How Do You Solve a System of Second Order ODEs with Matrix Methods?

In summary: I can solve for C and D. In summary, the conversation discusses solving a system of differential equations by reducing it to a first order system and finding the general solution using eigenvalues and eigenvectors. The final solution includes constants A and B, which can be solved for using the given boundary conditions.
  • #1
c0der
54
0

Homework Statement



Solve:

[ d^2y1/dx^2 ] = [ a -a ] [ y1 ]
[ d^2y2/dx^2 ] [ -a a ] [ y2 ]

A = [ a -a ]
[ -a a ]

Homework Equations



Everything required is in (1) above

The Attempt at a Solution



Reduce to 1st order system

M = [ 0 I ]
[ A 0 ]

Hence, M =
[ 0 0 1 0 ]
[ 0 0 0 1 ]
[ a -a 0 0 ]
[ -a a 0 0 ]

The eigenvalues of M are 0, 0, √2a and -√2a
The eigenvectors are [ 1 1 ], [ 1 1 ], [ 1 -1] and [ 1 -1]

Hence the general solution is (for y only):

y(x) = A*e^(0x)*[ 1 1 ]T + B*e^(0x)*[ 1 1 ]T + C*e^(√2a*x) [ 1 -1 ]T +
D*e^(-√2a*x) [ 1 -1 ]T

So:

y1 = A + B + C*cosh(√(2a)*x) + D*sinh(√(2a)*x)
y2 = A + B - C*cosh(√(2a)*x) - D*sinh(√(2a)*x)

Is this correct?
 
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  • #2
The way to check if that's correct is to plug your answer back into the ODE system.
 
  • #3
c0der said:

Homework Statement



Solve:

[ d^2y1/dx^2 ] = [ a -a ] [ y1 ]
[ d^2y2/dx^2 ] [ -a a ] [ y2 ]

A = [ a -a ]
[ -a a ]

Homework Equations



Everything required is in (1) above

The Attempt at a Solution



Reduce to 1st order system

M = [ 0 I ]
[ A 0 ]

Hence, M =
[ 0 0 1 0 ]
[ 0 0 0 1 ]
[ a -a 0 0 ]
[ -a a 0 0 ]

The eigenvalues of M are 0, 0, √2a and -√2a
The eigenvectors are [ 1 1 ], [ 1 1 ], [ 1 -1] and [ 1 -1]

Hence the general solution is (for y only):

y(x) = A*e^(0x)*[ 1 1 ]T + B*e^(0x)*[ 1 1 ]T + C*e^(√2a*x) [ 1 -1 ]T +
D*e^(-√2a*x) [ 1 -1 ]T

Do you not recall that the solution of [itex]u'' = 0[/itex] is [itex]u(x) = Ax + B[/itex]?


So:

y1 = A + B + C*cosh(√(2a)*x) + D*sinh(√(2a)*x)
y2 = A + B - C*cosh(√(2a)*x) - D*sinh(√(2a)*x)

Is this correct?
 
  • #4
Yes, ok. I see you mean because I have repeated roots, it's Ae^0(x) + Bxe^(0x). Even the above without A and B constants is a solution to the system. However, since I have 4 boundary conditions, I thought I'd keep them in there
 

Related to How Do You Solve a System of Second Order ODEs with Matrix Methods?

1. What is a system of second order ODEs?

A system of second order ODEs (ordinary differential equations) is a set of equations that describe the behavior of multiple dependent variables as a function of one independent variable. These equations involve the second derivative of the dependent variables and can be used to model various physical, chemical, and biological systems.

2. How is a system of second order ODEs solved?

Solving a system of second order ODEs involves finding a set of functions that satisfy all the equations in the system. This can be done analytically using techniques such as separation of variables, substitution, or series solutions. Alternatively, numerical methods such as Euler's method or Runge-Kutta methods can be used to approximate solutions.

3. What are the applications of systems of second order ODEs?

Systems of second order ODEs have a wide range of applications in various fields such as physics, engineering, and biology. They can be used to model complex systems such as electrical circuits, mechanical systems, population dynamics, and chemical reactions.

4. Can systems of second order ODEs have multiple solutions?

Yes, systems of second order ODEs can have multiple solutions. This is because the solutions of these equations depend on initial conditions, and different sets of initial conditions can lead to different solutions. However, for well-posed problems, there is usually a unique solution that satisfies all the equations in the system.

5. What is the difference between a system of second order ODEs and a system of first order ODEs?

The main difference between a system of second order ODEs and a system of first order ODEs is the presence of second derivatives in the former. This means that a system of second order ODEs requires two initial conditions for each dependent variable, whereas a system of first order ODEs only requires one initial condition per variable. Additionally, the methods used to solve these systems may differ, with systems of first order ODEs often being easier to solve analytically.

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