$\renewcommand{\vec}[1]{\underaccent{\tilde}{#1}}$ I'm trying to do a three part mechanics question that goes as follows:
i) If $\frac{du}{dt}+uf(t)=g(t)$ and $F'(t)=f(t)$, by considering $\frac{d}{dt}\left(ue^{F(t)}\right)$ show that $\displaystyle u=e^{-F(t)}\int e^{F(t)}g(t) \;dt$
ii) A projectile is fired with an initial velocity vector $\underset{\sim}{v}=\left(16\underset{\sim}{i}+12\underset{\sim}{j}\right)\text{ms}^{-1}$ from a point $O$ and travels in a vertical plane with position vector $\underset{\sim}{r}(t)=\left(x\underset{\sim}{i}+y\underset{\sim}{j}\right)$ metres at time $t$ seconds. The acceleration due to gravity is $10 \text{ms}^{-2}$ and a resistance force acts against motion such that $\underset{\sim}{\ddot{r}}(t)=\left[-\frac{1}{10}\dot{x}\underset{\sim}{i}-\left(\frac{1}{10}\dot{y}+10\right)\underset{\sim}{j}\right]\text{ms}^{-2}$.
By writing $\frac{d}{dt}\left(\dot{y}\right)+\frac{1}{10}\dot{y}=-10$ and using part i), find $\dot{y}$ as a function of $t$ and hence the maximum height attained by the projectile.
iii) Assuming the projectile is free to fall indefinitely after reaching its max height, find the limiting motion of the particle as $t\rightarrow\infty$ by considering expressions of $\underset{\sim}{r}(t)$ and $\underset{\sim}{\dot{r}}(t)$ as functions of $t$.
My attempt:
For i): $\frac{d}{dt}\left(ue^{F(t)}\right)=\frac{du}{dt}e^{F(t)}+uf(t)e^{F(t)}$
=$[g(t)-uf(t)]e^{F(t)}+uf(t)e^{F(t)}$
=$g(t)e^{F(t)}$
Therefore integrating both sides and dividing by $e^{F(t)}$ gives $\displaystyle u=e^{-F(t)}\int e^{F(t)}g(t)\;dt$.
For part ii) I'm starting to struggle. I assume that from the equation $\frac{d}{dt}\left(\dot{y}\right)+\frac{1}{10}\dot{y}=-10$, I can use part i) with $u=\dot{y}$, $f(t)=\frac{1}{10}$ and $g(t)=-10$.
That gives me $\displaystyle\dot{y}=e^{-\frac{t}{10}}\int e^{\frac{t}{10}}(-10)\;dt$
=$\displaystyle e^{-\frac{t}{10}}(-100)e^{\frac{1}{10}t}+c$
=$-100+c$, which is a constant.
iii) I assume the way you would do this is find $\dot{x}$ in terms of t, then integrate and integrate. To find $\dot{x}$ I solved the separable differential equation $\frac{d}{dt}(\dot{x})=\frac{-1}{10}\dot{x}$ to get $\dot{x}=12e^{-\frac{1}{10}t}$ after plugging in my initial value of $t=0, \dot{x}=12$. However I cannot progress further from here as I can't find $\dot{y}$ from the previous part, could someone help me do that?