Let $p(t),q(t)$, and $r(t)$ be continuous functions on an open interval $I$ and let $t_0\in I$. Assume that there exists a positive constant $M$ such that $$|p(t)|+|q(t)|+|r(t)|<M$$ for all $t\in I$. Let $f$ be a three times differentiable function which is a solution of a differential equation $$f'''(t) + p(t)f''(t) + q(t)f'(t) + r(t)f(t) = 0$$ on $I$. Finally, define $$\psi(t) := |f(t)|^2 + |f'(t)|^2 + |f''(t)|^2$$ I want to show that
(1) $|\psi'(t)|\leq 2(1+M)\psi(t)$
(2) $\psi(t_0)e^{-2(1+M)|t-t_0|} \leq \psi(t) \leq \psi(t_0)e^{2(1+M)|t-t_0|}$
If the inequality in (1) is shown, (2) can be proven using an integrating factor. But I cannot figure out how to prove (1). $$\psi'(t) = 2f(t)f'(t) + 2f'(t)f''(t) + 2f''(t)f'''(t)$$ The form of $\psi'(t)$ seems to require the Cauchy-Schwarz inequality to prove the inquality, but where to use the bounds of $p,q,r$? Another try was to substitute $-pf'' - qf' - rf$ in the place of $f'''$, but I have no idea how to relate its upper bound with the function $\psi$ itself.
Thanks in advance for any form of help, hint, or solution!
First use the Cauchy-Schwarz inequality, the differential equation, and the bounds on $p, q, r$ to estimate $f'''(t)^2$: $$ \begin{align} f'''(t)^2 &= \bigl(p(t)f''(t) + q(t)f'(t) + r(t)f(t)\bigr)^2 \\ &\le \bigl(f''(t)^2+f'(t)^2+f(t)^2\bigr) \cdot \bigl(p(t)^2+q(t)^2+r(t)^2\bigr) \\ &\le \bigl(f''(t)^2+f'(t)^2+f(t)^2\bigr) \cdot \bigl(|p(t)|+|q(t)|+|r(t)|\bigr)^2 \\ &\le \psi(t) M^2 \, . \end{align} $$ Then use the Cauchy-Schwarz inequality again to estimate $\psi'(t)^2$: $$ \begin{align} \psi'(t)^2 &= \bigl(2f(t)f'(t) + 2f'(t)f''(t) + 2f''(t)f'''(t)\bigr)^2 \\ &\le 4 \bigl(f(t)^2+f'(t)^2+f''(t)^2\bigr) \cdot \bigl(f'(t)^2+f''(t)^2+f'''(t)^2\bigr) \\ &\le 4 \psi(t) \cdot (\psi(t) + \psi(t) M^2) \\ &= 4 \psi(t)^2 (1 + M^2) \\ &\le 4 \psi(t)^2 (1+M)^2 \, . \end{align} $$