Rewrite the 2nd oder non linear D.E as a series of 1st order equations

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Homework Statement



Rewrite the 2nd oder non linear D.E [itex]\frac{d^2x}{dt^2}+x^2+x=0[/itex]
as a series of 1st order equations

Homework Equations



[itex]a\frac{d^2x}{dt^2}+b\frac{dx}{dt}+cx=0[/itex]

[itex]\frac{dx}{dt}=y[/itex]

[itex]\frac{dy}{dt}=-\frac{c}{a}x-\frac{b}{a}y[/itex]


The Attempt at a Solution



a=1, b=0 , c=1

SO

[itex]\frac{dx}{dt}=y[/itex]

[itex]\frac{dy}{dt}=-\frac{1}{1}x-\frac{0}{1}y[/itex] = [itex]-x[/itex]

Is that right?
 
on Phys.org
You have two linear equations there, where is the non-linear term?
 
beetle2 said:

Homework Statement



Rewrite the 2nd oder non linear D.E [itex]\frac{d^2x}{dt^2}+x^2+x=0[/itex]
as a series of 1st order equations

Homework Equations



[itex]a\frac{d^2x}{dt^2}+b\frac{dx}{dt}+cx=0[/itex]
This is a linear equation and so not a good format to use for your problem.

[itex]\frac{dx}{dt}=y[/itex]

[itex]\frac{dy}{dt}=-\frac{c}{a}x-\frac{b}{a}y[/itex]


The Attempt at a Solution



a=1, b=0 , c=1

SO

[itex]\frac{dx}{dt}=y[/itex]

[itex]\frac{dy}{dt}=-\frac{1}{1}x-\frac{0}{1}y[/itex] = [itex]-x[/itex]

Is that right?
Don't just try to match up with memorized formulas (especially when the formula doesn't fit the problem).

You have defined y to be dx/dt so you know that [itex]dy/dt= d^2x/dt^2[/itex]. Replace [itex]d^2x/dt^2[/itex] in [itex]\frac{d^2x}{dt^2}+x^2+x=0[/itex] with [itex]dy/dx[/itex] to get
[tex]\frac{dy}{dt}+ x^2+ x= 0[/tex]
or
[tex]\frac{dy}{dt}= -x^2- x[/tex].
 

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