# Riccati Equation for falling particle.

I'am trying to solve the differential equation for a falling particle of mass 1 with air resistance proportional to $v^2$ (v is velocity): $$v'=g-v^2$$ This is a Riccati-Equation with stationary solution $\hat{v}=\sqrt{g}$, so we can transform it to the Bernoulli Equation $$u'=-u^2-2\sqrt{g}u$$ Substituting $z=u^{-1}$ gives \begin{align*}z'=2\sqrt{g}z+1\\ \Rightarrow z(t)=Ce^{2\sqrt{g}t}-\frac1{2\sqrt{g}}\\ \Rightarrow u(t)=\frac{1}{Ce^{2\sqrt{g}t}-\frac1{2\sqrt{g}}}\\ \Rightarrow v(t)=\frac{1}{Ce^{2\sqrt{g}t}-\frac1{2\sqrt{g}}}+\sqrt{g}\end{align*}

The problem is that this solution does not make physical sense to me, since its decreasing and does not match my reference solution. Can you explain to me where I have gone wrong?

• You might want to tell us your reference solution so we can see what you were expecting... – Chappers Apr 16 '15 at 14:30
• @Chappers The one Dr. Sonnhard Graubner postet below. This is intuitive because it aproaches the stationary solution. – Achilles Apr 16 '15 at 14:33
• Well, so does your answer. It is increasing if $C>0$. You can get from one solution to the other by choosing the right value of $C$ (or perhaps easier is to start from the tanh and fiddle about with $c_1$). – Chappers Apr 16 '15 at 14:40
• For $C=1$(and g=9) Wolfram Alpha plots this – Achilles Apr 16 '15 at 14:48
• Glad you fixed the spelling of the name Riccati; in Italian, ricatti is the plural of ricatto, which means blackmail. ;-) – egreg Apr 16 '15 at 15:35

$$\alpha \tanh \left(c_1 \alpha+\alpha t\right)=\alpha\frac{Be^{2t\alpha}-1}{Be^{2t\alpha}+1},\text{ with } B=e^{2c_1\alpha}$$
$$\frac{1}{Ce^{2t\alpha}-\frac{1}{2\alpha}}+\alpha=\alpha\left(\frac{2}{2C\alpha e^{2t\alpha}-1}+1\right)=\alpha\frac{1+2\alpha Ce^{2t\alpha}}{2C\alpha e^{2t\alpha}-1}=\alpha\frac{-1-2\alpha Ce^{2t\alpha}}{-2C\alpha e^{2t\alpha}+1}$$
Now put $C:=-\frac{B}{2\alpha}$ and...