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!

  • $\begingroup$ 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

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 $$ \int \frac{dv}{v(F(v)-2)} = \ln(t) + C $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.