Nonlinear DE Reduction for \ddot{y} = - \dot{y} - y -sin(y)

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Homework Help Overview

The discussion revolves around the nonlinear differential equation \(\ddot{y} = - \dot{y} - y - \sin(y)\). Participants are exploring methods to reduce the order of this equation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find a solution to reduce the order of the differential equation, considering \(y_1 = 0\) as a potential solution but questioning its applicability. Other participants suggest substituting \(\frac{dy}{dx} = t\) as a method for reduction. There is also a discussion about the correct formulation of derivatives in this context.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for substitution and confirming each other's approaches. There is a collaborative atmosphere as they explore different methods to tackle the equation.

Contextual Notes

There is an indication of confusion regarding the correct application of derivatives, and assumptions about the form of the solution are being questioned. The participants are navigating the complexities of the equation without reaching a definitive conclusion.

sunrah
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Homework Statement



[itex]\ddot{y} = - \dot{y} - y -sin(y)[/itex]

Homework Equations


The Attempt at a Solution


to reduce the order I need to find a solution y1. it seems to me the only obvious solution is y1 = 0 but i can't use this to do a reduction can i
 
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Hi sunrah! :smile:

You should be able to reduce the given DE to first order by substituting [itex]\frac{dy}{dx}= t[/itex]
 
Infinitum said:
Hi sunrah! :smile:

You should be able to reduce the given DE to first order by substituting [itex]\frac{dy}{dx}= t[/itex]

indeed, thank you!
 
sunrah said:
indeed, thank you!


I hope you didn't write [tex]\frac{d^2y}{dt^2} = t'[/tex] :wink:
 
i used
[itex]\frac{d^{2}y}{dx^{2}} = \frac{dt}{dx}\frac{dy}{dy} = t\frac{dt}{dy}[/itex]
 
sunrah said:
i used
[itex]\frac{d^{2}y}{dx^{2}} = \frac{dt}{dx}\frac{dy}{dy} = t\frac{dt}{dy}[/itex]

Yep, perfect! :approve:
 

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