Let $\alpha,\beta,a,b$ be real constants. Show that the differential equation given by:
$y''= ay' + by \\ y(0)=\alpha\\ y'(0)=\beta$
has an unique solution and this solution is analytic in $\mathbb{R}$
Solution:
I am trying to bound $\frac{f^{(n)}(x)}{n!}$ by some $M$, to show it is analytic. I know that if $y$ is an analytic solution of this function, then it must be infinitely differentiable and, hence, must be continuous. Then, for $x$ in $[-R,R]$, $y$ must be bounded by some $M$.
After that, I am trying to show by induction that $$y^{(n+2)}= (y'')^{n}=(\alpha y' + by)^{n}= \alpha y^{(n+1)} + by^{(n)}.$$
So I need to show that $ y^{(n+1)} $ and $ y^{(n)} $ are bounded (which is what I am trying to show by induction). However, I am having trouble with the first step $n=1$ of the induction. Moreover, I am not using the hypothesis given regarding the behavior of the derivatives in $x=0$.
Thanks!
I could solve it. $y'(x)$ must be continuous because $y''(x)$ exists. As $y'(x)$ is on a bounded interval, then it must be bounded.
So I take $t=max(|a|,|b|)$, and $M=max(M1,M2)$ which are the bounds for the first derivative and the original function.
Finally, $y^{(n+2)}= (y'')^{n}=(\alpha y' + by)^{n}= \alpha y^{(n+1)} + by^{(n)} \leq |a|Mt^{n+1}(n+1)!+|b|Mt^{n}n! \leq Mt^{n}n!(2t^2(n+1)) \leq Mt^{n+2}(n+2)!$
Then in an interval $(-R,R)$ all the derivatives are bounded and the function is analytic.