General solution of a linear system (differential equations)

[wtrow]

Homework Statement

x''+13y'-4x=6sint , y''-2x'-9y=0

The Attempt at a Solution

I am not really sure how to solve this completely, but I have done this so far:

(D^2-4)x + 13Dy - 6sint = 0 , (D^2-9)y - 2Dx = 0

then I hit a brick wall. Any help would be appreciated, thanks.

 

[Mark44]

Do you know about eigenvectors, eigenvalues, and matrix diagonalization? The system you have here is an example of a coupled linear system. With suitable substitutions it can be converted from a system of two second-order (nonhomogeneous) differential equations into a system of four first-order differential equations, also nonhomogeneous.

Let's ignore the 6sint term for a while, which makes the system homogeneous. This substitution can be used to get to the four first-order equations:
y₁ = y
y₂ = y'
y₃ = x
y₄ = x'

With these substitutions, your system of two equations can be rewritten as:
y₁' = y₂
y₂' = 4y₁ - 13 y₄
y₃' = y₄
y₄' = 2y₂ + 9y₃

This system of equations can be written in matrix form as y' = Ay,
with
$$A~=~\left[<br /> \begin{array}{c c c c}<br /> 0&1&0&0\\<br /> 4&0&0&-13\\<br /> 0&0&0&1\\<br /> 0&2&9&0\\<br /> \end{array}<br /> \right]$$

The solution of the matrix differential equation y' = Ay is y = e^tAc, where c is a vector of constants. Where the eigenvalues, eigenvectors, and matrix diagonalization come in, is that it is much easier to evaluate e raised to a matrix power if the matrix is diagonal.

I hope some of these ideas are familiar to you. Yours is not a simple problem, and there is still the question of dealing with the nonhomogeneous system, which is not that more complicated if you understand what I've laid out here.

 

[HallsofIvy]

Homework Statement

x''+13y'-4x=6sint , y''-2x'-9y=0

The Attempt at a Solution

I am not really sure how to solve this completely, but I have done this so far:

(D^2-4)x + 13Dy - 6sint = 0 , (D^2-9)y - 2Dx = 0

then I hit a brick wall. Any help would be appreciated, thanks.

Mark44 is completely correct but, but from what you have written, I suspect you may not be ready for that method.

Treat these equations as algebraic equation for x and y and eliminate one of them. For example, if you "multiply" the first equation by 2D (actually you are differentiating the equation and multiplying by 2) you get 2D^2(D^2- 4)x+ 26D^2y= 6D(sin t)= 6cos t.
If you "multiply" the second equation by D^2- 4 (actually Differentiate the entire equation twice and subtract the equation from that) you get (D^2-4)(D^2-9)y- 2D(D^2-4)x= 0.

Adding the two equations eliminates x and gives you a fourth order equation for y.

 

[Mark44]

Mark44 is completely correct but, but from what you have written, I suspect you may not be ready for that method.

Treat these equations as algebraic equation for x and y and eliminate one of them. For example, if you "multiply" the first equation by 2D (actually you are differentiating the equation and multiplying by 2) you get 2D^2(D^2- 4)x+ 26D^2y= 6D(sin t)= 6cos t.
If you "multiply" the second equation by D^2- 4 (actually Differentiate the entire equation twice and subtract the equation from that) you get (D^2-4)(D^2-9)y- 2D(D^2-4)x= 0.

Adding the two equations eliminates x and gives you a fourth order equation for y.

HallsOfIvy, Thanks for jumping in on this with a simpler (and therefore better) approach.

To clarify, it looks like you are "multiplying" the first equation by the 2D² operator.