First Order Linear DE System

  • #1

Homework Statement


I need to solve this DE system for a lab:

[tex]q_1'=2-\frac{6}{5}q_1+q_2[/tex]
[tex]q_2'=3+\frac{3}{5}q_1-\frac{3}{2}q_2[/tex]


Homework Equations





The Attempt at a Solution



I know how to use the method of elimination to solve such systems, but this is non homogeneous because of the added constant. I've tried to do a change of variables for q1 and q2 so that it will become homogeneous, but I can't think of any substitution that would do that. How do I solve this problem? Thanks.
 

Answers and Replies

  • #2
vela
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Try a substitution of the form
\begin{align*}
p_1 &= q_1 + c_1 \\
p_2 &= q_2 + c_2
\end{align*}Rewrite the original equations in terms of the p's. Then solve for the c's so that the constant terms cancel out.
 
  • #3
if you know some linear algebra you could turn it into a matrix equation and use eigenvalue decomposition by setting a vector v=(q1,q2,1), v'=(q1',q2',1) related by a matrix A=((-6/5,1,2),(3/5,-3/2,3),(0,0,1))
that's how I'd solve it anyway
 
  • #4
HallsofIvy
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I don't see why being nonhomogeneous would matter here. If you differentiate the first equation you get
[tex]q_1''= -\frac{6}{5}q_1'+ q_2'[/tex]
From the second equation,
[tex]q_2'= 3+ \frac{3}{5}q_1- \frac{3}{2}q_2[/tex]
so that
[tex]q_1''= -\frac{6}{5}q_1'+ 3+ \frac{3}{5}q_1- \frac{3}{2}q_2[/tex]

From the first equation, again,
[tex]q_2= q_1'+ \frac{6}{5}q_1- 2[/tex]
Putting that in,
[tex]q_1''= -\frac{6}{5}q_1'+ 3+ \frac{3}{5}q_1- \frac{3}{2}(q_1'+ \frac{6}{5}q_1- 2)[/tex]
[tex]q_1''= -\frac{27}{10}q_1'- \frac{6}{5}q_1+ 3[/tex]
a linear equation with constant coefficients for [itex]q_1[/itex]

Or set it up as a matrix equation as genericusrnme suggests. The only thing the "2" and "3" add is that instead of "x'= Ax" you get "x'= Ax+ B". The same ideas apply. Find the eigenvalues and eigenvectors of A so that you have a matrix P such that [itex]P^{-1}AP= D[/itex] where D is the diagonal (or Jordan normal form) matrix with the eigenvalues on the diagonal. With "x'= Ax", you would then write the equation as "[itex]P^{-1}x'= P^{-1}AP(P^{-1}x)[/itex] or [itex]y'= Dy[/itex] with [itex]y= P^{-1}x[/itex]. With "x'= Ax+ B", it becomes [itex]P^{-1}x'= P^{-1}APP^{-1}y+ P^{-1}B[/itex] or [itex]y'= Dy+ C[/itex] with [itex]C= P^{-1}B[/itex].
 
  • #5
thanks, I figured it out
 

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