# First Order Linear DE System

1. Dec 7, 2011

1. The problem statement, all variables and given/known data
I need to solve this DE system for a lab:

$$q_1'=2-\frac{6}{5}q_1+q_2$$
$$q_2'=3+\frac{3}{5}q_1-\frac{3}{2}q_2$$

2. Relevant equations

3. 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.

2. Dec 7, 2011

### vela

Staff Emeritus
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. Dec 8, 2011

### genericusrnme

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. Dec 8, 2011

### HallsofIvy

Staff Emeritus
I don't see why being nonhomogeneous would matter here. If you differentiate the first equation you get
$$q_1''= -\frac{6}{5}q_1'+ q_2'$$
From the second equation,
$$q_2'= 3+ \frac{3}{5}q_1- \frac{3}{2}q_2$$
so that
$$q_1''= -\frac{6}{5}q_1'+ 3+ \frac{3}{5}q_1- \frac{3}{2}q_2$$

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

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 $P^{-1}AP= D$ 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 "$P^{-1}x'= P^{-1}AP(P^{-1}x)$ or $y'= Dy$ with $y= P^{-1}x$. With "x'= Ax+ B", it becomes $P^{-1}x'= P^{-1}APP^{-1}y+ P^{-1}B$ or $y'= Dy+ C$ with $C= P^{-1}B$.

5. Dec 8, 2011

thanks, I figured it out