Let's solve the initial value problem: \[ \frac{d\mathbf{x}}{dt} = A\mathbf{x}, \quad \mathbf{x}(0) = \mathbf{x}_0 \] where \[ A = \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}, \quad \mathbf{x}_0 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] We will do this in a step-by-step way: --- ### **Step 1: Find Eigenvalues and Eigenvectors of \(A\)** The characteristic equation is: \[ \det(A - \lambda I) = 0 \] \[ A - \lambda I = \begin{bmatrix} 4-\lambda & -2 \\ 3 & 1-\lambda \end{bmatrix} \] \[ \det(A - \lambda I) = (4-\lambda)(1-\lambda) - (-2)(3) = (4-\lambda)(1-\lambda) + 6 \] \[ = 4 - 4\lambda - \lambda + \lambda^2 + 6 = \lambda^2 - 5\lambda + 10 \] So, \[ \lambda^2 - 5\lambda + 10 = 0 \] Solving the quadratic: \[ \lambda = \frac{5 \pm \sqrt{25 - 40}}{2} = \frac{5 \pm i\sqrt{15}}{2} \] --- ### **Step 2: Write the General Solution** For a system with complex eigenvalues \(\lambda = \alpha \pm i\beta\), the general solution is: \[ \mathbf{x}(t) = c_1 \Re(\mathbf{v} e^{\lambda_1 t}) + c_2 \Im(\mathbf{v} e^{\lambda_1 t}) \] where \(\lambda_1 = \frac{5}{2} + i\frac{\sqrt{15}}{2}\) and \(\mathbf{v}\) is the corresponding eigenvector. Let's find \(\mathbf{v}\): Solve \((A - \lambda_1 I)\mathbf{v} = 0\): \[ A - \lambda_1 I = \begin{bmatrix} 4-\lambda_1 & -2 \\ 3 & 1-\lambda_1 \end{bmatrix} \] Let’s pick the first row: \[ 4 - \lambda_1 = 4 - \left(\frac{5}{2} + i\frac{\sqrt{15}}{2}\right) = \frac{3}{2} - i\frac{\sqrt{15}}{2} \] So, \[ \begin{bmatrix} \frac{3}{2} - i\frac{\sqrt{15}}{2} & -2 \\ 3 & -\frac{3}{2} - i\frac{\sqrt{15}}{2} \end{bmatrix} \] Set up the first row: \[ \left(\frac{3}{2} - i\frac{\sqrt{15}}{2}\right)v_1 - 2v_2 = 0 \] \[ v_1 = \frac{2}{\frac{3}{2} - i\frac{\sqrt{15}}{2}} v_2 = \frac{4}{3 - i\sqrt{15}} v_2 \] Multiply numerator and denominator by the conjugate: \[ \frac{4}{3 - i\sqrt{15}} \cdot \frac{3 + i\sqrt{15}}{3 + i\sqrt{15}} = \frac{4(3 + i\sqrt{15})}{9 + 15} = \frac{4(3 + i\sqrt{15})}{24} = \frac{1}{6}(3 + i\sqrt{15}) \] So, we can take \(v_2 = 1\), \(v_1 = \frac{1}{6}(3 + i\sqrt{15})\). So, \[ \mathbf{v} = \begin{bmatrix} \frac{1}{6}(3 + i\sqrt{15}) \\ 1 \end{bmatrix} \] --- ### **Step 3: Write Solution in Real Form** Let \(\lambda = \alpha + i\beta = \frac{5}{2} + i\frac{\sqrt{15}}{2}\), \(\mathbf{v} = \mathbf{p} + i\mathbf{q}\), where: - \(\mathbf{p} = \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix}\) - \(\mathbf{q} = \begin{bmatrix} \frac{\sqrt{15}}{6} \\ 0 \end{bmatrix}\) The general solution is: \[ \mathbf{x}(t) = e^{\alpha t}[c_1 (\mathbf{p} \cos{\beta t} - \mathbf{q} \sin{\beta t}) + c_2 (\mathbf{p} \sin{\beta t} + \mathbf{q} \cos{\beta t})] \] But more directly: \[ \mathbf{x}(t) = e^{\alpha t} [A \cos{\beta t} + B \sin{\beta t}] \] where \(A, B\) are determined by the initial conditions. --- ### **Step 4: Apply Initial Condition** Let’s write the real-valued fundamental solutions: Let \[ \mathbf{x}_1(t) = e^{\alpha t}(\mathbf{p} \cos{\beta t} - \mathbf{q} \sin{\beta t}) \] \[ \mathbf{x}_2(t) = e^{\alpha t}(\mathbf{p} \sin{\beta t} + \mathbf{q} \cos{\beta t}) \] So, \[ \mathbf{x}(t) = c_1 \mathbf{x}_1(t) + c_2 \mathbf{x}_2(t) \] At \(t=0\), \[ \mathbf{x}(0) = c_1 \mathbf{p} + c_2 \mathbf{q} = \mathbf{x}_0 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] So, \[ c_1 \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} \frac{\sqrt{15}}{6} \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] This gives: - First row: \( \frac{1}{2}c_1 + \frac{\sqrt{15}}{6}c_2 = 0 \) - Second row: \( c_1 = 1 \) So \(c_1 = 1\), and \[ \frac{1}{2} + \frac{\sqrt{15}}{6}c_2 = 0 \implies c_2 = -\frac{3}{\sqrt{15}} \] --- ### **Step 