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Using power series method solve $$tx''(t) - tx'(t) - x(t) = 0 , \\ x(0)=0, \\ x'(0)=1$$

We can take $$x(t) = \sum_{n=0}^{\infty}a_nt^n$$ Furthermore we have $$x'(t) = \sum_{n=0}^{\infty}na_nt^{n-1} \\ x''(t) = \sum_{n=0}^{\infty}n(n-1)a_nt^{n-2}$$

Then I substituted it to $tx''(t) - tx'(t) - x(t) = 0$ and I received $$\sum_{n=2}^{\infty} n(n-1)a_nt^{n-2} - \sum_{n=1}^{\infty}n a_nt^{n} - \sum_{n=0}^{\infty}a_nt^{n} = 0$$

What can I do next? Thanks in advance.

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Hints: Get all the powers in the summations the same, Get all of the ranges of the summations the same, Write the general terms for the constants, Solve for those terms –  Amzoti Sep 2 '13 at 17:25
    
You will discover the recurrence relation between the coefficients. –  dot dot Sep 2 '13 at 17:25
    
It looks like you have a typo in the first series in your last equation; it should be $t^{n-1}$ and not $t^{n-2}$? –  user2566092 Sep 2 '13 at 17:26
1  
I would recommend writing $x(t) = a_0 + a_1t + a_2t^2+a_3t^3+a_4t^4+\cdots$ to start with. Do the differentiations, make the substitutions, apply the initial conditions and then simplify. Most questions only ask for the first few terms (the $k$-jet) of $x$ anyway. If you're asked for the whole thing, then starting with finite terms will help you see any patterns. –  Fly by Night Sep 2 '13 at 17:26
    
You should get a series that is the expansion of $x(t) = t~e^t$. –  Amzoti Sep 2 '13 at 17:36

1 Answer 1

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If $$ x(t)=\sum_{n=0}^\infty a_nt^n $$ is a solution of the DE $$ tx''-tx'-x=0, $$ then we have $$ 0=t\sum_{n=2}^\infty n(n-1)a_nt^{n-2}-t\sum_{n=1}^\infty na_nt^{n-1}-\sum_{n=0}^\infty a_nt^n=\sum_{n=0}^\infty(n+1)\left[na_{n+1}-a_n\right]t^n. $$ It follows that $$ na_{n+1}-a_n=0 \quad \forall n \ge 0 $$ with $$ a_0=x(0)=0,\ a_1=x'(0)=1. $$ Thus for every $n \ge 2$ we have $$ a_n=\frac{1}{n-1}\ldots\frac{1}{2-1}a_{2-1}=\frac{1}{(n-1)!}. $$ We then deduce that $$ x(t)=t+\sum_{n=2}^\infty\frac{1}{(n-1)!}t^n=t+t\sum_{n=1}^\infty\frac{1}{n!}t^n=t+t(e^t-1)=te^t. $$

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