Solve 1st Order ODE from Transcendental Equation

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SUMMARY

The discussion centers on solving a transcendental equation of the form y(x)² = 1/(x² Log²(A(x) y(x)²)) by transforming it into a first-order ordinary differential equation (ODE). The participant questions whether it is valid to express y(x) as y(x) = 1/(x Log(A(x) y(x)²)) and differentiate both sides to derive a first-order ODE. The conversation highlights the complexity of modifying differential equations and the implications of such transformations on the dependency of the resulting equations on A(x).

PREREQUISITES
  • Understanding of transcendental equations
  • Knowledge of first-order ordinary differential equations (ODEs)
  • Familiarity with logarithmic properties and transformations
  • Basic calculus skills, particularly differentiation
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  • Research methods for solving transcendental equations
  • Learn about first-order ODE techniques and their applications
  • Study the properties of logarithms in mathematical transformations
  • Explore the implications of variable dependencies in differential equations
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Mathematicians, engineering students, and anyone interested in solving complex equations involving transcendental functions and differential equations.

gonadas91
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It is a general doubt about the following equation: Imagine I want to calculate an unknown function y(x), and my starting equation is of the type

y(x)^{2}=\frac{1}{x^{2}Log^{2}(A(x)y(x)^{2})}

, then, am I allowed to start with the equation

y(x)=\frac{1}{xLog(A(x)y(x)^{2})}

and differenciate in both sides of the equation, to obtain a first order diferential equation to get y(x)?

(Note that the beginning equation is a trascendental equation, but why not trying to solve the first order ODE?)Thank you
 
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I don't see why this wouldn't be allowed but I'd assume setting ##f(x)=y(x)^2## and solving for ##f(x)## would be simpler. You can ofc use ##log(a)^2=log(a)*log(a)=log(2a)## to remove the square on the logarithmic function.
 
Mmm the last property of the logs is not valid. My question actually has to be with the modifications that are allowed on a differential equation. Also, the resulting differential equation using that is not depending on A(g) which annoys me somehow...
 
oh yeah ofc it isn't haha, too late for this here
 

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