Absolute Values in Separable Differential Equations

Click For Summary
SUMMARY

The discussion focuses on the treatment of absolute values in the context of solving separable differential equations. The equation ln|v-49| = -t/5 + C is transformed into |v-49| = e^{-t/5+C}, leading to v = 49 + ce^{-t/5}, where the constant c absorbs the ± from the absolute value. This transformation is due to the exponential function being always positive, allowing for the simplification of the absolute value into a single constant term. The participants confirm that the new constant c effectively incorporates the sign ambiguity.

PREREQUISITES
  • Understanding of separable differential equations
  • Familiarity with logarithmic and exponential functions
  • Knowledge of constants in mathematical equations
  • Basic calculus concepts related to differential equations
NEXT STEPS
  • Study the properties of logarithmic functions in differential equations
  • Explore the concept of integrating factors in differential equations
  • Learn about the implications of absolute values in mathematical transformations
  • Investigate advanced techniques for solving separable differential equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone seeking to deepen their understanding of the manipulation of absolute values in mathematical contexts.

patrickbotros
Messages
34
Reaction score
1
When solving a separable differential equation, my textbook says this:
ln|v-49|=-t/5+C→
|v-49|=e-t/5+C→
v=49+ce-t/5
What happened to the absolute values? I think it has something to do with the exponential always being positive.
 
Physics news on Phys.org
patrickbotros said:
When solving a separable differential equation, my textbook says this:
ln|v-49|=-t/5+C→
|v-49|=e-t/5+C→
v=49+ce-t/5
What happened to the absolute values? I think it has something to do with the exponential always being positive.
The equation ##\displaystyle\ |v-49|=e^{-t/5+C}\ ## is equivalent to
##\displaystyle\ v-49=\pm e^{-t/5+C}\ ## → ##\displaystyle\ v-49=c\, e^{-t/5}\ ##, where ##\ c = \pm\ln(C) ##​
So the new constant, c, (lower case) absorbs the ± .
 
SammyS said:
The equation ##\displaystyle\ |v-49|=e^{-t/5+C}\ ## is equivalent to
##\displaystyle\ v-49=\pm e^{-t/5+C}\ ## → ##\displaystyle\ v-49=c\, e^{-t/5}\ ##, where ##\ c = \pm\ln(C) ##​
So the new constant, c, (lower case) absorbs the ± .
Ohhhh. HAHA!:oldlaugh: That was dumb :)
 

Similar threads

Undergrad Determining equilibrium solutions to differential equations
  • · Replies 16 ·
Replies
16
Views
4K
Undergrad Why Lagrange’s method of solving Pp + Qq=R works?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Separation of Variables for Partial Differential Equations
  • · Replies 18 ·
Replies
18
Views
5K
Undergrad Solving Differential Equation Using Reduction of Order
  • · Replies 2 ·
Replies
2
Views
4K
Undergrad Solve Int. Eq.: Exponential Growth Diff. Eq.
  • · Replies 7 ·
Replies
7
Views
4K
Undergrad On vibrating string and differentiating infinite sum
  • · Replies 7 ·
Replies
7
Views
3K
MHB Solve Differential Eq: xe^-1/(k+e^-1) for x, k, t
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Can Non-Separable ODEs Be Solved with Coordinate Transformations?
  • · Replies 13 ·
Replies
13
Views
3K
Undergrad Separation of Variables, but not equal to constant
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad A question about variables of integration
  • · Replies 3 ·
Replies
3
Views
7K