Consider the power series $$ f(t)=\sum_{n=0}^{\infty} \frac{1}{\Gamma\left(n+\frac{1}{2}\right)} t^{n}, \quad(\Gamma \text { is the Gamma function) } $$
Determine the set of convergent $D$ of the power series
Show that for all $t \in D$ $$ 2 t f^{\prime}(t)-(2 t+1) f(t)+\frac{1}{\sqrt{\pi}}=0 $$
Deduce the closed form of $f(t)$.
My answer that I have tried.
For (1), I find the radius of convergent by using $$R=\lim_{n\to+\infty}\frac{|a_n|}{|a_{n+1}|}=+\infty$$ Then the set of convergent $D=\mathbb{R}$
For (2) I have tried by finding:
$$f^{\prime}(t)=\sum_{n=1}^{\infty}\frac{n}{\Gamma\left(n+1/2\right)}t^{n-1}$$ $$2tf^{\prime}(t)=\sum_{n=1}^{\infty}\frac{2n}{\Gamma\left(n+1/2\right)}t^{n}$$ $2tf^{\prime}(t)-(2t+1)f(t)= \sum_{n=1}^{\infty}\frac{2n}{\Gamma\left(n+1/2\right)}t^{n}-(2t+1) \sum_{n=0}^{\infty}\frac{1}{\Gamma\left(n+1/2\right)}t^{n} $ $$2tf^{\prime}(t)-(2t+1)f(t)= \sum_{n=0}^{\infty}\frac{2n-2t-1}{\Gamma\left(n+1/2\right)}t^{n} $$ Then I stuck on this part. Could you give me some recommendations to do this ?
Note that $1/\sqrt\pi=f(0)$, so that \begin{align*} 2tf'(t)-f(t)+f(0)&=\sum_{n=1}^\infty\frac{2n-1}{\Gamma(n+1/2)}t^n \\\color{gray}{\big[\Gamma(z+1)=z\Gamma(z)\big]}\quad&=2\sum_{n=1}^\infty\frac{t^n}{\Gamma(n-1/2)} \\\color{gray}{\big[n\mapsto n+1\big]}\quad&=2\sum_{n=0}^\infty\frac{t^{n+1}}{\Gamma(n+1/2)}=2tf(t). \end{align*} Regarding the closed form, we just solve the resulting ODE: $$f(t)=e^t\sqrt{t}\left(\textrm{const.}-\int\frac{e^{-t}\,dt}{2t\sqrt{\pi t}}\right);$$ using integration by parts and the condition $f(0)=1/\sqrt\pi$, we obtain $$f(t)=\frac1{\sqrt\pi}\left(1+e^t\sqrt{t}\int_0^t\frac{e^{-x}}{\sqrt{x}}\,dx\right).$$ This can also be written as $f(t)=f(0)+e^t\sqrt{t}\operatorname{erf}\sqrt{t}$.