5: Write Final Solution** So, \[ \mathbf{x}(t) = \mathbf{x}_1(t) - \frac{3}{\sqrt{15}} \mathbf{x}_2(t) \] Write explicitly: \[ \mathbf{x}_1(t) = e^{\frac{5}{2}t} \left[\begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} \cos\left(\frac{\sqrt{15}}{2}t\right) - \begin{bmatrix} \frac{\sqrt{15}}{6} \\ 0 \end{bmatrix} \sin\left(\frac{\sqrt{15}}{2}t\right) \right] \] \[ \mathbf{x}_2(t) = e^{\frac{5}{2}t} \left[\begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} \sin\left(\frac{\sqrt{15}}{2}t\right) + \begin{bmatrix} \frac{\sqrt{15}}{6} \\ 0 \end{bmatrix} \cos\left(\frac{\sqrt{15}}{2}t\right) \right] \] Therefore, \[ \boxed{ \mathbf{x}(t) = e^{\frac{5}{2}t} \left( \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} \cos\left(\frac{\sqrt{15}}{2}t\right) - \begin{bmatrix} \frac{\sqrt{15}}{6} \\ 0 \end{bmatrix} \sin\left(\frac{\sqrt{15}}{2}t\right) - \frac{3}{\sqrt{15}} \left[ \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} \sin\left(\frac{\sqrt{15}}{2}t\right) + \begin{bmatrix} \frac{\sqrt{15}}{6} \\ 0 \end{bmatrix} \cos\left(\frac{\sqrt{15}}{2}t\right) \right] \right) } \] You can simplify further if desired, but this is the **complete step-by-step solution**. If you want the components written out explicitly, just ask!

Let's carefully solve the initial value problem \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\) with \(\mathbf{x}(0) = \mathbf{x}_0\), where \[ A = \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}, \quad \mathbf{x}_0 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] --- ### **Step 1: Find Eigenvalues of \(A\)** The characteristic equation is: \[ \det(A - \lambda I) = 0 \] Compute: \[ A - \lambda I = \begin{bmatrix} 4 - \lambda & -2 \\ 3 & 1 - \lambda \end{bmatrix} \] Determinant: \[ (4 - \lambda)(1 - \lambda) - (-2)(3) = (4 - \lambda)(1 - \lambda) + 6 \] Expand: \[ (4 - \lambda)(1 - \lambda) = 4 \times 1 - 4 \lambda - \lambda \times 1 + \lambda^2 = 4 - 4\lambda - \lambda + \lambda^2 = 4 - 5 \lambda + \lambda^2 \] Add 6: \[ 4 - 5 \lambda + \lambda^2 + 6 = \lambda^2 - 5 \lambda + 10 \] Set equal to zero: \[ \lambda^2 - 5 \lambda + 10 = 0 \] Solve quadratic: \[ \lambda = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 1 \times 10}}{2} = \frac{5 \pm \sqrt{25 - 40}}{2} = \frac{5 \pm \sqrt{-15}}{2} \] Since discriminant is negative, eigenvalues are complex: \[ \boxed{ \lambda = \frac{5}{2} \pm i \frac{\sqrt{15}}{2} } \] --- ### **Step 2: Find Eigenvectors** Choose \(\lambda = \frac{5}{2} + i \frac{\sqrt{15}}{2}\). Set up: \[ (A - \lambda I)\mathbf{v} = 0 \] Compute \(A - \lambda I\): \[ A - \lambda I = \begin{bmatrix} 4 - \lambda & -2 \\ 3 & 1 - \lambda \end{bmatrix} \] Calculate each element: \[ 4 - \lambda = 4 - \left(\frac{5}{2} + i \frac{\sqrt{15}}{2}\right) = \frac{8}{2} - \frac{5}{2} - i \frac{\sqrt{15}}{2} = \frac{3}{2} - i \frac{\sqrt{15}}{2} \] \[ 1 - \lambda = 1 - \left(\frac{5}{2} + i \frac{\sqrt{15}}{2}\right) = \frac{2}{2} - \frac{5}{2} - i \frac{\sqrt{15}}{2} = -\frac{3}{2} - i \frac{\sqrt{15}}{2} \] So, \[ A - \lambda I = \begin{bmatrix} \frac{3}{2} - i \frac{\sqrt{15}}{2} & -2 \\ 3 & -\frac{3}{2} - i \frac{\sqrt{15}}{2} \end{bmatrix} \] Focus on the first row: \[ \left(\frac{3}{2} - i \frac{\sqrt{15}}{2}\right) v_1 - 2 v_2 = 0 \] Express \(v_1\) in terms of \(v_2\): \[ v_1 = \frac{2 v_2}{\frac{3}{2} - i \frac{\sqrt{15}}{2}} \] Multiply numerator and denominator by the conjugate \(\frac{3}{2} + i \frac{\sqrt{15}}{2}\): \[ v_1 = 2 v_2 \times \frac{\frac{3}{2} + i \frac{\sqrt{15}}{2}}{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{15}}{2}\right)^2} \] Calculate