# System of Linear Differential Equations with Constant Coefficients

• Feb 27th 2011, 06:01 AM
rebghb
System of Linear Differential Equations with Constant Coefficients
Hello Everyone,

Now I am familiar, roughly, in solving systems of DEs with constant coefficients, granted that these equations are homogenous, but how can we solve such a system of equations:

$\displaystyle y_1'=ay_1+by_2 y_2'=cy_1+dy_2+f(x)$

Thanks a lot!
• Feb 27th 2011, 06:12 AM
FernandoRevilla
You have to find a particular solution of the complete system and add all the solutions of the homogeneous. There is a well known theorem about it. Have you studied the corresponding theory?
• Feb 27th 2011, 06:25 AM
rebghb
I am afraid not, I konly know how to solve linear ODEs (not necessarily homogenous) and homogenous systems. I tried searching for an answer but didn't know what to search for.
• Feb 27th 2011, 10:18 AM
FernandoRevilla
Quote:

Originally Posted by rebghb
I am afraid not, I konly know how to solve linear ODEs (not necessarily homogenous) and homogenous systems. I tried searching for an answer but didn't know what to search for.

I suppose you know how to compute the exponential of a matrix. A particular solution for the system

$\displaystyle Y'=AY+b(x)$

is

$\displaystyle Y(x)=\displaystyle\int_0^xe^{tA}b(t)dt$
• Feb 27th 2011, 11:11 AM
HallsofIvy
Quote:

Originally Posted by rebghb
Hello Everyone,

Now I am familiar, roughly, in solving systems of DEs with constant coefficients, granted that these equations are homogenous, but how can we solve such a system of equations:

$\displaystyle y_1'=ay_1+by_2 y_2'=cy_1+dy_2+f(x)$

Thanks a lot!

Well, no, the second equaton is NOT homogeneous.

Here is a much more elementary method:

Differentiate both sides of the first equation to get
$\displaystyle y_1''= ay_1'+ by_2'$
From the second equation, $\displaystyle by_2'= b(cy_1+ dy_2+ f(x))= bcy_1+ bdy_2+ bf(x)$ so we can write that as
$\displaystyle y_1''= ay_1'+ bcy_1+ bdy_2+ bf(x)$
and now, from the first equation again, $\displaystyle by_2= y1'- ay_1$ so we have
$\displaystyle y_1''= ay_1'+ bcy_1+ d(y_1'- ay_1)+ bf(x)$
$\displaystyle y_1''= (a+ d)y_1'+ (a-ad)y_1+ bf(x)$
a single, non-homogeneous second order linear equation with constant coefficients. Once you have solved for $\displaystyle y_1$ you can use $\displaystyle by_2= y_1'- ay_1$ to find =$\displaystyle y_2$
• Feb 28th 2011, 01:45 AM
rebghb
Thank you, I thought of that, honestly, but what I was looking for was something to solve a bigger problem, say 10 equations!
As I have posted earlier, we need to find the eigenvalues for the matrix, a step towards solving the system, shall the system be homogenous; but what if it isn't? Is there a particular procedure? If yes I'd like to know it's name, I'll try to search for it and learn it on my own...
• Feb 28th 2011, 02:00 AM
Ackbeet
Quote:

Originally Posted by FernandoRevilla
I suppose you know how to compute the exponential of a matrix. A particular solution for the system

$\displaystyle Y'=AY+b(x)$

is

$\displaystyle Y(x)=\displaystyle\int_0^xe^{tA}b(t)dt$

Did you mean

$\displaystyle \displaystyle Y(x)=e^{tA}\int_{0}^{x}e^{-tA}b(x)\,dx?$

[EDIT]: See Fernando's next post for a correction in the variables.
• Feb 28th 2011, 03:29 AM
FernandoRevilla
Quote:

Originally Posted by Ackbeet
Did you mean

$\displaystyle \displaystyle Y(x)=e^{tA}\int_{0}^{x}e^{-tA}b(x)\,dx?$

No, I meant:

$\displaystyle Y(x)=e^{xA}\displaystyle\int_0^xe^{-tA}b(t)\;dt$

:)
• Feb 28th 2011, 04:51 AM
Ackbeet
Yeah, that. Right.