Solving a Non-Homogeneous First Order Linear DE System

[themadhatter1]

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

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.

 

[vela]

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

 

[HallsofIvy]

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.

 

[themadhatter1]

thanks, I figured it out