General solution of a linear system (differential equations)

In summary: This would give you the following: D^2(D^2- 4)x+ 26D^2y= 6D(sin t)+ 6cos t. Next, you would "differentiate" this equation to get: D(sin t)+D cos t=-6. Finally, you would add these two equations together to get: y=6cos t.
  • #1
wtrow
1
0

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.
 
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  • #2
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
[tex]
A~=~\left[
\begin{array}{c c c c}
0&1&0&0\\
4&0&0&-13\\
0&0&0&1\\
0&2&9&0\\
\end{array}
\right]
[/tex]

The solution of the matrix differential equation y' = Ay is y = etAc, 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.
 
  • #3
wtrow said:

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.
 
  • #4
HallsofIvy said:
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.
 

1. What is a general solution of a linear system?

A general solution of a linear system is a set of equations that satisfy the given differential equations and include all possible solutions. It is a combination of particular solutions and a set of constants that represent the arbitrary values of the system.

2. How is a general solution different from a particular solution?

A particular solution is a specific set of values that satisfies the given differential equations. It is obtained by substituting the initial conditions into the general solution. On the other hand, a general solution includes all possible solutions and can be obtained by adding a set of arbitrary constants to the particular solution.

3. Can a linear system have multiple general solutions?

Yes, a linear system can have multiple general solutions. This is because the system can have different sets of arbitrary constants that can be added to the particular solution to form a general solution. These arbitrary constants can represent different initial conditions or boundary conditions.

4. How do you determine the arbitrary constants in a general solution?

The arbitrary constants in a general solution can be determined by substituting the initial conditions or boundary conditions into the equations. This will result in a system of equations that can be solved to find the values of the arbitrary constants.

5. Can a general solution be used to find a specific solution for a given initial condition?

Yes, a general solution can be used to find a specific solution for a given initial condition. By substituting the initial condition into the general solution and solving for the arbitrary constants, a particular solution can be obtained that satisfies the given initial condition.

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