Parametric Adventure

$\begin{cases} \cfrac{dy}{dx}=2t \\ \cfrac{d^3y}{dx^3}=t^3 \end{cases}$

Relevant wikis: Parametric Derivative , Chain Rule

$\cfrac{dy}{dx}=\cfrac{dy}{dt}\cdot\cfrac{dt}{dx}$

$\therefore \cfrac{dy}{dt}\cdot\cfrac{dt}{dx}=2t$

$\cfrac{d^3y}{dx^3}=\cfrac{d}{dx}(\cfrac{d^2y}{dx^2})=\cfrac{d}{dt}(\cfrac{d^2y}{dx^2})\cdot\cfrac{dt}{dx}$

$\therefore \cfrac{d}{dt}(\cfrac{d^2y}{dx^2})\cdot\cfrac{dt}{dx}=t^3$

$\begin{cases} \cfrac{dy}{dt}\cdot\cfrac{dt}{dx}=2t \\ \cfrac{d}{dt}(\cfrac{d^2y}{dx^2})\cdot\cfrac{dt}{dx}=t^3 \end{cases}$

$\cfrac{\cfrac{d}{dt}(\cfrac{d^2y}{dx^2})\cdot\cfrac{dt}{dx}}{\cfrac{dy}{dt}\cdot\cfrac{dt}{dx}}=\cfrac{t^3}{2t}$

$\cfrac{\cfrac{d}{dt}(\cfrac{d^2y}{dx^2})}{\cfrac{dy}{dt}}=\cfrac{t^3}{2t}$

$\cfrac{\cfrac{d}{dt}(\cfrac{d^2y}{dx^2})}{2t\cdot\cfrac{dx}{dt}}=\cfrac{t^3}{2t}$

$\cfrac{\cfrac{d}{dt}(\cfrac{d^2y}{dx^2})}{\cfrac{dx}{dt}}=t^3$

$\cfrac{d^2y}{dx^2}=\cfrac{d}{dx}(\cfrac{dy}{dx})=\cfrac{d}{dt}(\cfrac{dy}{dx})\cdot\cfrac{dt}{dx}=\cfrac{d}{dt}(2t)\cdot\cfrac{dt}{dx}=2\cfrac{dt}{dx}$

$\therefore \cfrac{\cfrac{d}{dt}(2\cfrac{dt}{dx})}{\cfrac{dx}{dt}}=t^3$

$\cfrac{\cfrac{d}{dt}(2\cdot\cfrac{1}{\cfrac{dx}{dt}})}{\cfrac{dx}{dt}}=t^3$

$\cfrac{\cfrac{-2}{(\cfrac{dx}{dt})^2}\cdot\cfrac{d^2x}{dt^2}}{\cfrac{dx}{dt}}=t^3$

Let $\cfrac{dx}{dt}=z$

$\Rightarrow \cfrac{d}{dt}({\cfrac{dx}{dt}})=\cfrac{dz}{dt}$

$\therefore \cfrac{d^2x}{dt^2}=\cfrac{dz}{dt}$

$\cfrac{-2\cfrac{dz}{dt}}{z^3}=t^3$

$\int \cfrac{-2}{z^3}dz=\int t^3dt$

$\cfrac{1}{z^2}=\cfrac{1}{4}\cdot{t^4}+c_{1}$

When $t=2$ , $\cfrac{dx}{dt}=-\cfrac{1}{2}$

When $t=2$ , $z=-\cfrac{1}{2}$

$\cfrac{1}{(-\cfrac{1}{2})^2}=\cfrac{1}{4}\cdot{2^4}+c_{1}$

$4=4+c_{1}$

$c_{1}=0$

$\therefore \cfrac{1}{z^2}=\cfrac{1}{4}\cdot{t^4}$

$|\cfrac{dx}{dt}|+\cfrac{dx}{dt}=0$

$|z|+z=0$

$\Rightarrow |z|=-z$

$\therefore z<0$

$\therefore \cfrac{1}{z}=\cfrac{-1}{2}\cdot{t^2}$

$\cfrac{1}{\cfrac{dx}{dt}}=\cfrac{-1}{2}\cdot{t^2}$

$\cfrac{dt}{dx}=\cfrac{-1}{2}\cdot{t^2}$

$\int\cfrac{-1}{t^2}dt=\cfrac{1}{2}\int1dx$

$\cfrac{1}{t}=\cfrac{x}{2}+c_{2}$

When $t=2$ , $x=1$

$\cfrac{1}{2}=\cfrac{1}{2}+c_{2}$

$c_{2}=0$

$\therefore \cfrac{1}{t}=\cfrac{x}{2}$

$\cfrac{dy}{dt}\cdot\cfrac{dt}{dx}=2t$

$\cfrac{dy}{dt}\cdot\cfrac{-1}{2}\cdot{t^2}=2t$

$\cfrac{dy}{dt}=2t\cdot{-2}\cdot\cfrac{1}{t^2}$

$\cfrac{dy}{dt}=\cfrac{-4}{t}$

$y=-4ln|t|+c_{3}$

When $t=1$ , $y=0$

$0=-4ln|1|+c_{3}$

$c_{3}=0$

$\therefore y=-4ln|t|$

$y=ln(\cfrac{1}{t})^4$

$y=ln(\cfrac{x}{2})^4$

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$\begin{cases} y=f(x) \\ e^y=x^2 \end{cases}$

$\Rightarrow \begin{cases} y=ln(\cfrac{x}{2})^4 \\ e^y=x^2 \end{cases}$

$\Rightarrow \begin{cases} e^y=(\cfrac{x}{2})^4 \\ e^y=x^2 \end{cases}$

$\therefore (\cfrac{x}{2})^4=x^2$

$\cfrac{x^4}{16}=x^2$

$x^4=16x^2$

$\because x>0, \therefore x\neq0$

$\therefore x^2=16$

$x>0,\therefore x=4$

When $x=4$ ,

$y=ln(\cfrac{4}{2})^4=ln(2)^4=ln16$

$\therefore p+q=4+16=20$