Variation of Parameters for ODEs

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Method of Variation of Parameters for Systems of ODEs Used to find a particular solution $\mathbf{x}_p(t)$ for non-homogeneous linear systems of ODEs. 1. System of First-Order Linear ODEs General Form $\mathbf{x}'(t) = \mathbf{A}(t)\mathbf{x}(t) + \mathbf{f}(t)$ where $\mathbf{x}(t)$ is an $n \times 1$ vector of unknown functions, $\mathbf{A}(t)$ is an $n \times n$ matrix of coefficients, and $\mathbf{f}(t)$ is an $n \times 1$ vector of forcing functions. Steps Find the homogeneous solution $\mathbf{x}_h(t)$: Solve the homogeneous system $\mathbf{x}'(t) = \mathbf{A}(t)\mathbf{x}(t)$. This yields $n$ linearly independent solutions $\mathbf{x}^{(1)}(t), \mathbf{x}^{(2)}(t), \dots, \mathbf{x}^{(n)}(t)$. Form the fundamental matrix $\mathbf{\Psi}(t)$ whose columns are these solutions: $$\mathbf{\Psi}(t) = \begin{pmatrix} \mathbf{x}^{(1)}(t) & \mathbf{x}^{(2)}(t) & \dots & \mathbf{x}^{(n)}(t) \end{pmatrix}$$ The homogeneous solution is then $\mathbf{x}_h(t) = \mathbf{\Psi}(t)\mathbf{c}$, where $\mathbf{c}$ is a constant vector. Assume a particular solution form: Assume $\mathbf{x}_p(t) = \mathbf{\Psi}(t)\mathbf{u}(t)$, where $\mathbf{u}(t)$ is a vector of unknown functions. Substitute into the non-homogeneous equation: Differentiate $\mathbf{x}_p(t)$: $\mathbf{x}_p'(t) = \mathbf{\Psi}'(t)\mathbf{u}(t) + \mathbf{\Psi}(t)\mathbf{u}'(t)$. Substitute into $\mathbf{x}'(t) = \mathbf{A}(t)\mathbf{x}(t) + \mathbf{f}(t)$: $\mathbf{\Psi}'(t)\mathbf{u}(t) + \mathbf{\Psi}(t)\mathbf{u}'(t) = \mathbf{A}(t)\mathbf{\Psi}(t)\mathbf{u}(t) + \mathbf{f}(t)$ Since $\mathbf{\Psi}(t)$ is a fundamental matrix for the homogeneous system, we know $\mathbf{\Psi}'(t) = \mathbf{A}(t)\mathbf{\Psi}(t)$. Substituting this into the equation: $\mathbf{A}(t)\mathbf{\Psi}(t)\mathbf{u}(t) + \mathbf{\Psi}(t)\mathbf{u}'(t) = \mathbf{A}(t)\mathbf{\Psi}(t)\mathbf{u}(t) + \mathbf{f}(t)$ This simplifies to: $\mathbf{\Psi}(t)\mathbf{u}'(t) = \mathbf{f}(t)$ Solve for $\mathbf{u}'(t)$: Since $\mathbf{\Psi}(t)$ is a fundamental matrix, it is invertible. Multiply by $\mathbf{\Psi}^{-1}(t)$ from the left: $\mathbf{u}'(t) = \mathbf{\Psi}^{-1}(t)\mathbf{f}(t)$ Integrate $\mathbf{u}'(t)$ to find $\mathbf{u}(t)$: $\mathbf{u}(t) = \int \mathbf{\Psi}^{-1}(t)\mathbf{f}(t) dt$ (component-wise integration) No constants of integration are needed here, as they would lead to terms already included in the homogeneous solution. Form the particular solution $\mathbf{x}_p(t)$: $\mathbf{x}_p(t) = \mathbf{\Psi}(t)\mathbf{u}(t)$ Write the general solution: The general solution is the sum of the homogeneous and particular solutions: $$\mathbf{x}(t) = \mathbf{x}_h(t) + \mathbf{x}_p(t) = \mathbf{\Psi}(t)\mathbf{c} + \mathbf{\Psi}(t)\int \mathbf{\Psi}^{-1}(t)\mathbf{f}(t) dt$$ Example 1: Constant Coefficient Matrix Problem Solve $\mathbf{x}' = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} e^t \\ t \end{pmatrix}$. Solution Steps Homogeneous Solution: Eigenvalues of $\mathbf{A} = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix}$ are $\lambda_1 = 1, \lambda_2 = -1$. Corresponding eigenvectors are $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$. Homogeneous solutions: $\mathbf{x}^{(1)}(t) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}e^t$, $\mathbf{x}^{(2)}(t) = \begin{pmatrix} 1 \\ 3 \end{pmatrix}e^{-t}$. Fundamental matrix: $$\mathbf{\Psi}(t) = \begin{pmatrix} e^t & e^{-t} \\ e^t & 3e^{-t} \end{pmatrix}$$ Inverse Fundamental Matrix: $\det(\mathbf{\Psi}(t)) = e^t(3e^{-t}) - e^{-t}(e^t) = 3 - 1 = 2$. $$\mathbf{\Psi}^{-1}(t) = \frac{1}{2} \begin{pmatrix} 3e^{-t} & -e^{-t} \\ -e^t & e^t \end{pmatrix}$$ Calculate $\mathbf{u}'(t)$: $$\mathbf{u}'(t) = \mathbf{\Psi}^{-1}(t)\mathbf{f}(t) = \frac{1}{2} \begin{pmatrix} 3e^{-t} & -e^{-t} \\ -e^t & e^t \end{pmatrix} \begin{pmatrix} e^t \\ t \end{pmatrix}$$ $$\mathbf{u}'(t) = \frac{1}{2} \begin{pmatrix} 3e^{-t}e^t - te^{-t} \\ -e^te^t + te^t \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 3 - te^{-t} \\ -e^{2t} + te^t \end{pmatrix}$$ Integrate $\mathbf{u}'(t)$ to find $\mathbf{u}(t)$: $u_1(t) = \int \frac{1}{2}(3 - te^{-t}) dt = \frac{3}{2}t - \frac{1}{2}\int te^{-t} dt$ Using integration by parts ($\int u dv = uv - \int v du$ with $u=t, dv=e^{-t}dt$): $\int te^{-t} dt = -te^{-t} - \int -e^{-t} dt = -te^{-t} - e^{-t}$. $u_1(t) = \frac{3}{2}t - \frac{1}{2}(-te^{-t} - e^{-t}) = \frac{3}{2}t + \frac{1}{2}te^{-t} + \frac{1}{2}e^{-t}$. $u_2(t) = \int \frac{1}{2}(-e^{2t} + te^t) dt = -\frac{1}{4}e^{2t} + \frac{1}{2}\int te^t dt$ Using integration by parts ($\int u dv = uv - \int v du$ with $u=t, dv=e^t dt$): $\int te^t dt = te^t - \int e^t dt = te^t - e^t$. $u_2(t) = -\frac{1}{4}e^{2t} + \frac{1}{2}(te^t - e^t)$. So, $$\mathbf{u}(t) = \begin{pmatrix} \frac{3}{2}t + \frac{1}{2}te^{-t} + \frac{1}{2}e^{-t} \\ -\frac{1}{4}e^{2t} + \frac{1}{2}te^t - \frac{1}{2}e^t \end{pmatrix}$$ Particular Solution: $$\mathbf{x}_p(t) = \mathbf{\Psi}(t)\mathbf{u}(t) = \begin{pmatrix} e^t & e^{-t} \\ e^t & 3e^{-t} \end{pmatrix} \begin{pmatrix} \frac{3}{2}t + \frac{1}{2}te^{-t} + \frac{1}{2}e^{-t} \\ -\frac{1}{4}e^{2t} + \frac{1}{2}te^t - \frac{1}{2}e^t \end{pmatrix}$$ $x_{p1}(t) = e^t(\frac{3}{2}t + \frac{1}{2}te^{-t} + \frac{1}{2}e^{-t}) + e^{-t}(-\frac{1}{4}e^{2t} + \frac{1}{2}te^t - \frac{1}{2}e^t)$ $x_{p1}(t) = \frac{3}{2}te^t + \frac{1}{2}t + \frac{1}{2} - \frac{1}{4}e^t + \frac{1}{2}t - \frac{1}{2} = \frac{3}{2}te^t + t - \frac{1}{4}e^t$. $x_{p2}(t) = e^t(\frac{3}{2}t + \frac{1}{2}te^{-t} + \frac{1}{2}e^{-t}) + 3e^{-t}(-\frac{1}{4}e^{2t} + \frac{1}{2}te^t - \frac{1}{2}e^t)$ $x_{p2}(t) = \frac{3}{2}te^t + \frac{1}{2}t + \frac{1}{2} - \frac{3}{4}e^t + \frac{3}{2}t - \frac{3}{2} = \frac{3}{2}te^t + 2t - \frac{3}{4}e^t - 1$. So, $$\mathbf{x}_p(t) = \begin{pmatrix} \frac{3}{2}te^t + t - \frac{1}{4}e^t \\ \frac{3}{2}te^t + 2t - \frac{3}{4}e^t - 1 \end{pmatrix}$$ General Solution: $$\mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix}e^t + c_2 \begin{pmatrix} 1 \\ 3 \end{pmatrix}e^{-t} + \begin{pmatrix} \frac{3}{2}te^t + t - \frac{1}{4}e^t \\ \frac{3}{2}te^t + 2t - \frac{3}{4}e^t - 1 \end{pmatrix}$$