Determining Order of Differential Equations

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SUMMARY

The discussion focuses on determining the order of differential equations (DEs), specifically analyzing the equation $\d{^2y}{x^2}+2\d{y}{x} \d{^3y}{x^3}+x=0$. The highest derivative present in this DE is third order, confirming that the order is three. Additionally, a substitution method is introduced where $y^{'} = u$ transforms the original DE into a second-order equation, $\displaystyle u^{\ '} + 2\ u\ u^{\ ''} + x = 0$. This highlights the potential to reduce the order of a DE through strategic substitutions.

PREREQUISITES
  • Understanding of differential equations and their orders
  • Familiarity with derivative notation and operations
  • Knowledge of substitution methods in solving differential equations
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the method of substitution in differential equations
  • Learn about higher-order differential equations and their solutions
  • Explore techniques for reducing the order of differential equations
  • Investigate specific examples of third-order differential equations
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations, as well as educators looking for effective teaching methods for explaining the order of DEs.

ineedhelpnow
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Hello. I can't seem to remember how to do these kind of problems. I need to determine the order of the differential equations. Can someone show how this is done so that I can understand how to do the rest?

$\d{^2y}{x^2}+2\d{y}{x} \d{^3y}{x^3}+x=0$
 
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The order of a DE is the order of the highest derivative present in the DE. What order is that?
 
Three. Got it. Thanks.
 
Last edited:
ineedhelpnow said:
Hello. I can't seem to remember how to do these kind of problems. I need to determine the order of the differential equations. Can someone show how this is done so that I can understand how to do the rest?

$\d{^2y}{x^2}+2\d{y}{x} \d{^3y}{x^3}+x=0$

The DE is of third order, of course... but with the substitution $\displaystyle y^{'} = u$ it becomes...

$\displaystyle u^{\ '} + 2\ u\ u^{\ ''} + x = 0\ (1)$

... which is of order two... solving (1) however is a different and not necessarly trivial task...

Kind regards

$\chi$ $\sigma$
 
ineedhelpnow said:
Three. Got it. Thanks.

You got it! Although:

chisigma said:
The DE is of third order, of course... but with the substitution $\displaystyle y^{'} = u$ it becomes...

$\displaystyle u^{\ '} + 2\ u\ u^{\ ''} + x = 0\ (1)$

... which is of order two... solving (1) however is a different and not necessarly trivial task...

Kind regards

$\chi$ $\sigma$

You see what chisigma is getting at? Essentially, you can reduce the order of the DE with a substitution. This is generally possible when the function itself, $y$, is not present in the DE.
 

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