Let's consider the differential equation $$at^2\ddot x(t)+bt\dot x(t)+cx(t)=0$$ where $a\neq 0$ and $a,b,c\in\mathbb R$.
Now it says in my book that if $\varphi: (0,\infty)\to\mathbb R$ is a solution, then so is $\tilde\varphi: (-\infty,0)\to\mathbb R, \tilde\varphi(t)=\varphi(-t)$ and I don't understand that.
Don't we have $\frac{d}{dt}\varphi(-t)=-\dot\varphi(t)$? And then $\frac{d^2}{dt^2}\varphi(-t)=\frac{d}{dt}(-\dot\varphi(t))=-\ddot\varphi(t)$?
No matter what I do, if I plug in the derivatives of $\tilde\varphi$, I don't see how it solves the equation above but I'm sure there is a very simple error in my train of thought.
Your approach is fine, but something went wrong with signs and parentheses.
Note that $\frac{d}{dt}\big(\varphi(-t)\big)=-\dot\varphi(-t)$ and $$\frac{d^2}{dt^2}\big(\varphi(-t)\big)=\frac{d}{dt}\big(-\dot\varphi(-t)\big)=\ddot\varphi(-t).$$ Moreover, since $\varphi(t)$ is a solution $$at^2\ddot \varphi(t)+bt\dot \varphi(t)+c\varphi(t)=0.$$ Thus, by replacing $t$ with $-t$, we get $$a(-t)^2\ddot \varphi(-t)+b(-t)\dot \varphi(-t)+c\varphi(-t)=0$$ that is $$at^2\frac{d^2}{dt^2}\big(\varphi(-t)\big)+bt\frac{d}{dt}\big(\varphi(-t)\big)+c\varphi(-t)=0$$ which means that $t\to \varphi(-t)$ is a solution.