91Ó°ÊÓ

The given system is, $$ \begin{aligned} &2 x^{\prime}+y^{\prime}+x+y=t^{2}+4 t \\\ &x^{\prime}+y^{\prime}+2 x+2 y=2 t^{2}-2 t \end{aligned} $$ We can write this as, $$\begin{pmatrix} (2\partial + 1) & (\partial + 1) \\ \partial & (\partial + 2)\\ \end{pmatrix}\begin{pmatrix} x \\ y \\ \end{pmatrix}=\begin{pmatrix} t^2+4t\\ 2t^2-2t\\ \end{pmatrix}$$ where \(\partial\) is the differential operator.

To find the inverse operator, we need to find the determinant and adjoint of the given matrix first. The determinant is, $$\det = (2\partial+1)(\partial+2)-(\partial)(\partial+1) = 2\partial^{2}+5\partial+2 - \partial^{2} - \partial = \partial^{2}+4\partial+2$$ Now, find the adjoint, $$\text{adj} = \begin{pmatrix} (\partial+2) & -(\partial+1)\\ -\partial & (2\partial+1)\\ \end{pmatrix}$$ The inverse operator is, $$\begin{pmatrix} x \\ y \\ \end{pmatrix}=\frac{1}{\det}\text{adj}\begin{pmatrix} t^2+4t\\ 2t^2-2t\\ \end{pmatrix}$$

Now, we multiply the inverse operator with the given functions to find the general solution. $$\begin{pmatrix} x \\ y \\ \end{pmatrix}=\frac{1}{\partial^{2}+4\partial+2}\begin{pmatrix} (\partial+2) & -(\partial+1)\\ -\partial & (2\partial+1)\\ \end{pmatrix}\begin{pmatrix} t^2+4t\\ 2t^2-2t\\ \end{pmatrix}$$ After multiplying, we get, $$\begin{pmatrix} x \\ y \\ \end{pmatrix}=\frac{1}{\partial^{2}+4\partial+2}\begin{pmatrix} -2t^2+8t+4t\\ 4t^2-6t-2t^2+4t\\ \end{pmatrix}$$ $$\begin{pmatrix} x \\ y \\ \end{pmatrix}=\frac{1}{\partial^{2}+4\partial+2}\begin{pmatrix} -2t^2+12t\\ 2t^2-2t\\ \end{pmatrix}$$

We have the expression for the inverse operator acting on the given functions. Now we will find the general solution by applying the inverse operator. $$x(t) = \frac{1}{\partial^{2}+4\partial+2} \left(-2t^2 + 12t\right)$$ $$y(t) = \frac{1}{\partial^{2}+4\partial+2} \left(2t^2 - 2t\right)$$ The final general solution is a combination of particular and homogeneous solutions, which you can find using standard methods like undetermined coefficients or variation of parameters depending on your course content. The final answer will be in the form: $$x(t) = x_p(t) + x_h(t)$$ $$y(t) = y_p(t) + y_h(t)$$ where \(x_p(t)\) and \(y_p(t)\) are the particular solutions, and \(x_h(t)\) and \(y_h(t)\) are the homogeneous solutions.