Question

V\(\begin{array}{lll}y_{1}^{\prime} & =-y_{1}+y_{2}, & y_{1}(0) & =1, \\\ y_{2}^{\prime} & =\quad y_{2}+t, \quad y_{2}(0) & =0\end{array}\)

Short answer

The general solution is: \[ \begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix} = e^{t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} \frac{1}{6}t^2 \\ -\frac{1}{6}t^2 + t \end{pmatrix}. \]

Step-by-step solution

Rewrite the system in matrix form

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 $$y_2' = y_2 + t$$.

Solve the corresponding homogeneous system

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.

Find a particular solution

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: \[ \begin{pmatrix} y_{1p} \\ y_{2p} \end{pmatrix} = \begin{pmatrix} \frac{1}{6}t^2 \\ - \frac{1}{6}t^2 + t \end{pmatrix}. \]

Determine the general solution and apply initial conditions

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: \[ \begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix} = e^{t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} \frac{1}{6}t^2 \\ -\frac{1}{6}t^2 + t \end{pmatrix}. \]

Key concepts

Matrix Form

When dealing with systems of linear differential equations, converting them into matrix form can simplify analysis and computations. In this particular exercise, the given system:

• \(y_1' = -y_1 + y_2\)
• \(y_2' = y_2 + t\)

is rewritten as a matrix equation:
\[\begin{pmatrix} y_1' \ y_2' \end{pmatrix} = \begin{pmatrix} -1 & 1 \ 0 & 1 \end{pmatrix}\begin{pmatrix} y_1 \ y_2 \end{pmatrix} + \begin{pmatrix} 0 \ t \end{pmatrix}.\]
In this form, it becomes easier to see that the system consists of a matrix of coefficients operating on the vector of functions \((y_1, y_2)\), plus a vector representing external influences or inputs (in this case, a function of \(t\)). Matrix representation is powerful, as it leads to more structured solutions and allows the use of techniques such as finding eigenvalues and eigenvectors.

Homogeneous System

To solve the linear system of differential equations, it is often helpful to start with the homogeneous system. The homogeneous system refers to setting the external influences to zero. This implies simplifying our matrix form equation to:
\[\begin{pmatrix} y_1' \ y_2' \end{pmatrix} = \begin{pmatrix} -1 & 1 \ 0 & 1 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix}.\]
This step is essential because it provides us with the core dynamic properties of the system without any external "noise." Solving the homogeneous equation enables us to find fundamental solutions that are then adjusted to include more complex external terms or forces, leading to a particular solution later on.

Eigenvalues and Eigenvectors

Finding the eigenvalues and eigenvectors of the coefficient matrix in a linear system is crucial for solving homogeneous systems. For the matrix \(A\), which is
\[\begin{pmatrix} -1 & 1 \ 0 & 1 \end{pmatrix},\]
we determine the eigenvalues \(\lambda\) by solving the characteristic equation:
\[\begin{vmatrix} -1 - \lambda & 1 \ 0 & 1 - \lambda \end{vmatrix} = \lambda^2 - 1 = (\lambda - 1)(\lambda + 1) = 0.\]
Thus, the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = -1\). Eigenvectors are found by substituting each eigenvalue back into the equation \((A - \lambda I)\mathbf{v} = \mathbf{0}\). These eigenvectors form part of the solution to the differential equation, offering distinct solution components that reflect inherent system behaviors.

Particular Solution

The next step after solving the homogeneous system is to find a particular solution that accounts for any non-homogeneous components in the original equation. For this system, we assumed a particular solution of the form:
\[\begin{pmatrix} y_{1p} \ y_{2p} \end{pmatrix} = \begin{pmatrix} At^2 + Bt + C \ Dt^2 + Et + F \end{pmatrix}.\]
By plugging this form into the original differential equation (with the non-homogeneous term \(\begin{pmatrix} 0 \ t \end{pmatrix}\)), we derive multiple equations to solve for the constants \(A, B, C, D, E,\) and \(F\). After simplification, we obtain the particular solution:
\[\begin{pmatrix} \frac{1}{6}t^2 \ -\frac{1}{6}t^2 + t \end{pmatrix}.\]
Combining this with the homogeneous solution leads to the complete solution, adjusted by the initial conditions to satisfy the full terms of the system.