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I am doing a physics project and I came up with the following equation: $$(A+B)\left(\frac{dh}{dt}\right)^2+C\frac{dh}{dt}t+Dt^2+Eh=0$$ Where A, B, C, D and E are constants.

From what I know, this is a first order non-linear ODE. I can't seem to find a way to solve this in part due to the fact that $\frac{dh}{dt}$ is squared. I am teaching myself differential equations at the moment so could you please explain any steps taken.

Thanks in advanced!

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  • $\begingroup$ Why have two terms with $(dh/dt)^2$? $\endgroup$ – Robert Israel Feb 6 '17 at 2:27
  • $\begingroup$ I could combine the coefficients, I'll edit the question. $\endgroup$ – Jonathan Densil Feb 6 '17 at 2:45
  • $\begingroup$ If I let $p = \frac{dh}{dt}$ then if I separate and integrate both sides I would get $h = pt - k$ where $k$ is the constant of integration. If I substitute all that into the equation I would get $(A+B)p^2+Cpt+Dt^2+Ept-Ek=0$. I don't know if that helps at all. $\endgroup$ – Jonathan Densil Feb 6 '17 at 2:57
  • $\begingroup$ It can actually simplify to $(A+B)p^2+(C+E)pt+Dt^2-Ek=0$ $\endgroup$ – Jonathan Densil Feb 6 '17 at 3:17
  • $\begingroup$ What do you mean? $h - t (dh/dt)$ is not constant. $\endgroup$ – Robert Israel Feb 6 '17 at 4:02
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Rewrite the d.e. as $$ \left(\frac{dh}{dt}\right)^2 + b \frac{dh}{dt} t + c t^2 + d h = 0$$ This does have a "closed-form" implicit solution (up to some integrations), but it's rather complicated. You might start by "completing the square", writing the differential equation as

$$ \left(\dfrac{dh}{dt} + \frac{b}{2} t\right)^2 + \left(c - \frac{b^2}{4}\right) t^2 + d h = 0 $$

and thus

$$\dfrac{dh}{dt} = -\frac{b}{2} t \pm \sqrt{\left(\frac{b^2}{4}-c \right)t^2 - d h}$$

Choosing $+$ or $-$, write this as

$$ \dfrac{dh}{dt} = -\tilde{b} t + \tilde{c} \sqrt{t^2 - \tilde{d} h}$$

This is "homogeneous of class G", i.e. we can write it as

$$ \dfrac{dh}{dt} = \frac{h}{t} F\left(\frac{h}{t^2}\right)$$ where in this case $$ F(v) = {\frac {\sqrt {1-\tilde{d}v}\;\tilde{c}-\tilde{b}}{v}}$$ and this has solutions of the form $h(t) = t^2 v$ where $$ \int \frac{dv}{v(F(v)-2)} = \ln(t) + C $$

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