Separable differential equations

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SUMMARY

This discussion focuses on the manipulation of separable differential equations, specifically transitioning from the equation -\frac{1}{5} \ln |1 - 5v^2| = \ln |x| + c to the subsequent form. The key point is the treatment of absolute values, where |1 - 5v^2| is simplified while |x| is retained to ensure that the constant C remains independent of x. The final expression derived is 1 - 5v^2 = \pm \left(\frac{e^{5c}}{|x|^5}\right), illustrating the importance of exponentiation in solving these equations.

PREREQUISITES
  • Understanding of separable differential equations
  • Familiarity with logarithmic properties and manipulation
  • Knowledge of absolute value functions in mathematical expressions
  • Basic skills in algebraic manipulation and exponentiation
NEXT STEPS
  • Study the method of solving separable differential equations in depth
  • Learn about the properties of logarithms and their applications in calculus
  • Research the implications of absolute values in mathematical equations
  • Practice problems involving exponentiation of both sides of equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, algebra, and calculus concepts. This discussion is beneficial for anyone seeking to deepen their understanding of the manipulation of mathematical expressions involving logarithms and absolute values.

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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.
 

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