System of Differential equations with a singular coefficient matrix, help?

[aeroegnr]

I tried using undetermined coefficients to solve this problem, but I know that I am missing something and i cannot find any reference material on this. If you help me, thank you.

The homogeneous equation for the system is:

y' = A*y

where y = \left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right]

and A = \left[ \begin{array}{cc} -2 & 1 \\ -1 & 0 \end{array} \right]

I end up with only one eigenvector of course, and I'm trying to use a solution that ends up as C1*V*e^t + c2*V*t*e^t where V is the only eigenvector of A, but that is not a complete solution.

What am I missing?

 

[Wong]

I suppose there is a general way to solve this type of linear constant coefficient differential equations.

I suppose you mean A is not diagonalisable? But A can always be reduced to a matrix in Jordan form. Do you know what is "matrix exponential"? The "fundamental solution" of the system is given by exp(tA). (The fundamental solution is a matrix whose columns are solutions of the system and linearly independent.) When A is reduced to a matrix in Jordan form, the "matrix exponential" is readily computable. So I think that is the solution you want?

And I think there is a way to solve this without using Jordan form? your equation reads
y1'=-2*y1+y2
y2'=-y1
That is
y1''=-2*y1'+y2'=-2*y1'-y1
which is readily solvable. From this you might obtain two independent solutions. Plug them into the other equation to obtain the solution to the system.

 

[aeroegnr]

I found out what the problem was, I had to create a new vector U instead of just using V.

Thanks Wong, but I'm taking an advanced math class where they are teaching us how to use linear algebraic methods to solve differential equations.

 

[eJavier]

I tried using undetermined coefficients to solve this problem, but I know that I am missing something and i cannot find any reference material on this. If you help me, thank you.

The homogeneous equation for the system is:

y' = A*y

where y = \left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right]

and A = \left[ \begin{array}{cc} -2 & 1 \\ -1 & 0 \end{array} \right]

I end up with only one eigenvector of course, and I'm trying to use a solution that ends up as C1*V*e^t + c2*V*t*e^t where V is the only eigenvector of A, but that is not a complete solution.

What am I missing?

Initial conditions perhaps?

 

[profuse007]

Initial conditions perhaps?

initital condition only solve for coefficient of the Y solution. like his c1, c2

first find ur eigenvalues then find ur eigenvectors and use that egeinvector in the Y solution.