resolve an non-homogeneous differential system

Solve the given non-homogeneous system:

$$\dfrac{dx}{dt}=3x + 3 y – 2 z + e^t$$
$$\dfrac{dy}{dt} = x+y+ 2 z$$
$$\dfrac{dz}{dt}=x+3y+e^t$$

Can you help me understand a simple method to resolve the system?

Thanks for any help.

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We are given the nonhomogeneous system:

$$\dfrac{dx}{dt}=3x + 3 y – 2 z + e^t$$

$$\dfrac{dy}{dt} = x+y+ 2 z$$

$$\dfrac{dz}{dt}=x+3y+e^t$$

We can write this as $X'(t) = A x(t) + F[t]$, where:

$$A = \begin{bmatrix} 3& 3 & -2 \\ 1 & 1 & 2 \\ 1 & 3 & 0 \end{bmatrix}, ~~ F[t] = \begin{bmatrix} e^t \\ 0 \\ e^t \end{bmatrix}$$

We have the homogeneous solution (you said you know how to find this):

$$X_h(t) = \begin{bmatrix} x_h(t) \\ y_h(t) \\ z_h(t) \end{bmatrix} = c_1e^{4t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix}1 \\ -1 \\ 1 \end{bmatrix} + c_3e^{2t} \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$$

Now we need to find the particular solution using the Fundamental Matrix approach, by solving:

$$X(t) = e^{At}X_0 + \int_{t_0}^t e^{A(t-s)}F(s)~ds$$

So, we we use a linear combination from the components of the homogeneous solution to write:

  • Write $\phi(t) = [x_1(t)~ | ~x_2(t)~ | ~x_3(t)] = \begin{bmatrix} e^{4t} & e^{-2t} & -e^{2t} \\ e^{4t} & -e^{-2t} & e^{2t} \\ e^{4t} & e^{-2t} & e^{2t} \end{bmatrix}$
  • Find $\phi^{-1}(t) = \begin{bmatrix} e^{-4t}/2 & e^{-4t}/2 & 0 \\ 0 & -e^{2t}/2 & e^{2t}/2 \\ -e^{-2t}/2 & 0 & e^{-2t}/2 \end{bmatrix}$
  • Find $\phi^{-1}(t) \cdot F(t) = \begin{bmatrix} e^{-3t}/2 \\ e^{3t}/2 \\ 0 \end{bmatrix}$
  • Now we integrate the previous result and this gives us: $w = \begin{bmatrix} -\dfrac{1}{6}e^{-3t} \\ \dfrac{1}{6}e^{3t} \\ 0 \end{bmatrix}$
  • Next, we find $X_p(t) = \begin{bmatrix} x_p(t) \\ y_p(t) \\ z_p(t) \end{bmatrix} = \phi(t) \cdot w = \begin{bmatrix} 0 \\ -e^{t}/3 \\ 0 \end{bmatrix}$

All that is left is to write the final solution as:

$$X(t) = X_h(t) + X_p(t)$$

Update

This is adding an example of using the VoP method for a $2×2$ matrix, so you can see the process. You can make up an initial condition for your $3×3$ and emulate this process as you can easily check the result by substituting it back into the system of ODEs.

$x’ = Ax(t) + f(t) = \begin{bmatrix} 6/7 & -15/14 \\ -5/7 & 37/14 \end{bmatrix}~x(t) + \begin{bmatrix} e^{2t} \\ e^{-t} \end{bmatrix}, x(0) = \begin{bmatrix} 4 \\ -1 \end{bmatrix}$

$e^{At} = \begin{bmatrix} \dfrac{6}{7}e^{t/2} + \dfrac{1}{7}e^{3t} & \dfrac{3}{7}e^{t/2}- \dfrac{3}{7}e^{3t} \\ \dfrac{2}{7}e^{t/2}- \dfrac{2}{7}e^{3t} & \dfrac{1}{7}e^{t/2} + \dfrac{6}{7}e^{3t} \end{bmatrix}$

$e^{At}x(0) = \begin{bmatrix} 3e^{t/2} + e^{3t} \\ e^{t/2}-2e^{3t} \end{bmatrix}$

$e^{A(t-s)} = \begin{bmatrix} \dfrac{6}{7}e^{(t-s)/2} + \dfrac{1}{7}e^{3(t-s)} & \dfrac{3}{7}e^{(t-s)/2}- \dfrac{3}{7}e^{3(t-s)} \\ \dfrac{2}{7}e^{(t-s)/2}- \dfrac{2}{7}e^{3(t-s)} & \dfrac{1}{7}e^{(t-s)/2} + \dfrac{6}{7}e^{3(t-s)} \end{bmatrix}$

$e^{A(t-s)} \cdot f(s) = \begin{bmatrix} \dfrac{6}{7}e^{(t-s)/2} + \dfrac{1}{7}e^{3(t-s)} & \dfrac{3}{7}e^{(t-s)/2}- \dfrac{3}{7}e^{3(t-s)} \\ \dfrac{2}{7}e^{(t-s)/2}- \dfrac{2}{7}e^{3(t-s)} & \dfrac{1}{7}e^{(t-s)/2} + \dfrac{6}{7}e^{3(t-s)} \end{bmatrix} \cdot \begin{bmatrix} e^{2s} \\ e^{-s} \end{bmatrix}$

$\displaystyle \int_0^t e^{A(t-s)} \cdot f(s)~ds = \begin{bmatrix} \dfrac{3}{7}e^{2t} -\dfrac{5}{28}e^{-t} -\dfrac{2}{7}e^{t/2} + \dfrac{1}{28}e^{3t} \\ \dfrac{10}{21}e^{2t} -\dfrac{13}{42}e^{-t} -\dfrac{2}{21}e^{t/2} – \dfrac{1}{14}e^{3t} \end{bmatrix}$

$\displaystyle x(t) = e^{At}x(0) + \int_0^t e^{A(t-s)} \cdot f(s)~ds = \begin{bmatrix} 3e^{t/2} + e^{3t} \\ e^{t/2}-2e^{3t} \end{bmatrix} + \begin{bmatrix} \dfrac{3}{7}e^{2t} -\dfrac{5}{28}e^{-t} -\dfrac{2}{7}e^{t/2} + \dfrac{1}{28}e^{3t} \\ \dfrac{10}{21}e^{2t} -\dfrac{13}{42}e^{-t} -\dfrac{2}{21}e^{t/2} – \dfrac{1}{14}e^{3t} \end{bmatrix} = \begin{bmatrix} \dfrac{3}{7}e^{2t} -\dfrac{5}{28}e^{-t} +\dfrac{19}{7}e^{t/2} + \dfrac{29}{28}e^{3t} \\ \dfrac{10}{21}e^{2t} -\dfrac{13}{42}e^{-t} +\dfrac{19}{21}e^{t/2} – \dfrac{29}{14}e^{3t} \end{bmatrix}$