Separation of variables , but for 2nd order

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SUMMARY

The discussion focuses on the application of separation of variables in solving second-order ordinary differential equations (ODEs). The specific equation analyzed is u''*sin(x) - 2u'*cos(x) = 0, which is transformed into the form u''/u' = -2cos(x)/sin(x). Participants explore various substitution methods, including letting w = u', to simplify the integration process. Ultimately, the discussion clarifies that the transformation leads to ln(w) = integration of the right-hand side, making the solution more accessible than initially perceived.

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  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with separation of variables technique
  • Knowledge of integration methods
  • Experience with substitution techniques in differential equations
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  • Study the method of reduction of order in ODEs
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Mathematics students, educators, and professionals dealing with differential equations, particularly those focusing on second-order ODEs and separation of variables techniques.

ericm1234
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"separation of variables", but for 2nd order

Ok, I know how to separate variables in solving an ODE. I am unable to understand a solution I have for a problem which was the result of reduction of order- we end up with u''*sinx-2u'*cosx=0
so turn this into u''/u'=-2cosx/sinx
At this point I don't understand the process. I've tried looking at it as making the left side a "u substitution", or rewriting it d(du/dt)/dt / du/dt and integrating both sides,
also, tried making a substition w=u', but this seems hard to get u back in the end.
Can someone explain thoroughly what the process looks like in solving this?
 
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hi ericm1234! :wink:
ericm1234 said:
u''/u'=-2cosx/sinx
…
also, tried making a substition w=u', but this seems hard to get u back in the end.

well, then that's w'/w = -2cosx/sinx,

so ln(w) = … ? :smile:​
 


Thanks, got it now, not as complicated as I was making it.
 

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