Variation of Parameters

Nonhomogenous Linear System

$${\vec{x}}^{\prime }=A(t)\vec{x}+\vec{f}(t)$$

Solution Form

$$\vec{x}(t)={\vec{x}}_p(t)+{\vec{x}}_c(t)$$

where ${\vec{x}}_p$ is a particular solution to the original (nonhomogenous system) where ${\vec{x}}_c$ are general solutions to the homogenous system of ${\vec{x}}_c^{\prime }=A(t){\vec{x}}_c$

Suppose that the solutions of the homogenous system are

$${\vec{x}}_c(t)=C_1{\vec{x}}_1(t)+\cdots +C_n{\vec{x}}_n(t)$$

The following matrix is called a fundamental matrix

$$M(t)=[{\vec{x}}_1(t)\dots {\vec{x}}_n(t)]$$

Then

$${\vec{x}}_c(t)=M(t)\vec{C}$$

Let $\vec{x}(t)=M(t)\vec{u}(t)$ Then ${\vec{u}}^{\prime }=M(t{)}^{-1}\vec{f}(t)$ Integrate both sides to get ${\vec{u}}_p=\int M(t{)}^{-1}\vec{f}(t)dt$ Then ${\vec{x}}_p=M(t){\vec{u}}_p(t)$ is a particular solution

Example

$${\vec{x}}^{\prime }=\left[\begin{array}{cc}1& 1\\ 4& 1\end{array}\right]\vec{x}+\left[\begin{array}{c}-1+21t+e^{3t}\tan (t)\\ 2+30t+2e^{3t}\tan (t)\end{array}\right]$$

Find ${\vec{x}}_c$ first, start with eigenstuff.

$${\vec{x}}_c(t)=C_1{e}^{-t}\left[\begin{array}{c}-1\\ 2\end{array}\right]+C_2{e}^{3t}\left[\begin{array}{c}1\\ 2\end{array}\right]$$

Find fundamental matrix

$$M(t)=\left[\begin{array}{cc}{\vec{x}}_1(t)& {\vec{x}}_2(t)\end{array}\right]=\left[\begin{array}{cc}-e^{-t}& e^{3t}\\ 2e^{-t}& 2e^{3t}\end{array}\right]$$

Set $\vec{x}(t)=M(t)\vec{u}(t)$, which implies ${\vec{u}}^{\prime }=M(t{)}^{-1}\vec{f}(t)$ Evaluate ${\vec{u}}^{\prime }$

$${\vec{u}}^{\prime }(t)={\left[\begin{array}{cc}-e^{-t}& e^{3t}\\ 2e^{-t}& 2e^{3t}\end{array}\right]}^{-1}\left[\begin{array}{c}-1+21t+e^{3t}\tan (t)\\ 2+30t+2e^{3t}\tan (t)\end{array}\right]$$

Recall ![[Linear Algebra#2-times-2-inverse-shortcut|$2times2$ Inverse Shortcut]]

$${\vec{u}}^{\prime }(t)=\frac{1}{-4e^{2t}}\left[\begin{array}{cc}2e^{3t}& -e^{3t}\\ -2e^{-t}& -e^{-t}\end{array}\right]\left[\begin{array}{c}-1+21t+e^{3t}\tan (t)\\ 2+30t+2e^{3t}\tan (t)\end{array}\right]$$ $${\vec{u}}^{\prime }(t)=\left[\begin{array}{c}(1-3t)e^t\\ 18t{e}^{-3t}+\tan (t)\end{array}\right]$$

Integrate ${\vec{u}}^{\prime }$:

$$\vec{u}(t)=\left[\begin{array}{c}(4-3t)e^t+C_1\\ (-2-6t)e^{-3t}-\ln (\cos (t))+C_2\end{array}\right]$$

Acquire ${\vec{x}}_p(t)$

$${\vec{x}}_p(t)=\left[\begin{array}{cc}-e^{-t}& e^{3t}\\ 2e^{-t}& 2e^{3t}\end{array}\right]\left[\begin{array}{c}(4-3t)e^t+C_1\\ (-2-6t)e^{-3t}-\ln (\cos (t))+C_2\end{array}\right]=\left[\begin{array}{c}-6-3t-e^{3t}\ln (\cos (t))\\ 4-18t-2e^{3t}\ln (\cos (t))\end{array}\right]$$

Final answer

$$\vec{x}(t)=C_1{e}^{-t}\left[\begin{array}{c}-1\\ 2\end{array}\right]+C_2{e}^{3t}\left[\begin{array}{c}1\\ 2\end{array}\right]+\left[\begin{array}{c}-6-3t-e^{3t}\ln (\cos (t))\\ 4-18t-2e^{3t}\ln (\cos (t))\end{array}\right]$$