Problem involving ordinary differential equation

  • Thread starter Thread starter chwala
  • Start date Start date
Click For Summary

SUMMARY

The discussion addresses the ordinary differential equation 10^y dy = (dx) / (x ln 10) and demonstrates the substitution y₁ = 10^y to simplify it to dy₁ = dx / x. The solution is explicitly given as y = log₁₀(ln(ax)) with domain restrictions |x| > e^(-C). The substitution generalizes to any base b > 0, b ≠ 1, preserving the simplified form dy₁ = dx / x. The conversation also explores asymptotic behavior and domain considerations for negative constants x and C.

PREREQUISITES

  • Separable ordinary differential equations (ODEs)
  • Logarithmic and exponential function properties, including change of base formulas
  • Substitution methods for solving ODEs
  • Asymptotic analysis of functions

NEXT STEPS

  • Study substitution techniques for ODEs involving arbitrary exponential bases b > 0, b ≠ 1
  • Analyze asymptotic behavior of solutions involving nested logarithmic functions
  • Explore domain restrictions and continuity conditions for solutions with negative parameters
  • Investigate generalizations to differential equations with logarithmic and exponential terms in other bases

USEFUL FOR

Mathematicians, applied mathematicians, and students working on solving nonlinear ordinary differential equations involving exponential and logarithmic functions, especially those interested in substitution methods and asymptotic solution behavior.

chwala
Gold Member
Messages
2,843
Reaction score
428
Homework Statement
Consider the ode:
which admits the solution which admits the solution

I am interested in 1. Any-non obvious substitution that simplifies the equation further. 2. Possible generalisations to other logarithmic / exponential bases. 3 Asymptotic behavior of solutions.
Relevant Equations
variable separable.
##10^y dy = \dfrac{dx}{x\ln 10}##
##\int 10^y dy = \int \dfrac{dx}{x\ln 10}##

...

##10^y = \ln x +C##
##10^y = \ln (ax)##
##y=\lg(\ln(ax))##
 
Last edited:
Physics news on Phys.org
$$\ln 10 e^{y\ln 10 }dy = \frac{dx}{x}$$
$$e^{y\ln 10 } = \ln |x |+ C$$
$$y= \frac{\ln(\ln |x| + C)}{\ln 10}$$defined for
$$|x| >e^{-C}$$

x,C < 0 cases are added to your result.
 
Last edited:
  • Like
Likes   Reactions: chwala
chwala said:
Homework Statement: Consider the ode:
which admits the solution which admits the solution

I am interested in 1. Any-non obvious substitution that simplifies the equation further. 2. Possible generalisations to other logarithmic / exponential bases. 3 Asymptotic behavior of solutions.
Relevant Equations: variable separable.
The substitution ## y_1=10^y ## simplifies the equation $$ 10^ydy=\frac{dx}{x\ln10} $$ to the equation ## dy_1=dx/x ##.
The general case can be obtained by replacing ## 10 ## in the substitution ## y_1=10^y ## with ## b ##, where ## b\gt0 ## and ## b\neq1 ##, while the equation ## dy_1=dx/x ## remains unchanged.
 
  • Like
  • Informative
Likes   Reactions: chwala and SammyS

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K