# First Order Linear DE System

## Homework Statement

I need to solve this DE system for a lab:

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

## 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.

Staff Emeritus
Homework Helper
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.

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

Homework Helper
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 $C= P^{-1}B$.