91Ó°ÊÓ

Let us rewrite the given system of linear differential equations as a matrix equation: \[ \begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} + \begin{pmatrix} 0 \\ t \end{pmatrix}. \] Here, we denoted $$y_1' = -y_1 + y_2$$ and $$y_2' = y_2 + t$$.

To solve the homogeneous system, we set the non-homogeneous term equal to zero: \[ \begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}. \] Now, we can find the eigenvalues and eigenvectors of the matrix A: \[ \begin{vmatrix} -1 - \lambda & 1 \\ 0 & 1 - \lambda \end{vmatrix} = \lambda^2 - 1 = (\lambda - 1)(\lambda + 1) = 0. \] The eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = -1\). The corresponding eigenvectors are: - For \(\lambda_1 = 1\): \((1, 0)^T\). - For \(\lambda_2 = -1\): \((-1, 1)^T\). Thus, the solution of the homogeneous system is given by: \[ \begin{pmatrix} y_{1h} \\ y_{2h} \end{pmatrix} = C_1 e^{t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}, \] where \(C_1\) and \(C_2\) are constants.

Let's look for a particular solution of the form: \[ \begin{pmatrix} y_{1p} \\ y_{2p} \end{pmatrix} = \begin{pmatrix} At^2 + Bt + C \\ Dt^2 + Et + F \end{pmatrix}. \] We will find the derivatives and plug them into the non-homogeneous system: \[ \begin{pmatrix} (2At + B)' \\ (2Dt + E)' \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} At^2 + Bt + C \\ Dt^2 + Et + F \end{pmatrix} + \begin{pmatrix} 0 \\ t \end{pmatrix}. \] This gives us the following relationships: - \(2At + B = -At^2 - Bt - C + Dt^2 + Et + F\). - \(2Dt + E = Dt^2 + Et + F + t\). Solving these equations will provide us with the values of \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\). After some algebra, we get: \[ A = \frac{1}{6}, \quad B = 0, \quad C = 0, \quad D = -\frac{1}{6}, \quad E = 1, \quad F = 0. \] So, the particular solution is given by: \[ \begin{pmatrix} y_{1p} \\ y_{2p} \end{pmatrix} = \begin{pmatrix} \frac{1}{6}t^2 \\ - \frac{1}{6}t^2 + t \end{pmatrix}. \]

The general solution of the non-homogeneous system is the sum of the homogeneous and particular solution: \[ \begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix} = \begin{pmatrix} y_{1h} + y_{1p} \\ y_{2h} + y_{2p} \end{pmatrix} = C_1 e^{t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} -1 \\ 1 \end{pmatrix} + \begin{pmatrix} \frac{1}{6}t^2 \\ -\frac{1}{6}t^2 + t \end{pmatrix}. \] Now, we will apply the initial conditions to find the values for \(C_1\) and \(C_2\). - For \(y_1(0) = 1\), we have: \(1 = C_1 - C_2\). - For \(y_2(0) = 0\), we have: \(0 = C_2\). From this, \(C_2 = 0\) and \(C_1 = 1\). Therefore, we have the following solution to our given system: \[ \begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix} = e^{t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} \frac{1}{6}t^2 \\ -\frac{1}{6}t^2 + t \end{pmatrix}. \]