Finding Particular Solution for System of Equations

In summary: Thanks for your great help!In summary, the conversation discusses finding the particular solution of a system of equations using eigenvalues and vectors. It also addresses the use of undetermined coefficients and the need for a particular solution of the form [At, Bt+C]. A clarification is made about the type of solutions (exponential or trigonometric) for a linear system.
  • #1
Benny
584
0
Hi, I'm having trouble finding the particular solution of the following system.

[tex]
\left[ {\begin{array}{*{20}c}
{\mathop x\limits^ \bullet } \\
{\mathop y\limits^ \bullet } \\
\end{array}} \right] = \left[ {\begin{array}{*{20}c}
1 & { - 1} \\
{ - 1} & 1 \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
x \\
y \\
\end{array}} \right] + \left[ {\begin{array}{*{20}c}
2 \\
{ - 5} \\
\end{array}} \right]
[/tex]

I found the complimentary function, it had some sines and cosines in it but I don't think that matters in terms of finding the particular solution. The independent variable t isn't present anywhere in the equation so I don't need to do any differentiation. If I set x' = y' = 0 (the prime denotes differentiation wrt t) then I end up with an 'inconsistent' set of equations, namely, x - y = 2 and x - y = 5. Can someone tell me how I can find the particular solution? Thanks.
 
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  • #2
I would be leary of your complimentary solution as a first order system should usually be exponential not trigonometric. I will help you to find the complete solution, (complimentary + particular)

you'll want to use eigenvalues and vectors to "decouple" the equations

the way I solved it was to let:
{x;y}={phi}e^nt (where phis is the matrix of eigenvectors) and find the eigenvalues of the matrix [1 -1;-1 1] (the variable n is the eigenvalue)

do this by subsituting into the original equation.

at this point I will apologize for my unwillingness to use latex.

the eigenvalues are n= 0 and 2

the eigenvectors are then [1;1] and [1;-1]

let psi be the fundamental matrix = [1 e^2t;1 -e^2t]

now your equation is of the form:

(x;y)'=A(x;y)+f(t)

where f(t)=(2;5)
at this point I will introduce the missing function u(t) needed to solve the equation.

the solution is x=psi*u(t)
where psi*u'(t)=f(t)

solve for u1' and u2' and integrate to find u1 and u2

then [x;y]=psi*[u1;u2]
or
x=(7/2)*t+c1+.75+c2*e^2t
y=(7/2)*t+c1-.75-c2*e^2t

where c1 and c2 are determined by y'(0) and x'(0)

this solution is correct, substitute it back into the original equation if you are unsure (it satisfies the heck out of it)

hope this helps.
 
  • #3
by the way

x=xc +xp

where xp=7/2*t+0.75
and xc=c1+c2e^2t

y of course is separated similarly.
 
  • #4
lance said:
I would be leary of your complimentary solution as a first order system should usually be exponential not trigonometric. I will help you to find the complete solution, (complimentary + particular)
Let me clarify that: A linear system can have either exponential or trigonometric (or polynomial or some combination) solutions. In this case, since the coefficient matrix is symmetric and so has real eigenvalues, the complementary solution is exponential, not trigonometric.

Benny, recalculate the eigenvalues! Here, the problem is that one of the eigenvalues is 0 so one of the solutions to the associated homogeneous equation is the constant solution. Try a solution of the form
[tex]\left[ \begin{array}{c}At \\ Bt \end{array} \right][/tex]
 
  • #5
Yep, the complimentary function should be in terms of exponentials, I mixed signs when I calculated the determinant.

Thanks for the suggestion HallsofIvy. Does the method of undetermined coefficients for single ODEs carry over to systems?

Edit: Hmm...I still get down to the problem of having an inconsistent system of equations. If I assume a solution of the form (At, Bt)^T and I plug it back into the matrix equation I end up with two equations.

A = At - Bt + 2
B = -At + Bt - 5

Rewriting this:

(A-B)t + (2-A) = 0...(1)
(B-A)t + (-5-B) = 0...(2)

Equating coefficients I get A - B = 0 so that A = B. But I also need A = 2 meaning that B = 2. But then -5-B wouldn't be zero.
 
Last edited:
  • #6
Sorry about that: you will need a particular solution of the form
[tex]\left[ \begin{array}{c}At \\ Bt+ C \end{array} \right][/tex]
 
  • #7
Ok thanks I'll try that.
 

1. What is a particular solution in a differential equation?

A particular solution in a differential equation is a specific function that satisfies the equation for certain values of the independent variables. It is often used to find the general solution of the equation by combining it with the complementary function.

2. How is a particular solution found?

A particular solution can be found by using the method of undetermined coefficients or variation of parameters. In the method of undetermined coefficients, we assume the particular solution to have a similar form as the non-homogenous term in the equation and solve for the coefficients. In variation of parameters, we use the complementary function and the non-homogenous term to find a particular solution.

3. Can a particular solution be unique?

No, a particular solution is not unique. There can be multiple particular solutions for a single differential equation, depending on the chosen method and assumptions. However, the general solution of the equation is unique and includes all possible particular solutions.

4. What is the role of initial conditions in finding a particular solution?

Initial conditions play a crucial role in finding a particular solution. They help us determine the specific values of the arbitrary constants in the general solution, which leads to a unique particular solution. Without initial conditions, the general solution remains incomplete.

5. Can a particular solution be used to solve all differential equations?

No, a particular solution can only be used to solve non-homogenous linear differential equations. It cannot be applied to solve higher-order or non-linear differential equations. For those types of equations, other methods such as Laplace transform or numerical methods are used to find solutions.

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