Question

Find the particular solution of the system ′ = 4 − 3, ′ = 6 − 7 that satisfies the initial conditions x(0) = 2, y(0) = −1.

Answer 1

To find the particular solution of the system
`\displaystyle\sf x' = 4x - 3y` and
`\displaystyle\sf y' = 6x - 7y` that satisfies the initial conditions
`\displaystyle\sf x(0) = 2` and
`\displaystyle\sf y(0) = -1`, we can use the method of solving a system of linear differential equations.

Let's first find the general solution of the system by solving the differential equations. We can rewrite the system in matrix form as follows:

`\displaystyle\sf \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 6 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}`.

The coefficient matrix
`\displaystyle\sf A` is
`\displaystyle\sf \begin{bmatrix} 4 & -3 \\ 6 & -7 \end{bmatrix}`, and the vector
`\displaystyle\sf X` is
`\displaystyle\sf \begin{bmatrix} x \\ y \end{bmatrix}`.

To find the general solution, we need to find the eigenvalues and eigenvectors of matrix
`\displaystyle\sf A`.

By solving
`\displaystyle\sf \det(A-\lambda I) = 0`, where
`\displaystyle\sf \lambda` is the eigenvalue and
`\displaystyle\sf I` is the identity matrix, we can find the eigenvalues.

Solving
`\displaystyle\sf \det(A-\lambda I) = \det\left(\begin{bmatrix} 4 & -3 \\ 6 & -7 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}\right) = 0`, we get
`\displaystyle\sf \lambda^(2) - (-3-6)\lambda + (4\cdot -7 - 3\cdot 6) = \lambda^(2) + 9\lambda - 54 = 0`.

Solving the quadratic equation, we find
`\displaystyle\sf \lambda = -12` and
`\displaystyle\sf \lambda = 3`.

Next, we find the corresponding eigenvectors by solving
`\displaystyle\sf (A-\lambda I)X = 0` for each eigenvalue.

For
`\displaystyle\sf \lambda = -12`, we have
`\displaystyle\sf (A-(-12)I)X = \begin{bmatrix} 16 & -3 \\ 6 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0`. By solving this system of equations, we find
`\displaystyle\sf X_(1) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}`.

For
`\displaystyle\sf \lambda = 3`, we have
`\displaystyle\sf (A-(3)I)X = \begin{bmatrix} 1 & -3 \\ 6 & -10 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0`. By solving this system of equations, we find
`\displaystyle\sf X_(2) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}`.

The general solution of the system is given by
`\displaystyle\sf X = c_(1)e^(-12t)X_(1) + c_(2)e^(3t)X_(2)`, where
`\displaystyle\sf c_(1)` and
`\displaystyle\sf c_(2)` are constants, and
`\displaystyle\sf X_(1)` and
`\displaystyle\sf X_(2)` are the eigenvectors corresponding to the eigenvalues -12 and 3, respectively.

To find the particular solution that satisfies the initial conditions
`\displaystyle\sf x(0) = 2` and
`\displaystyle\sf y(0) = -1`, we substitute these values into the general solution and solve for the constants
`\displaystyle\sf c_(1)` and
`\displaystyle\sf c_(2)`.

Substituting
`\displaystyle\sf t = 0`,
`\displaystyle\sf x(0) = 2`, and
`\displaystyle\sf y(0) = -1` into the general solution, we have:

`\displaystyle\sf 2 = c_(1)e^(-12\cdot 0)(3) + c_(2)e^(3\cdot 0)(1)`,

`\displaystyle\sf -1 = c_(1)e^(-12\cdot 0)(2) + c_(2)e^(3\cdot 0)(2)`.

Simplifying these equations, we get:

`\displaystyle\sf 2 = 3c_(1) + c_(2)`,

`\displaystyle\sf -1 = 2c_(1) + 2c_(2)`.

Solving this system of equations, we find
`\displaystyle\sf c_(1) = (1)/(5)` and
`\displaystyle\sf c_(2) = (7)/(5)`.

Therefore, the particular solution of the system that satisfies the initial conditions is:

`\displaystyle\sf X = (1)/(5)e^(-12t)\begin{bmatrix} 3 \\ 2 \end{bmatrix} + (7)/(5)e^(3t)\begin{bmatrix} 1 \\ 2 \end{bmatrix}`.