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 April 1st, 2014, 10:38 AM #1 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 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? Tags differential, equation, returning Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jeremymcbreairty Computer Science 22 May 31st, 2013 01:30 PM trueblue Academic Guidance 3 March 21st, 2013 03:43 PM peewster Algebra 5 October 4th, 2012 10:51 AM noah++ Academic Guidance 1 May 5th, 2010 02:51 PM cc2008 New Users 3 March 21st, 2008 01:02 PM

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