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Decoupling ODEs

  • Thread starter Tony11235
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  • #1
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In order to solve this pde that I'm on, I must solve this system of odes, [tex] \frac{dx}{dt} = -y [/tex] and [tex] \frac{dy}{dt} = x [/tex] , which doesn't look bad, but I haven't had a second semester of ode yet where systems of differential equations are covered. How is this solved?
 

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  • #2
lurflurf
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Tony11235 said:
In order to solve this pde that I'm on, I must solve this system of odes, [tex] \frac{dx}{dt} = -y [/tex] and [tex] \frac{dy}{dt} = x [/tex] , which doesn't look bad, but I haven't had a second semester of ode yet where systems of differential equations are covered. How is this solved?
-The solutions are well known functions.
-There exist EODE class that does not cover linear systems in one semester?
If decoupling the odes is what you whant to do just solve for each variable (trivial) and substitute one into the other.
solve
[tex]x=\frac{dy}{dt}[/tex]
[tex]y=-\frac{dx}{dt}[/tex]
substitute
[tex]x=\frac{dy}{dt}=\frac{d}{dt}\left(-\frac{dx}{dt}\right)=-\frac{d^2x}{dt^2}[/tex]
[tex]y=-\frac{dx}{dt}=-\frac{d}{dt}\left(\frac{dy}{dt}\right)=-\frac{d^2y}{dt^2}[/tex]
Decoupled
Now as for finding soulutions I heard somewhere that some special functions having something to do with cirlces and triangles has something to do with it.
Also be aware you have introduced extraneous solutions so eliminate them be substituting into the original equation.
 
  • #3
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lurflurf said:
-The solutions are well known functions.
-There exist EODE class that does not cover linear systems in one semester?
Ok I think we did. I'm just slow I guess. Maybe I should just not be so quick to ask questions and actually THINK. Sorry I should have recognized the two odes.:grumpy: Simple substitution! I don't know why I was thinking laplace transformations.
 
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