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I have a first order differential equation question with initial value.

$$\frac{dv}{dt} + \frac{v^2}{5} = 9.8 \text{ with } v(0)=0$$

I tried to solve it but just didn't know how to remove the absolute sign in my solution.

Thank you.

Please let me know if whole question is needed to remove the absolute sign inside the solution.

Edit 1: Part where I got stuck:

After integration and since C = 0:

$$\frac{5ln(lv+7l)}{14} - \frac{5ln(lv-7l)}{14} = t $$

My absolute signs look weird.

Edit 2: I think my solution is wrong..

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    $\begingroup$ Can you add the point in your solution where you got stuck. ... This is the classic resistance is proportional to the square of the velocity. The projectile will tend to a terminal velocity. $\endgroup$ – Donald Splutterwit Apr 28 '17 at 11:22
  • $\begingroup$ I added the point where I got stuck. $\endgroup$ – alex then Apr 28 '17 at 11:31
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I'll write $9.8$ as $g$ for convenience. $$\frac{dv}{dt} = g - \frac{1}{5}v^2 = \frac{1}{5} (5g - v^2) $$ $$ \frac{dv}{5g-v^2} = \frac{dt}{5} $$ $$ \int \frac{dv}{5g-v^2} = \int \frac{dt}{5} = \frac{t}{5} + c$$ We know that $ \int \frac{1}{a^2-x^2} dx = \frac{1}{a} \tanh^{-1} (\frac{x}{a} ) $ so $$ \frac{1}{\sqrt{5g}} \cdot \tanh^{-1} \frac{v}{\sqrt{5g}} = \frac{t}{5} + c $$ $$ \tanh^{-1} \frac{v}{\sqrt{5g}} = \sqrt{\frac{g}{5}} t + c $$ $$ v = \sqrt{5g} \cdot \tanh(\sqrt{\frac{g}{5}}t+c) $$

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  • $\begingroup$ Looks like my solution is wrong.... $\endgroup$ – alex then Apr 28 '17 at 11:32
  • $\begingroup$ Could you change "I'll right" to "I'll write" for esthetic reasons? $\endgroup$ – LutzL Apr 28 '17 at 12:08
  • $\begingroup$ Sure, I didn't even notice the typo. Thanks! $\endgroup$ – Ofek Gillon Apr 28 '17 at 12:13

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