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 April 1st, 2014, 09:38 AM #1 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra 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}}} \\ \end{align*} 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 $$u = \frac{\dot{y}}{y} = \begin{cases} \sqrt{\frac{k}{3}}\cdot\frac{e^{t\sqrt{3k}} - C_1}{e^{t\sqrt{3k}} + C_1} &\text{where \left|u\right| \lt \sqrt{\frac{k}{3}},\, C_1 \gt 0} \\ \sqrt{\frac{k}{3}}\cdot\frac{e^{t\sqrt{3k}} + C_1}{e^{t\sqrt{3k}} - C_1} &\text{where \left|u\right| \gt \sqrt{\frac{k}{3}},\, C_1 \gt 0} \\ \end{cases}$$ 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?

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