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How can I prove that

$$\ v_1(t) = \sqrt{\frac{mg}{k}} \tanh\left(\sqrt{\frac{kg}{m}}*t\right) $$

is a solution of the differential equation: $$\ \frac{d}{dt}v(t)= -g + \frac{k}{m} v(t)^2 $$

My idea was to integrate over $\frac{d}{dt}v(t)$ so I get $\ v_2(t) $. If $ v_1(t) $ is a particular integral of $v_2(t)$ then $v_1(t)$ a solution...

But somehow that approach led me to nowhere...

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up vote 0 down vote accepted

Note that the equation may be written as

$$\int \frac{dv}{\frac{k}{m} v^2 -g} = \int dt $$

Make the substitution $v = \sqrt{(m g)/k} \cosh{u}$ and integrate both sides to get

$$ \sqrt{\frac{m}{g k}} \int du \; \mathrm{csch}{u} = \int dt $$

Use the fact that

$$ \int du \; \mathrm{csch}{u} = \tanh{\frac{u}{2}} $$

and properties of the hyperbolic functions to get

$$ \log{\left (\sqrt{\frac{1-\sqrt{\frac{k}{m g}} v}{1+\sqrt{\frac{k}{m g}} v}}\right )} = \sqrt{\frac{g k}{m}} t + C $$

where $C$ is a constant of integration. Assume that $v(0) = 0$ so that $C=0$. Then we get the solution by solving this equation for $v$:

$$ v(t) = \sqrt{\frac{m g}{k}} \tanh{\left (\sqrt{\frac{g k}{m}} t \right )}$$

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It isn't quite right. The $t$ shouldn't be inside the square root, and there should be a negative sign in front.

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Try to substitute $v_1$ in the equation and obtain: $\ \frac{d}{dt}\Bigl(\sqrt{\frac{mg}{k}}\tanh\Bigl(\sqrt{\frac{kg}{m}}t\Bigr)\Bigr)= -g + \frac{k}{m}\Bigl(\sqrt{\frac{mg}{k}}\tanh\Bigl(\sqrt{\frac{kg}{m}}t\Bigr)\Bigr)^2$

Your function $v_1$ is a solution if and only if this equality is an identity.

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If it was the correct solution, this would indeed be the way to do it. – Robert Israel Jan 20 '13 at 19:38

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