How Do You Solve a System of Linear ODEs with Equal Second Derivatives?

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SUMMARY

The discussion focuses on solving a system of linear ordinary differential equations (ODEs) defined by the equations x''(t) = x + y and y''(t) = x + y. The solution approach involves setting x''(t) equal to y''(t) and deriving relationships between x and y, leading to a vector equation representation. The final solution will include four arbitrary constants, which arise from the nature of the second-order ODEs. The use of eigenvalues and eigenvectors is also suggested as a method to simplify the solution process.

PREREQUISITES
  • Understanding of second-order ordinary differential equations (ODEs)
  • Familiarity with eigenvalues and eigenvectors
  • Knowledge of vector equations in differential equations
  • Basic calculus and differential calculus concepts
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  • Study the method of solving second-order linear ODEs with constant coefficients
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Students and professionals in mathematics, engineering, and physics who are working with systems of linear ordinary differential equations and require a deeper understanding of solution methods and techniques.

s_j_sawyer
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Homework Statement



Solve this system of linear ODEs:

1) x''(t) = x + y
2) y''(t) = x + y


Just fyi, this is part of a much larger problem but I need to solve this system!

Homework Equations



See above.

The Attempt at a Solution



Okay so I think the most logical way to solve this would be to set x''(t) = y''(t).

Then
x''(t) = y''(t)
x' = y' + c1
x = y + c1t + c2

which implies

3) x'' = 2x - c1t - c2
4) y'' = 2y + c1t + c2

But I am not sure what to do from here. Apparently there should be 4 arbitrary constants in the final answer. But wouldn't solving x'' for x give you 2 and then solving y'' for y give you two more?

Thanks for any help.
 
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Did you find the eigenvalues and eigenvectors? What is the general solution to this linear system of ODEs? Yes you will end up with 4 constants but that's the minor step
 
You could also write it as a vector equation
<br /> \vec x&#039;&#039;(t) = \left(\begin{array}{cc} 1&amp;1\\1&amp;1\end{array}\right)\vec x(t)<br />
and go on from there.
 

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