During lecture while going through an ODE question there was a specific part of the proof that requires the observation that:
$$ \lim_{t \to \infty} \frac{\left(e^{(\lambda_1-\lambda_2)t}(x(t)e^{-\lambda_1t})'\right)'}{e^{-\lambda_2t}} =\lim_{t \to \infty} \frac{e^{(\lambda_1-\lambda_2)t}(x(t)e^{-\lambda_1t})'}{e^{-\lambda_2t}} =0$$
But I'm confused as to why the anti-derivative of the denominator isn't
$$\frac{e^{-\lambda_2t}}{-\lambda_2} $$
Don't know if you require the full question but I'll just add it in w the suggested solution just in case:
Question:
Let $ p,q \in \Bbb{R} $. Suppose $ \lambda_1, \lambda_2$ are roots of the equation $ \lambda^2 + p\lambda +q = 0$ where $ Re(\lambda_1) < 0 $ and $ Re(\lambda_2)<0 $. Assume there exists $ x(t) \in C^2([0,\infty))$ satisfying $ \lim_{t \to \infty} (x''(t) + px'(t) +qx(t) =0 $. Prove that $ \lim_{ t \to \infty} x(t) = 0 $.
Suggested hint:
By vietas' formula, $ \lambda_1 +\lambda_2 = -p , \lambda_1\lambda_2= q $;
Then
\begin{align}
x''(t) + px'(t) +qx(t) &= x''(t) – \lambda_1x'(t) – \lambda_2x'(t) + \lambda_1\lambda_2x(t)
\\&= (x'(t) – \lambda_1x(t))' – \lambda_2[x'(t)-\lambda_1x(t)] \\&= e^{\lambda_2t} \cdot e^{-\lambda_2t}\left((x'(t) – \lambda_1x(t))' – \lambda_2[x'(t)-\lambda_1x(t)]\right) \\&= e^{\lambda_2t} \left( e^{-\lambda_2t}[x'(t) – \lambda_1x(t)]\right)' \\&= e^{\lambda_2t} \left( e^{-\lambda_2t}\cdot e^{\lambda_1t} \cdot e^{-\lambda_1t}[x'(t) -\lambda_1x(t)]\right)'
\\&= e^{\lambda_2t} \left( e^{(\lambda_1-\lambda_2)t}\left( x(t)e^{-\lambda_1t}\right)'\right)'
\end{align}
which brings me to my question above
edit: do we have to use the information that $ Re(\lambda_1) < 0 $ and $ Re(\lambda_2)<0 $ for the L'hopital part?
Thank you for reading my question!
Best Answer
As you've noted, it should read$$\lim_{t \to \infty} \frac{\left(e^{(\lambda_1-\lambda_2)t}(x(t)e^{-\lambda_1t})^\prime\right)^\prime}{e^{-\lambda_2t}} =-\lambda_2\lim_{t \to \infty} \frac{e^{(\lambda_1-\lambda_2)t}(x(t)e^{-\lambda_1t})^\prime}{e^{-\lambda_2t}} =0$$or$$\lim_{t \to \infty} \frac{e^{(\lambda_1-\lambda_2)t}(x(t)e^{-\lambda_1t})^\prime}{e^{-\lambda_2t}} =0\implies\lim_{t \to \infty} \frac{\left(e^{(\lambda_1-\lambda_2)t}(x(t)e^{-\lambda_1t})^\prime\right)^\prime}{e^{-\lambda_2t}} =0$$but the author's notation was careless.