# Nonhomogeneous Linear ODE

$$x' =\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)x + \left(\begin{array}{rr}t^{-3} \\ -t^{-2}\end{array}\right), t>0$$

To find the general solution of the given system above, my solution is:

$det(\left(\begin{array}{rr}4-r & 8 \\ -2 & -4-r\end{array}\right)) = r^2$ , eigenvalues are $r_{1,2}=0$

$\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)\left(\begin{array}{rr}v_{11} \\ v_{12}\end{array}\right) = \left(\begin{array}{rr}0 \\ 0\end{array}\right)$ Therefore $v_{1} =\left(\begin{array}{rr}-2 \\ 1\end{array}\right)$

$\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)\left(\begin{array}{rr}v_{21} \\ v_{22}\end{array}\right) = \left(\begin{array}{rr}-2 \\ 1\end{array}\right)$ From there, I have and equation $2v_{21}+4v_{22}=-1$

What should I do now? Thanks.

• Can you find a solution to the homogeneous system $x'=Ax?$ – Git Gud Dec 19 '13 at 22:30
• Actually I got stuck there because I couldn't understand how to write $v_{2}$, should I pick arbitrary numbers for $v_{21}$ and $v_{22}$? – cecemelly Dec 19 '13 at 22:32
• No. You want to find a generalised eigenvector, $\xi$, such that $A\cdot \xi = \psi$, where $\psi$ is your eigenvector; note that it is not scalable. – Chris K Dec 19 '13 at 22:33
• You get the freedom to choose one of $v_{21}$ or $v_{22}$, just like how you had freedom for $v_{11}$ or $v_{12}$ – Euler....IS_ALIVE Dec 19 '13 at 22:34
• Hint - General solution of your problem is $$x = e^{At} x(0) + \int_0^t e^{A(t-\tau)} f(\tau)\ d\tau$$ so you just need to find those matrix exponents. – Kaster Dec 19 '13 at 22:43

Since $\begin{pmatrix} 4 & 8\\ -2 & -4 \end{pmatrix}\cdot \binom{-1/2}{0} = \binom{-2}{1}$ works, we get a homogeneous solution to the ODE of:
$c_{1}\binom{-2}{1}+c_{2}(t\binom{-2}{1}+\binom{-1/2}{0})$. Now it suffices to find a particular solution.
You can use the variation of parameters formula for multidimensional problems. It says, that if $\dot x =Ax +f(t)$ is your ode, than the solution is $x(t) = e^{At}\cdot a + \int_0^t e^{A(t-\tau)} f(\tau) \mathrm{d}\tau$ where $x(0)=a$. Put your matrix $A$ in Jordan normal form to determine $e^{At}$.