The difference equation and the boundary value are:

y(n)-10.1y(n-1) y(n-2)=-1.35n,y(-1)=0,y(100)=101/6

I have managed to solve for y(n) with eigenvector to obtain the homogeneous solution and particular solution and solve the coefficients using the boundary value. However, I do not know how to reformulate the whole difference equation into a matrix equation. Can anyone provide me with an idea or hint?

CodePudding user response：

Hint: You need a Y vector with 2 entries y(n), y(n-1)

A recurrence relation of the form:

y(n) = a_{n-1}y(n-1) a_{n-2}y(n-2) .. a_0y(0)

can be extended with n-1 identities of the form: y(n-k)=y(n-k) to obtain a linear system:

Y(n) = [y(n),y(n-1),..,y(1)]^T = A * [y(n-1),y(n-2),..,y(0)]^T = A * Y(n-1)