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I am an A level student and am stuck on this differential equation. I know it is simple for most of the brilliant minds here but I have been trying for an hour with no good result.

$$\frac{dh}{dt}=\frac{5}{h^2}-\frac{1}{20}$$

We need to solve the differential equation to obtain an equation for $t$ in terms of $h$.

Also at time $t=0$, $h=0$ in case we need to find the constant.

Thank you for the help

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  • $\begingroup$ Your initial conditions imply an infinite slope at $t=0$, can you check that they are really correct? What solution method have you tried or thought of to try or suspect that might work based on other problems solved recently? How far did you get? $\endgroup$ – LutzL Apr 20 at 9:55
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$$\frac{dh}{dt}=\frac{5}{h^2}-\frac{1}{20}$$ or $$20h^2\frac{dh}{dt}=100-h^2$$ $$\frac{20h^2}{100-h^2}dh=dt$$ $$\frac{20h^2}{(10-h)(10+h)}dh=dt$$ $$20(-1+\frac{5}{10-h}+\frac{5}{10+h})dh=dt$$

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Indeed, this equation is (more or less) equivalent to $$20\int_0^h\frac{u^2du}{100-u^2}=t$$ and the LHS can be easily computed. Now, it is necessary to do all the calculations to see whether you can obtain an explicit formula for $h$.

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  • $\begingroup$ Note that the aim is not a formula for $h(t)$ but a formula for $t(h)$. $\endgroup$ – LutzL Apr 20 at 9:55
  • $\begingroup$ i am sorry but i dont understand what is meant by u in this equation..i would really appreciate it if you would give me an exact equation with relevant steps..i am trying to understand the method so i can do it again for my papers...Thank You $\endgroup$ – Moeez Demon Apr 20 at 10:01
  • $\begingroup$ E.H.E clarified what I suggested, I think. It now remains for you to integrate the equation he found, which yields $\endgroup$ – Morien Pierre-Luc Apr 20 at 12:25

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