System of Differential Equations

#math

Preface

Consider $$y'' + p(x)y' + q(x)y = 0$$Consider a change in variable as $y_1=y$ and $y_2=y^{\prime }$
Thus $y_1^{\prime }=y_2$ and $y_2^{\prime }=-(q(x)y_1+p(x)y_2)$.
Thus we can reduce this to a system of linear equations represented by $$\begin{pmatrix} y_{1}'\ y_{2}'\end{pmatrix} = \begin{bmatrix} 0 & 1\ -q(x) & -p(x)\end{bmatrix}\cdot \begin{pmatrix}y_{1}\y_{2}\end{pmatrix}$$$$\implies Y' = MY$$
Generalizing the coefficients, $$\begin{pmatrix}y_{1}'\y_{2}'\end{pmatrix} = \begin{bmatrix}f_{1}(x, y_{1}, y_{2})\f_{2}(x,y_{1},y_{2})\end{bmatrix} \implies Y' = F(Y)$$
This can be generalized to a n-th order differential equation, where $$Y = [y_{1}\ \ y_{2}\ \ y_{3}\ \ \dots\ \ y_{n}]^T$$

Constant $M$ matrix

Suppose the equation was such that $$\begin{align}Y' &= \begin{bmatrix}a & b\c&d\end{bmatrix}Y \tag{1}\ \ \mathbf{y'} &= A\mathbf{y}\end{align}$$
This implies a case similar to constant coefficients, where can take $e^{\lambda x}$ as trial solution.
So let $$\mathbf{y} = \mathbf{v}e^{\lambda x} \tag{2}$$be a solution, where $\mathbf{v}$ is an arbitrary vector of appropriate dimensions
Substituting in $(1)$, we get $$\begin{align}\lambda \mathbf{v}e^{\lambda x} &= \begin{bmatrix}a&b\c&d\end{bmatrix}\mathbf{v}e^{\lambda x}\ \ A\mathbf{v} &= \lambda \mathbf{v} \end{align}$$
We get something akin to eigen-values and eigen-vectors, $$\lambda^2 - (a+d)\lambda + (ad-bc) = 0$$a quadratic equation with two roots, signalling towards two distinct variations of $(2)$.
In case of an n-th order differential ofc, we will get an n-th degree polynomial with a maximum of $n$ different eigen-values

Thus, the general solution can be written as $$\begin{pmatrix}y_{1}\y_{2}\end{pmatrix} = C_{1}\mathbf{v}{1}e^{\lambdax} + C_{2}\mathbf{v}{2}e^{\lambdax}$$where ${\mathbf{v}}_1,{\mathbf{v}}_2$ are eigen-vectors corresponding to ${\lambda }_1,{\lambda }_2$ respectively.
The above works well for distinct roots.

Roots as complex conjugates

When the roots are complex conjugates, we obtain two eigen values $\alpha \pm i\beta$. For one of the eigen-values, let corresponding eigen vector $\mathbf{v}=\mathbf{a}+i\mathbf{b}$. Thus the complex solution corresponding to this, $$\mathbf{y} = \mathbf{v}e^{(\alpha + i\beta)x} \implies (\mathbf{a} + i\mathbf{b})e^{\alpha x}(\cos(\beta x) + i\sin(\beta x))$$
Distributing, $$\mathbf{y} = e^{\alpha x}[(\mathbf{a}\cos(\beta x) - \mathbf{b}\sin(\beta x)) + i(\mathbf{a}\sin(\beta x) + \mathbf{b}\cos(\beta x))]$$
Since this solution $\mathbf{y}$ satisfies a linear differential equation, $\mathrm{Re}(\mathbf{y})$ and $\mathrm{Im}(\mathbf{y})$ must be the LI solutions of $(1)$

Notice we used one eigen value and one eigen vector to find a basis for solution set. Using the other one would also give you the same solution set, albeit with a different basis.

Repeated Roots

When an eigen value has multiplicity $m>1$ but provides only one eigen-vector, $\mathbf{v}$, we guess that we need another term and that the solution is of the form $$\mathbf{y}(x) = \mathbf{v}xe^{\lambda x} + \mathbf{w}e^{\lambda x} $$where $\mathbf{w}$ is a generalized eigen vector to be determined.
Omitting the proof, we obtain $\mathbf{w}$ to be that which satisfies $$(A-\lambda I)\mathbf{w} = \mathbf{v}$$for the same eigen value $\lambda$ which gave us $\mathbf{v}$

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