System of linear differential equation

Question

How to solve the system of linear differential equation of the form $$x' = Ax + b$$ I can solve the homogeneous form by finding the eigenvalues and respective eigenvectors, but how to find the particular solution part. Also is there any limitation from getting eigenvalues positive, negative or complex. Any other different method involving matrix algebra is also welcome. Thank you.

Answer 1

Take Laplace transform therefore $$sX(s)-x(0)=AX(s)+b\\(sI-A)X(s)=b+x(0)$$or$$X(s)=(sI-A)^{-1}(b+x(0))$$but we can't proceed any further. We need to know the characteristics of the system.

Answer 2

If $b$ is the columnspace of $A$, define $y$ so that $Ay = -b$. Then note that $x(t) = e^{At}x(0) + y$ solves the differential equation since $x' = A e^{At}x(0)$ and $Ax = A e^{At}x(0) - b$.

Answer 3

Being $b$ a fixed constant (suppose $\det A\neq0$), we can choose a new variable $x_*$ that solves $$Ax_* + b = 0 \Rightarrow x_* = -A^{-1}b$$ The now we can solve the equation in the fixed point $z = x-x_* = x+A^{-1}b$. This variable solves the following equation $$\dot{z} = Az$$ In fact, form the definition of $z$ $$\dot{z}=\dot{x}\\ Az = A(x+A^{-1}b) = Ax+b$$ So we're saying that, if $x$ is a solution to $\dot{x}=Ax+b$ then the variable $z = x+A^{-1}b$ is a solution to the simpler equation $\dot{z}=Az$.

The solution to this simpler equation is just $$z(t)= [e^{At}]z(0)$$ where $[e^{At}]$ is the exponential of the matrix $A$. At this point you can go back to find $x(t)$ simply by using the definition of $z$: $$x(t) = z(t)-A^{-1}b = [e^{At}](x(0)+A^{-1}b)-A^{-1}b = [e^{At}]x(0)+A^{-1}b([e^{At}]-1)$$ where we have used the fact that $$z(0) = x(0) + A^{-1}b$$

Easier way

You can even use this simpler method: given the system of equations $$\dot{x} = Ax+f(t)$$ the solution to this is as follows $$x(t) = [e^{At}]x(0)+\int_{0}^t e^{A(t-t')}f(t')dt'$$ in your case the function is just a constant function $f(t)=b$ for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix