trouble with non-homogeneous ODE system... which method shall I use?

Question

I am an undergrad statistics student and I am having troubles with non-homogeneous ODE systems. During my classes I went over just three methods for solving odes: Laplace transform, Fourier transform and power series; so I should use one of those to solve such systems and now I am having trouble with a particular one. I would be glad if you could show me the solution! The system is the following: $$\begin{align} x'(t)&=x(t)-3y(t)+2\\ y'(t)&=5x(t)+y(t)-t \end{align}$$

As I told you I don't know how to solve it and it would be great if you can time me a detailed solution. Thank you very much!!!

Answer 1

As Paul stated, you can solve this using Laplace transform, however in this case it is also possible to convert this system of equations to a homogeneous ODE system.

For a system of $N$ linear differential equations with polynomial functions in time, with maximum order $M$, which makes it non-homogeneous, it is possible to find an equivalent ODE which is homogeneous by defining new variables. Thus the system of differential equations can be written as,

$$ \dot{x}_i = \sum_{j=1}^{N}a_{i,j}\ x_j + \sum_{k=0}^{M}b_{i,k}\ t^k \quad \forall\ i\in \{1,\dots,N\}. \tag{1} $$

The new variables are defined as,

$$ x_i = y_i + \sum_{k=0}^{M}c_{i,k}\ t^k \quad \forall\ i\in \{1,\dots,N\}, \tag{2} $$

such that the added polynomial coefficients satisfy,

$$ (k+1)\ c_{i,k+1} = \sum_{j=1}^{N}a_{i,j}\ c_{j,k}+b_{i,k} \quad \forall\ \big\{(i,k)|i\in\{1,\dots,N\}\land k\in\{0,\dots,M\}\big\}, \tag{3} $$

where $c_{i,M+1}=0$ for every $i$, because the added polynomials are of maximum order $M$.

This will simplify equations $(1)$ to a homogeneous ODE,

$$ \dot{y}_i = \sum_{j=1}^{N}a_{i,j}\ y_j. \tag{4} $$

This system of equations can be solved relative easy. These solutions can then be substituted into equations $(2)$ to find the solutions for $x_i(t)$.

In order to find the polynomial coefficients, $c_{j,k}$, you can write it as a vector $\textbf{w}$, whose components are defined as $w_{j+Nk}=c_{j,k}$, such that it can be written as a $N(M+1)\times N(M+1)$ matrix equation,

$$ \textbf{A}\textbf{w}=\textbf{v}. $$

The components of the matrix $\textbf{A}$ can then be found with $k+1=A_{i+Nk,\ i+N(k+1)}$ and $-a_{i,j}=A_{i+Nk,\ j+Nk}$ and the components of the vector $\textbf{v}$ can then be found with $v_{i+Nk}=b_{i,k}$.

For example when $N=2$ and $M=2$ the polynomial coefficients, $c_{j,k}$, can be found by solving,

$$ \begin{bmatrix} -a_{1,1} & -a_{1,2} & 1 & 0 & 0 & 0 \\ -a_{2,1} & -a_{2,1} & 0 & 1 & 0 & 0 \\ 0 & 0 & -a_{1,1} & -a_{1,2} & 2 & 0 \\ 0 & 0 & -a_{2,1} & -a_{2,1} & 0 & 2 \\ 0 & 0 & 0 & 0 & -a_{1,1} & -a_{1,2} \\ 0 & 0 & 0 & 0 & -a_{2,1} & -a_{2,1} \end{bmatrix} \begin{bmatrix} c_{1,0} \\ c_{2,0} \\ c_{1,1} \\ c_{2,1} \\ c_{1,2} \\ c_{2,2} \end{bmatrix} = \begin{bmatrix} b_{1,0} \\ b_{2,0} \\ b_{1,1} \\ b_{2,1} \\ b_{1,2} \\ b_{2,2} \end{bmatrix} $$

The determinant of such a matrix will be equal to $|\textbf{a}|^{M+1}$, thus if the homogeneous ODE does not have any eigenvalues/poles at zero then you will be able to solve for the polynomial coefficients.

Answer 2

look for a particular solution in the form $$x = at + b, y = ct + d$$ subbing this in $$ \begin{align} x'&=x-3y+2\\ y'&=5x+y-t \end{align}$$ you find that

$$ \begin{align} a &=(a-3c)t + b - 3d + 2\\ c&= (5a + c-1)t +5b+d \end{align}$$

equating the coefficients of $t$ we have $$a = 3c, 5a + c = 1 \to c = 1/16, a = 3/16$$ equating the constants gives you $$b- 3d = -29/16\\ 5b+d = 1/16 \tag 1$$ solving $1) should give you $b,d$ and we are done with finding a particular solution.

for the homogeneous solution you will need the eigenvalues of $\pmatrix{1&-3\\5&1} .$ they are given by the roots of $$\lambda^2 - 2\lambda + 16 = 0 \to \lambda=1\pm \sqrt {15} i $$

now you can find the eigenvectors corresponding to these eigenvalues and compute the $e^{At}.$