Third order ODE

Question

Hi everyone !

In the context of a linear algebra course, I need to solve this complex third order ODE using matrices:

$$y''' - iy'' +y'-iy = e^{it} + t$$

I already managed to write the matrix form for the homogeneous equation:

$$ \frac{d}{dt} \begin{pmatrix} y \\ y'\\ y''\\ \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ i & -1 & i \\ \end{pmatrix} \begin{pmatrix} y\\ y'\\ y''\\ \end{pmatrix} $$

I'm stuck at this stage: I don't understand how I need to proceed now...

Thanks a lot for your help !

Answer 1

Solve the characteristic equation first: $$r^3-ir^2+r-i=0$$ You can rewrite it as $$r^2(r-i)+(r-i)=(r-i)(r^2+1)=(r-i)^2(r+i)=0$$ Note that the solutions of this equation are similar to the power of the particular solution, so the particular solution must contain a polynomial multiplied with $e^{it}$.

Answer 2

A good way to solve this linear matrix DE

$$ \dot y(t) = A\cdot y(t) + b(t) $$

with $y(t) = (y_1(t),y_2(t),y_3(t))^{\dagger},\ \ b(t) = (0,0,e^{i t}+t)^{\dagger}$ is using the Laplace Transform. According to it we have

$$ s Y(s)-y(0) = A Y(s) + B(s) $$

and then

$$ (sI-A)Y(s) = B(s)+y(0) $$

and also

$$ Y(s) = (sI-A)^{-1}(B(s)+y(0)) $$

here

$$ s I-A = \left( \begin{array}{ccc} s & -1 & 0 \\ 0 & s & -1 \\ -i & 1 & s-i \\ \end{array} \right) $$

and

$$ (s I-A)^{-1} = \frac{1}{(s-i)^2(s+i)}\left( \begin{array}{ccc} s (s-i)+1 & s-i & 1 \\ i & s (s-i) & s \\ i s & i-s & s^2 \\ \end{array} \right) $$

and $B(s) = (0,0,\frac{1}{s^2}+\frac{1}{s-i})$ etc.

NOTE

$(sI-A)^{-1}$ gives the determinant of $sI-A$ at the denominator which contains the main expansion factors for the inversion.