# Triangle inequality and reverse triangle inequality problem

Alright, so I am delving into the behavior of an non-homogeneous underdamped equation, for a project in Differential Equations, with the equation $$y^{''}+py^{'}+qy=f(t)$$ where $p^{2}<4q$ and $p$ and $q$ are both positive.

Right now, I'm to the point where I have already shown $$y(t)=c_1e^{\alpha t}cos\beta t+c_2e^{\alpha t}sin\beta t-\frac 1\beta e^{\alpha t}cos\beta t\int_0^t f(v)e^{-\alpha v}sin\beta v\space dv+\frac 1\beta e^{\alpha t}sin\beta t\int_0^t f(v)e^{-\alpha v}cos\beta v\space dv$$ is a general solution to the D.E. and I have to get from there to here: $$\lvert y(t) \rvert \le (\lvert c_1 \rvert + \lvert c_2\rvert)e^{\alpha t}+ \frac {2K}{\lvert\alpha\rvert\beta}(1-e^{\alpha t})$$ using the properties of abosolute values. I also know $\lvert f(v)\rvert\le K$ for all $v\ge 0$. I have looked at this multiple ways and the first half seems fairly straight forward to use the triangle inequality, and I'm pretty sure you have to use the reverse triangle inequality for the second half but I keep getting 0 if I do that. If I use the regular triangle inequality then my signs are opposite for some reason. Some insight would be much appreciated.

First, we have $p$ is positive, so $\alpha$ is negative. Next, by applying the triangle inequality, we get $$\vert y(t)\vert\leq (\lvert c_1 \rvert + \lvert c_2\rvert)e^{\alpha t}+\frac{1}{\vert \beta \vert} e^{\alpha t}\int_0^t \vert f(v)\vert e^{-\alpha v}\space dv+\frac{1}{\vert \beta \vert} e^{\alpha t}\int_0^t \vert f(v)\vert e^{-\alpha v}\space dv$$ Since $f$ is bounded on $\mathbb{R}_+$, ie, $\exists K\geq 0,\ s.t.\ \vert f(t)\vert \leq K$, $\quad\forall t\geq0$, so