MHB Determining Order of Differential Equations

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.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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