
    How am I supposed to get this? I really cant understand, also the function with $e$     – Math_begineer Jun 12 '22 at 21:24

1 Answers1

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The function is given by finding the general solution, using characteristic equation of the ODE which leads to the general solution f(x)= a$e^{x}+bxe^{x}$,Setting f(0)=0 we get a=0 and setting f '(0)=1 we get b+b.o=1 hence b=1. So the function you are seeking is f(x)=$xe^{x}$ which is x($\sum_{0}^{\infty }\frac{x^{n}}{n!}$) in power series!

  • Hi, how did you think of such a thing? how did you think I need to reach that f(x)? I dont know what is ODE, I have never learned that thing, isn't there any way with power series? since this is the subject we learn. – Math_begineer Jun 12 '22 at 05:04
  • ODE we will learn soon, we have not learned it yet:\ – Math_begineer Jun 12 '22 at 05:04
  • Hi, I reached that point: $a_n=2a_{n+1}\left(n+1\right)-a_{n+2}\left(n+1\right)\left(n+2\right)$ How do I reach your f(x)? – Math_begineer Jun 12 '22 at 21:14