denominator: \[ \left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{15}}{2}\right)^2 = \frac{9}{4} + \frac{15}{4} = \frac{24}{4} = 6 \] Therefore: \[ v_1 = 2 v_2 \times \frac{\frac{3}{2} + i \frac{\sqrt{15}}{2}}{6} = \frac{v_2}{3} \left( 3 + i \sqrt{15} \right) \] Choose \(v_2 = 1\): \[ v_1 = \frac{1}{3} (3 + i \sqrt{15}) = 1 + i \frac{\sqrt{15}}{3} \] Thus, an eigenvector corresponding to \(\lambda = \frac{5}{2} + i \frac{\sqrt{15}}{2}\): \[ \boxed{ \mathbf{v} = \begin{bmatrix} 1 + i \frac{\sqrt{15}}{3} \\ 1 \end{bmatrix} } \] --- ### **Step 3: Write the General Real Solution** Since the eigenvalues are complex conjugates, the real solution is expressed as: \[ \mathbf{x}(t) = e^{\alpha t} \left[ C_1 \cos{\beta t} \ \mathbf{Re}(\mathbf{v}) - C_2 \sin{\beta t} \ \mathbf{Im}(\mathbf{v}) \right] \] where: \[ \alpha = \frac{5}{2}, \quad \beta = \frac{\sqrt{15}}{2} \] and \[ \mathbf{v} = \mathbf{p} + i \mathbf{q} \] with \[ \mathbf{p} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] and \[ \mathbf{q} = \begin{bmatrix} \frac{\sqrt{15}}{3} \\ 0 \end{bmatrix} \] --- ### **Step 4: Find Constants \(C_1, C_2\) from Initial Conditions** At \(t=0\): \[ \mathbf{x}(0) = C_1 \mathbf{p} + C_2 \mathbf{q} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] which gives a system: \[ \begin{cases} C_1 \times 1 + C_2 \times \frac{\sqrt{15}}{3} = 0 \\ C_1 \times 1 + C_2 \times 0 = 1 \end{cases} \] From the second: \[ C_1 = 1 \] Plug into the first: \[ 1 + \frac{\sqrt{15}}{3} C_2 = 0 \Rightarrow C_2 = - \frac{3}{\sqrt{15}} \] --- ### **Step 5: Write the Final Solution** The solution: \[ \boxed{ \mathbf{x}(t) = e^{\frac{5}{2} t} \left[ \cos{\left(\frac{\sqrt{15}}{2} t\right)} \cdot \mathbf{p} - \sin{\left(\frac{\sqrt{15}}{2} t\right)} \cdot \mathbf{q} \right] } \] with \[ \mathbf{p} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \mathbf{q} = \begin{bmatrix} \frac{\sqrt{15}}{3} \\ 0 \end{bmatrix} \] and the constants: \[ C_1 = 1, \quad C_2 = - \frac{3}{\sqrt{15}} \] Expressed explicitly: \[ \boxed{ \mathbf{x}(t) = e^{\frac{5}{2} t} \left[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \cos{\left(\frac{\sqrt{15}}{2} t\right)} - \frac{3}{\sqrt{15}} \begin{bmatrix} \frac{\sqrt{15}}{3} \\ 0 \end{bmatrix} \sin{\left(\frac{\sqrt{15}}{2} t\right)} \right] } \] Simplify the second term: \[ \frac{3}{\sqrt{15}} \times \frac{\sqrt{15}}{3} = 1 \] Thus, \[ \mathbf{x}(t) = e^{\frac{5}{2} t} \left[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \cos{\left(\frac{\sqrt{15}}{2} t\right)} - \begin{bmatrix} 1 \\ 0 \end{bmatrix} \sin{\left(\frac{\sqrt{15}}{2} t\right)} \right] \] --- ### **Final Answer:** \[ \boxed{ \boxed{ \mathbf{x}(t) = e^{\frac{5}{2} t} \left[ \begin{bmatrix} \cos{\left(\frac{\sqrt{15}}{2} t\right)} - \sin{\left(\frac{\sqrt{15}}{2} t\right)} \\ \cos{\left(\frac{\sqrt{15}}{2} t\right)} \right] \right] } \] This is the explicit solution satisfying the initial condition \(\mathbf{x}(0) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\). --- **Note:** This detailed derivation involves complex eigenvalues, eigenvectors, and converting into real-valued solutions using sine and cosine functions.

The solution to the initial value problem \(\frac{d\mathbf{x}}{dt} = A \mathbf{x}\), with \(\mathbf{x}(0) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\), where \[ A = \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix} \] is: \[ \boxed{ \mathbf{x}(t) = e^{\frac{5}{2} t} \left[ \begin{bmatrix} \cos{\left(\frac{\sqrt{15}}{2} t\right)} - \sin{\left(\frac{\sqrt{15}}{2} t\right)} \\[6pt] \cos{\left(\frac{\sqrt{15}}{2} t\right)} \end{bmatrix} \right] } \]