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So I have this differential equation

$-0.4 \cdot 9.81+\frac{1}{100}v^2=0.4 v'$

I was able to solve it which gives me

$\ln(\frac{v+20}{v-20})=t+c$

My problem is I can't isolate $v$ after that i get it in this form and also when I try to find the constant $c$ knowing that $v(0) = 0$ I get

$c=\ln(\frac{20}{-20})$

Would I be able to say that

$c=\ln(\frac{20}{-20}) = 0$

If anyone could point me in the right direction to be able to get a function like $v(t) = ...$

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  • $\begingroup$ use the exponential function $\endgroup$ Commented Jun 11, 2017 at 16:21
  • $\begingroup$ @Dr.SonnhardGraubner The OP solved it wrong! $\endgroup$ Commented Jun 11, 2017 at 16:26

2 Answers 2

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Well, in general:

$$\text{a}+\frac{\text{v}\left(t\right)^2}{\text{b}}=\text{c}\cdot\text{v}'\left(t\right)\space\Longleftrightarrow\space\int\frac{\text{c}\cdot\text{v}'\left(t\right)}{\text{a}+\frac{\text{v}\left(t\right)^2}{\text{b}}}\space\text{d}t=\int1\space\text{d}t\tag1$$

So, for the integrals:

  • Substitute $\text{u}:=\text{v}\left(t\right)$: $$\int\frac{\text{c}\cdot\text{v}'\left(t\right)}{\text{a}+\frac{\text{v}\left(t\right)^2}{\text{b}}}\space\text{d}t=\frac{\text{c}}{\text{a}}\int\frac{1}{1+\frac{\text{u}^2}{\text{a}\cdot\text{b}}}\space\text{d}\text{u}\tag2$$
  • Substitute $\text{p}:=\frac{\text{u}}{\sqrt{\text{a}}\cdot\sqrt{\text{b}}}$: $$\frac{\text{c}}{\text{a}}\int\frac{1}{1+\frac{\text{u}^2}{\text{a}\cdot\text{b}}}\space\text{d}\text{u}=\frac{\text{c}\cdot\sqrt{\text{b}}}{\sqrt{\text{a}}}\int\frac{1}{1+\text{p}^2}\space\text{d}\text{p}=\frac{\text{c}\cdot\sqrt{\text{b}}}{\sqrt{\text{a}}}\cdot\arctan\left(\text{p}\right)+\text{K}_1\tag3$$
  • $$\int1\space\text{d}t=t+\text{K}_2\tag4$$

So, we get:

$$\frac{\text{c}\cdot\sqrt{\text{b}}}{\sqrt{\text{a}}}\cdot\arctan\left(\frac{\text{v}\left(t\right)}{\sqrt{\text{a}}\cdot\sqrt{\text{b}}}\right)=t+\text{K}\tag5$$

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  • 1
    $\begingroup$ Oh i was sure i had the correct solution because when I use the TI Nspire that's what it gives me. I will try it this way thank you :) $\endgroup$
    – I.B
    Commented Jun 11, 2017 at 16:31
  • $\begingroup$ @CNuts You're welcome, I'm glad that I could help you! $\endgroup$ Commented Jun 11, 2017 at 16:31
  • $\begingroup$ I don't understand how the c/a became c*sqr(b)/sqr(a) on step (3). $\endgroup$
    – I.B
    Commented Jun 11, 2017 at 16:48
  • $\begingroup$ @CNuts Look at the substitution I made and try to find: $$\frac{\text{d}\text{p}}{\text{d}\text{u}}$$ $\endgroup$ Commented Jun 11, 2017 at 16:51
  • $\begingroup$ Oh forgot about that it makes sense now, thanks :) $\endgroup$
    – I.B
    Commented Jun 11, 2017 at 16:55
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The full equation for free fall under air friction reads $$ m\dot v=-c|v|v-mg $$ which has only one stationary point $v_\infty=-\sqrt{\frac{mg}c}$. Then the equation can be reformulated in new constants as $$ \dot v = -b(|v|v+v_\infty^2) $$ For $v\le 0$ partial fraction decomposition and integration leads to an expression similar to the one you got that should properly read $$ \ln\left|\frac{v(t)+v_\infty}{v(t)-v_\infty}\right|=-2bv_\infty\, t+C=2b|v_\infty|\,t+C $$ and for $v(0)=0$ you get $C=0$. Exponentiation and algebraic manipulation then leads to \begin{align} \frac{v_\infty+v(t)}{v_\infty-v(t)}&=e^{2b|v_\infty|\,t} \\~\\ v(t)(e^{2b|v_\infty|\,t}+1)&=v_\infty(e^{2b|v_\infty|\,t}-1)\\~\\ v(t)&=\frac{e^{2b|v_\infty|\,t}-1}{e^{2b|v_\infty|\,t}+1}v_\infty =\tanh(b|v_\infty|\,t)\,v_\infty \end{align}

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  • $\begingroup$ Thanks for your answer, but how did you go from the formula with the two v(t) too the one with only one? $\endgroup$
    – I.B
    Commented Jun 11, 2017 at 21:46
  • $\begingroup$ I figured it out thank you so much I think this is the actual solution since the $a$ is negative. $\endgroup$
    – I.B
    Commented Jun 11, 2017 at 22:13

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