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Let f be infinitely differentiable on R, and exist constants $a,b$ s.t.:

$f''(x)+af'(x)+bf(x)=0$ for all x, I want to show: $f(x)=\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ for all x.

2 Answers 2

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One slightly mundane way is to actually solve the differential equation; we get $$c_1 e^{\frac{1}{2} x \left(-\sqrt{a^2+4 b}-a\right)}+c_2 e^{\frac{1}{2} x \left(\sqrt{a^2+4 b}-a\right)}$$
Now, we know that $\exp\left[f(x)\right]$ admits a Power Series, so we are pretty much done. All that is left is to solve for the radius of convergence.

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The differential equation $f''(x)+af'(x)+bf(x)=0$ has the char. polynomial $p(t)=t^2+at+b$. Let $t_{1/2}$ the roots of $p$.

Case 1: $t_1=t_2.$ Then $f$ has the form

$f(x)=c_1e^{t_1 x}+c_2xe^{t_1 x}$ with constants $c_1,c_2$. Hence $f(x)=\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ for all $x$.

Case 2: $t_1 \ne t_2$ and $t_1,t_2 \in \mathbb R$. Then $f$ has the form

$f(x)=c_1e^{t_1 x}+c_2e^{t_2 x}$ with constants $c_1,c_2$. Hence $f(x)=\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ for all $x$.

Case 3: $t_1,t_2 \in \mathbb C \setminus \mathbb R$.

Let $t_1=u+iv$ with $u \in \mathbb R$ and $v \in \mathbb R \setminus \{0\}$

Then $f$ has the form

$f(x)=c_1e^{u x}\cos(v x)+c_2e^{ u x} \sin(v x)$ with constants $c_1,c_2$. Hence $f(x)=\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ for all $x$.