Separable differential equations

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The discussion focuses on the manipulation of separable differential equations, specifically the transition from one equation to another involving absolute values. The participant is confused about why the absolute value is removed from one term, (1-5v^2), while it is retained for x. The manipulation involves logarithmic properties and exponentiation, leading to the conclusion that C, a constant, should not depend on x. The reasoning for maintaining the absolute value for x is to ensure C remains independent of x. The thread highlights the importance of understanding logarithmic transformations in solving differential equations.
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Homework Statement



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Homework Equations


The Attempt at a Solution



I've highlighted two equations on the screenshot. How did it proceed from the first to the second? I'm actually confused with the absolute values. What is the idea behind getting rid of the first absolute value(1-5v^2) while keeping the second one(x)?
 
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So for starters, the manipulation is something like:
##-\frac 15 \ln | 1 - 5v^2 | = \ln |x| + c ##
##\begin{align*}
\ln | 1 - 5v^2 | &= -5(\ln |x| +c)\\
&= -5( \ln |x| - \ln e^{c} )\\
&=-5 ( \ln \frac{|x|}{e^{c} })\\
&= \ln \left(\frac{|x|}{e^{c} }\right)^{-5} \\
&= \ln \left(\frac{e^{5c} }{|x|^5}\right) \end{align*}##
Then, removing the absolute value from the left gives: ##1-5v^2 = \pm \left(\frac{e^{5c} }{|x|^5} \right) ##
So ##C = \pm e^{5c} ##
There only reason not to take out the absolute value from x that I can see is so that C is not dependent on x.
 
You get rid of the ln by exponentiating both sides.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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