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April 1st, 2014, 10:38 AM   #1
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Returning to a Differential Equation

Somebody recently posted the following equation $$\frac{\ddot{y}}{\dot{y}}=\frac{1}{2}\left( k \frac{y}{\dot{y}} - \frac{\dot{y}}{y} \right)$$

I set $$u = \frac{\dot{y}}{y}$$ (noting that the original equation is invalid when $y=0$). This leads to $$ \int{\frac{du}{k-3u^2}} = \frac{1}{2}\int{dt} \;\text{where $u^2 \ne \frac{k}{3}$} $$

First I will note that if we look at the excluded case \begin{align*}
u^2 &= \frac{k}{3} \\
\frac{\dot{y}}{y} &= \pm \sqrt{\frac{k}{3}} \\
\log{\left|y\right|} &= \pm t\sqrt{\frac{k}{3}} + \log{c} \\
\left|y\right| &= C_1 e^{t\sqrt{\frac{k}{3}}} + C_2 e^{-t\sqrt{\frac{k}{3}}} \\
for constants $C_1, C_2$.

Back to the equation in $u$, which we decompose by the method of partial fractions to get $$ \int{\frac{\sqrt{3}}{\sqrt{k}+u\sqrt{3}} - \frac{-\sqrt{3}}{\sqrt{k}-u\sqrt{3}}du} = \sqrt{3k}\int{dt}$$ which leads to $$ C_1 \left| \frac{\sqrt{k}+u\sqrt{3}}{\sqrt{k}-u\sqrt{3}} \right| = e^{t\sqrt{3k}} \; \text{where $C_1 > 0$} $$

Rearranging these, we get $$u = \frac{\dot{y}}{y} = \begin{cases}
\sqrt{\frac{k}{3}}\cdot\frac{e^{t\sqrt{3k}} - C_1}{e^{t\sqrt{3k}} + C_1} &\text{where $\left|u\right| \lt \sqrt{\frac{k}{3}},\, C_1 \gt 0$} \\
\sqrt{\frac{k}{3}}\cdot\frac{e^{t\sqrt{3k}} + C_1}{e^{t\sqrt{3k}} - C_1} &\text{where $\left|u\right| \gt \sqrt{\frac{k}{3}},\, C_1 \gt 0$} \\
which we solve by setting $v = e^{t\sqrt{3k}}$ which yields $$\int{\frac{1}{y}dy} = \begin{cases}
\frac{1}{3}\int{\frac{v - C_1}{v\left(v + C_1\right)}dv} \\
\frac{1}{3}\int{\frac{v + C_1}{v\left(v - C_1\right)}dv} \\
\end{cases}$$and again we decompose by partial fractions and integrate to get $$\left|y\right| = \begin{cases}
\left(C_2\left|\frac{\left(v + C_1\right)^2}{v}\right|\right)^{\frac{1}{3}} \\
\left(C_2\left|\frac{\left(v - C_1\right)^2}{v}\right|\right)^{\frac{1}{3}} \\
\end{cases} \;\text{where $C_2 > 0$}$$
Here, the definition of $v$ gives $v \gt 0 \, \forall t$, and allow negative values of $C_1$ so we can write $$y = \left(C_2\frac{\left(e^{t\sqrt{\frac{k}{3}}} + C_1\right)^2}{e^{t\sqrt{\frac{k}{3}}}}\right)^{\fr ac{1}{3}} \;\text{where $C_1 \ne 0, C_2 \ne 0$}$$

Finally, we note that $C_1 = 0$ gives us $y = C_2 e^{t\sqrt{\frac{k}{3}}}$ so we have
$$y = \left(C_2\frac{\left(e^{t\sqrt{\frac{k}{3}}} + C_1\right)^2}{e^{t\sqrt{\frac{k}{3}}}}\right)^{\fr ac{1}{3}} \;\text{where $C_2 \ne 0$}$$

So, my question is: shouldn't I have a similar expression that for a suitable choice of constants would yield $y = C_2 e^{-t\sqrt{\frac{k}{3}}}$?

Where have I lost that family of solutions?
v8archie is offline  

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