ordinary differential equations - Power series of the solution of $2t^2x'' + tx' -(t+1)x=0$

Use the method of Frobenius, with the larger root of the indicial
equation, to find the first three terms of the power series of the solution to $$2t^2x'' +
tx' -(t+1)x=0.$$

My work:

Note that $t=0$ is a regular singular point. On solving the indicial equation, we get $r=1$ or $r=-\frac12$. We proceed by letting $$x(t) = t^r \sum_{k=0}^{\infty} a_k t^k$$ which, after plugging into the equation, yields $$(ra_1 - a_0)t^{r+1} - a_0 t^r + \sum_{n=r+2}^{\infty} [(2n^2-n-1)a_{n-r} - a_{n-r-1}]t^n=0.$$ However, I cannot proceed. The above with $r=1$ would imply that all coefficients are zero. The calculation should be fine as several of my friends also get the same answer. Should we conclude that the first three terms are all zero?

Answer

you have $$2t^2x''+tx' - (t+1)x= 0 $$ now sub $$x = t^k + a_1t^{k+1} + a_2 t^{k+2} + \cdots\\
x' = kt^{k-1} + (k+1)a_1t^k + (k+2)a_2t^{k+1} + \cdots\\
x'' = (k-1)kt^{k-2} + k(k+1)a_1t^{k-1} + (k+1)(k+2)a_2t^{k} + \cdots$$ we get
$$2t^2\left( (k-1)kt^{k-2} + k(k+1)a_1t^{k-1} + (k+1)(k+2)a_2t^{k} + \cdots\right) + t\left( kt^{k-1} + (k+1)a_1t^k + (k+2)a_2t^{k+1} + \cdots\right)- (t+1)\left( t^k + a_1t^{k+1} + a_2 t^{k+2} + \cdots\right) =0$$

equating the coefficient of $t^{k}$ we have

$$f(k)=2(k-1)k +k-1=(k-1)(2k+1)=0 \to k = 1, k = -\frac 12.$$

equating the coefficient of $t^{k+1}$ we have

$$2k(k+1)a_1+(k+1)a_1 -a_1-1=0\to a_1 = \frac 1{f(k+1)} $$
in the same way you find $$a_2 = \frac {a_1}{f(k+2)}, \cdots, a_{n+1}=\frac {a_n}{f(n+k)}, = n = 1, 2, \cdots $$

let us look at the series for the case $k = 0$

$$a_1 = \frac 1{f(2)} = \frac 1{1 \cdot 5}\\
a_2 = \frac {a_1}{f(3)}=\frac 1{1\cdot 2 \cdot 5 \cdot 7}\\

a_3 = \frac{a_2} {f(4)}=\frac 1{1\cdot 2 \cdot 3 \cdot 5 \cdot 7 \cdot 9} \\
\vdots$$