Solving $\frac{dx}{dt} = 2p \text{ } \text{ , } \text{ } \frac{dp}{dt} = - e^x$

Question

I am having trouble in solving this problem :

If $x$ and $p$ are functions of $t$ i.e. $x=x(t)$ and $p=p(t)$, then how to solve the following problem :

$$\frac{dx}{dt} = 2p \text{ } \text{ , } \text{ } \frac{dp}{dt} = - e^x$$

My attempt :

$$\frac{d^2 x}{dt^2}= \frac{2dp}{dt} = -2e^x \ldots (*)$$

Now my question does it follow from $(*)$ that , $$(\frac{dx}{dt})^2 = -4e^x \implies \frac{dx}{dt} = 2ie^{\frac{x}{2}}$$ and then I might be able to solve it.

And similarly, for $p(t)$.

Am I correct ? Please correct my mistakes.

Answer 1

Just use chain rule, no need to even think about a second order d.e.

$$\frac{dx}{dp} = \frac{dx}{dt}\cdot \frac{dt}{dp} = \frac{\frac{dx}{dt}}{\frac{dp}{dt}} = -2pe^{-x}$$

Separate variables and integrate,

$$\int e^x dx = \int -2p dp$$

$$e^x = -p^2 + c$$

where $c$ is an arbitrary constant.

If all you need is a relationship between $x$ and $p$, you're done. However, if you want to find $p(t)$ and $x(t)$, proceed as follows.

From the above and the second equation,

$$\frac{dp}{dt} = p^2 - c$$

And you can find $p(t)$ by separation of variables. Using that on the first equation allows you to find $x(t)$.

Answer 2

Hint $$\frac{d^2 x}{dt^2}= \frac{2dp}{dt} = -2e^x \ldots (*)$$ $$x''= -2e^x$$ $$x''x'= -2e^xx'$$ $$\frac {(x'^2)'}2=-2(e^x)'$$ Integrate $$x'^2=-4e^x+K$$ $$x'=\pm \sqrt{K-4e^x}$$ $$\int \frac {dx}{\sqrt{K-4e^x}}=\pm t+K_2$$ $$....$$