$$ m\frac{dv}{dt}=\frac{dm}{dt}\cdot 282g-gm $$ So this is the initial differential equation for the thrust of a rocket where g is the gravitational constant. This is basically the net force of gravity and thrust on the rocket. My goal is to solve for displacement by solving firstly the diff equation and then integrating the velocity again to get to displacement.
$$ \frac{dv}{dt}=\frac{dm}{dt}\cdot 282\frac{\:3.98332⋅10^{24}}{\left(6371+\right)2}-\frac{\:3.98332⋅10^{24}}{\left(6371+\right)2}m $$ So I defined g in terms of displacement and then I integrate the function twice once to get to velocity and once to get to displacement. However i can not figure out how x and t are related because it is basically referencing itself. $$ \int \left(\int \:\frac{282}{m}\cdot \:\frac{3.98332\cdot \:\:10^{24}}{\left(6371+x\right)^2}dm\right)dt-\int \left(\int \:\frac{3.98332\cdot \:10^{24}}{\left(6371+x\right)^2}dt\right)dt=x $$ I think it is implicit and therefore can not be solved.
I am almost completely sure it has no solution and that you need to introduce another equation and create a system of equations. I have though about using kinetic and potential energy to get a second equation that can help me solve it but I haven't figured out how to actually do it. Can anyone guide me in the right direction?
Thanks
There is actually a simple, intuitive explanation why your equation is underspecified.
Imagine a liquid-fueled rocket with pumps that pump fuel from the tanks into the engine. Moreover, we have some control over the pumps; they can run at their maximum speed, transfering some amount of fuel into the engines each second, or we can reduce the speed of the pumps so that less fuel delivered to the engines each second. Or we can shut the pumps off altogether. Whatever fuel is delivered to the engines will be burned and used to produce thrust.
Essentially the pump control allows us to set $\frac{dm}{dt}$ to some predetermined maximum value, or cut $\frac{dm}{dt}$ in half, or reduce $\frac{dm}{dt}$ to zero. While we are manipulating the pump control, at all times the rocket continues to fly according to the equation $$ m\frac{dv}{dt} = 282g \frac{dm}{dt} - mg. \tag1 $$
But it should be intuitively obvious that the rocket will fly quite differently if we reduce the fuel flow by half just after liftoff, and then cut it altogether after one quarter of the fuel is burned, than it would if we simply let the pumps run at full speed from liftoff until all the fuel is burned.
So the missing information is what is $\frac{dm}{dt}$. A simple answer could be that $\frac{dm}{dt}$ is equal to some constant $r_0$ until all the fuel is burned, after which $\frac{dm}{dt} = 0.$ But that is information that we cannot deduce merely by looking at Equation $(1)$.
So what you need to to first is to find out what $\frac{dm}{dt}$ will be, on the understanding that this is additional information that you have to get from some source other than the equation you have already presented. If $\frac{dm}{dt}$ is a known function of time (as it is, for example, in the case where we burn the fuel at a constant rate until it is all gone), then you can write fuel flow as $\dot m_f(t)$, and you can work out mass as a function of time, $m(t),$ simply by integrating $\dot m_f(t)$:
$$ m(t) = m(0) - \int_0^t \dot m_f(t) \,\mathbf dt. $$
Since you have decided to model the decrease in gravity with height, you can write gravity as a function of height, which in turn is a function of time, $$ g(x(t)) = \frac{g_0}{(R + x(t))^2} $$ where $g_0$ is the gravity at the Earth's surface, $R$ is the radius of the Earth, and $x(t)$ is the height above the surface at time $t.$
You can then write a system of differential equations,
\begin{align} \frac{d}{dt} v(t) &= \left(282 \frac{\dot m_f(t)}{m(t)} - 1\right)g(x(t)), \\ \frac{d}{dt} x(t) &= v(t). \end{align}
This can be solved numerically. (I very much doubt it can be solved exactly in a conventional "closed form".)
See Solving a second order non-linear ODE (rocket height function) $h''(t)=\frac{F}{m+k-r\cdot t}-\frac{G\cdot M}{(R+h)^{2}}$ for some further discussion. (I note that the answer to that question includes an approximation that assumes constant gravity and shows that within a few thousand meters of the Earth's surface this is not a bad approximation.)