Questions: Solve the system: [x', y'] = (1/t) * [[5, 1], [3, 3]] * [x, y] + [5, 5] with x(1) = [0, 2], X(t) = [[t^2, t^6], [-3t^2, t^6]]

Solve the system:  
[x', y'] = (1/t) * [[5, 1], [3, 3]] * [x, y] + [5, 5] with x(1) = [0, 2], X(t) = [[t^2, t^6], [-3t^2, t^6]]
Transcript text: Solve the system: \[ \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]=\frac{1}{t}\left[\begin{array}{ll} 5 & 1 \\ 3 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]+\left[\begin{array}{l} 5 \\ 5 \end{array}\right] \text { with } \mathbf{x}(1)=\left[\begin{array}{l} 0 \\ 2 \end{array}\right], \mathbf{X}(t)=\left[\begin{array}{cc} t^{2} & t^{6} \\ -3 t^{2} & t^{6} \end{array}\right] \]
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Solution

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Solve the system of differential equations using the method of variation of parameters.

Identify the system and initial conditions.

The system is given by:
\[ \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]=\frac{1}{t}\left[\begin{array}{ll} 5 & 1 \\ 3 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]+\left[\begin{array}{l} 5 \\ 5 \end{array}\right] \]
with initial condition \(\mathbf{x}(1)=\left[\begin{array}{l} 0 \\ 2 \end{array}\right]\). The fundamental matrix is \(\mathbf{X}(t)=\left[\begin{array}{cc} t^{2} & t^{6} \\ -3 t^{2} & t^{6} \end{array}\right]\).

Apply the method of variation of parameters.

Let \(\mathbf{x}(t) = \mathbf{X}(t) \mathbf{u}(t)\), where \(\mathbf{u}(t)\) is a vector function to be determined. Then \(\mathbf{x}^{\prime}(t) = \mathbf{X}^{\prime}(t) \mathbf{u}(t) + \mathbf{X}(t) \mathbf{u}^{\prime}(t)\). Substituting into the system, we have:
\[ \mathbf{X}^{\prime}(t) \mathbf{u}(t) + \mathbf{X}(t) \mathbf{u}^{\prime}(t) = \frac{1}{t} \mathbf{A} \mathbf{X}(t) \mathbf{u}(t) + \mathbf{g}(t) \]
where \(\mathbf{A} = \left[\begin{array}{ll} 5 & 1 \\ 3 & 3 \end{array}\right]\) and \(\mathbf{g}(t) = \left[\begin{array}{l} 5 \\ 5 \end{array}\right]\).

Simplify the equation using the properties of the fundamental matrix.

Since \(\mathbf{X}(t)\) is a fundamental matrix, \(\mathbf{X}^{\prime}(t) = \frac{1}{t} \mathbf{A} \mathbf{X}(t)\). Therefore, the equation simplifies to:
\[ \mathbf{X}(t) \mathbf{u}^{\prime}(t) = \mathbf{g}(t) \]
Thus, \(\mathbf{u}^{\prime}(t) = \mathbf{X}^{-1}(t) \mathbf{g}(t)\).

Calculate the inverse of the fundamental matrix \(\mathbf{X}(t)\).

The determinant of \(\mathbf{X}(t)\) is \(t^{2} \cdot t^{6} - (-3 t^{2}) \cdot t^{6} = t^{8} + 3 t^{8} = 4 t^{8}\). Therefore,
\[ \mathbf{X}^{-1}(t) = \frac{1}{4 t^{8}} \left[\begin{array}{cc} t^{6} & -t^{6} \\ 3 t^{2} & t^{2} \end{array}\right] = \frac{1}{4} \left[\begin{array}{cc} t^{-2} & -t^{-2} \\ 3 t^{-6} & t^{-6} \end{array}\right] \]

Find \(\mathbf{u}^{\prime}(t)\) and integrate to find \(\mathbf{u}(t)\).

\[ \mathbf{u}^{\prime}(t) = \mathbf{X}^{-1}(t) \mathbf{g}(t) = \frac{1}{4} \left[\begin{array}{cc} t^{-2} & -t^{-2} \\ 3 t^{-6} & t^{-6} \end{array}\right] \left[\begin{array}{l} 5 \\ 5 \end{array}\right] = \frac{1}{4} \left[\begin{array}{l} 0 \\ 20 t^{-6} \end{array}\right] = \left[\begin{array}{c} 0 \\ 5 t^{-6} \end{array}\right] \]
Integrating, we get:
\[ \mathbf{u}(t) = \int \mathbf{u}^{\prime}(t) dt = \int \left[\begin{array}{c} 0 \\ 5 t^{-6} \end{array}\right] dt = \left[\begin{array}{c} c_{1} \\ -t^{-5} \end{array}\right] \]
where \(c_{1}\) is a constant of integration.

Find the homogeneous and particular solutions.

The homogeneous solution is:
\[ \mathbf{x}(t) = \mathbf{X}(t) \mathbf{u}(t) = \left[\begin{array}{cc} t^{2} & t^{6} \\ -3 t^{2} & t^{6} \end{array}\right] \left[\begin{array}{c} c_{1} \\ -t^{-5} \end{array}\right] = \left[\begin{array}{c} c_{1} t^{2} - t \\ -3 c_{1} t^{2} - t \end{array}\right] \]
The particular solution is:
\[ \mathbf{x}_p(t) = \left[\begin{array}{cc} t^{2} & t^{6} \\ -3 t^{2} & t^{6} \end{array}\right] \left[\begin{array}{c} 0 \\ -t^{-5} \end{array}\right] = \left[\begin{array}{c} -t \\ -t \end{array}\right] \]

Combine solutions and apply initial conditions.

The general solution is:
\[ \mathbf{x}(t) = \left[\begin{array}{c} c_{1} t^{2} + c_{2} t^{6} - t \\ -3 c_{1} t^{2} + c_{2} t^{6} - t \end{array}\right] \]
Apply the initial condition \(\mathbf{x}(1) = \left[\begin{array}{l} 0 \\ 2 \end{array}\right]\):
\[ \left[\begin{array}{c} c_{1} + c_{2} - 1 \\ -3 c_{1} + c_{2} - 1 \end{array}\right] = \left[\begin{array}{l} 0 \\ 2 \end{array}\right] \]
Solve the system of equations:
\(c_{1} + c_{2} = 1\)
\(-3 c_{1} + c_{2} = 3\)
Subtracting the first equation from the second, we get \(-4 c_{1} = 2\), so \(c_{1} = -\frac{1}{2}\).
Then \(c_{2} = 1 - c_{1} = 1 - (-\frac{1}{2}) = \frac{3}{2}\).

\(\boxed{\left[\begin{array}{c} -\frac{1}{2} t^{2} + \frac{3}{2} t^{6} - t \\ \frac{3}{2} t^{2} + \frac{3}{2} t^{6} - t \end{array}\right]}\)

The solution to the system of differential equations is:
\(\boxed{\left[\begin{array}{c} -\frac{1}{2} t^{2} + \frac{3}{2} t^{6} - t \\ \frac{3}{2} t^{2} + \frac{3}{2} t^{6} - t \end{array}\right]}\)

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