Variation of Parameters Example

As confusing as this formula looks its actually pretty straightforward. For example take this:

The complementary solution is very normal and should be the exact same formula as complementary & particular below:

The particular solution is the stuff on the other end of the addition sign. In this example

Then you just add them to the solution:

Complementary & Particular Solutions Example

Complementary & Particular solutions are reasonably straightforward to find - consisting of just finding the eigenpair and then solving a simple system:

1. Take this example $\frac{d}{dx}\vec{y}=\left[\begin{array}{cc}2& 4\\ 4& 2\end{array}\right]\vec{y}-\left[\begin{array}{c}26\\ 22\end{array}\right]$
2. Find a solution that looks like $C_1{e}^{{\lambda }_{1}x}\left[\begin{array}{c}\overrightarrow{V_1}\end{array}\right]+C_2{e}^{{\lambda }_{2}x}\left[\begin{array}{c}\overrightarrow{V_2}\end{array}\right]$ by paying attention to only the first $2\times 2$ matrix. This just means finding the eigenvalues and then eigenvectors & then plugging them into the formula above.
3. Find the particular by setting the $2\times 2=0$ and add in the given particular section:
   1. $\left[\begin{array}{c}0\\ 0\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 4& 2\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]-\left[\begin{array}{c}26\\ 22\end{array}\right]$
   2. This solves to $a_1=3$ and $a_2=5$.
4. Then just add that back to the original and and plug everything in:
   1. $y(x)=C_1{e}^{{\lambda }_{1}x}\left[\begin{array}{c}\overrightarrow{V_1}\end{array}\right]+C_2{e}^{{\lambda }_{2}x}\left[\begin{array}{c}\overrightarrow{V_2}\end{array}\right]+\left[\begin{array}{c}3\\ 5\end{array}\right]$
   2. Solution: $y(x)=C_1{e}^{6x}\left[\begin{array}{c}1\\ 1\end{array}\right]+C_2{e}^{-2x}\left[\begin{array}{c}-1\\ 1\end{array}\right]+\left[\begin{array}{c}3\\ 5\end{array}\right]$

An Imaginary Root Gives Two Real-Valued Solutions

Take eigenvalue ${\lambda }_2=2+3i$ with an eigenvector $\left[\begin{array}{c}-5+3i\\ 3+3i\\ 2\end{array}\right]$.

The eigenvalue can be turned into $e^{(2+3i)x}$ which can be decomposed into $e^{2x}{e}^{3ix}$ which using Euler's formula can be turned into $e^{2x}(\cos (3x)+i\sin (3x))$. You then can foil the decomposed eigenvalue with the the values in the eigenvector. I will show the first one as an example:

Or as the top row of the vector:

If the eigenvalues are purely imaginary there might be no $e^x$ terms.

Stability

Suppose P is a real invertible $2\times 2$ matrix. Then $\frac{d}{dx}\vec{y}=P\vec{y}$ has a unique equilibrium point $\vec{y}e=\vec{0}$ which is:

1. Asymptotically stable if all the eigenvalues of P have negative real parts
2. Non-asymptotically stable if the eigenvalues of P. Are purely imaginary
3. Unstable if at least one of the eigenvalues has positive real part

Defective Matrix

1. This is a $2\times 2$ matrix example. First solve for the eigenvalues - if you get only one eigenvalue with algebraic multiplicity 2
2. Find the corresponding eigenvector
3. Solve this: $\left[\begin{array}{cc}{a}_1-\lambda & b_1\\ a_2& b_2-\lambda \end{array}\right]=\left[\begin{array}{c}\text{The eigenvector it corresponds to}\end{array}\right]$
4. Lets call the solution to that $\vec{c}$
5. Then the general solution is $\vec{y}(x)=C_1{e}^{\lambda x}\left[\begin{array}{c}\overrightarrow{\text{Original Eigenvector}}\end{array}\right]+C_2(x{e}^{\lambda x}\left[\begin{array}{c}\overrightarrow{\text{Original Eigenvector}}\end{array}\right]+e^{\lambda x}\left[\begin{array}{c}\vec{c}\end{array}\right])$

Linearization

Very simple: Just the Jacobian matrix $\times$ some linearization factor: