General Solution of a DE in Matrix Form - example

Determine the Real General Solution

We are given the system of differential equations:

$${\mathbf{q}}^{\prime }=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\mathbf{q}+\left[\begin{array}{c}0\\ 1\end{array}\right]$$

Step 1: Solve the Homogeneous System

The homogeneous part is:

$${\mathbf{q}}^{\prime }=A\mathbf{q},\text{where }A=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$$

Finding Eigenvalues:

The characteristic equation is:

$$\det (A-\lambda I)=\left|\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right|=(-\lambda )(-\lambda )-(1)(-4)={\lambda }^2+4=0$$

Solving for $\lambda$:

$${\lambda }_{1,2}=\pm 2i$$

Finding Eigenvectors:

For $\lambda =2i$:

$$(A-2iI)\mathbf{v}=0$$ $$\left[\begin{array}{cc}-2i& 1\\ -4& -2i\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$

Solving:

$$-2i{v}_1+v_2=0\Rightarrow v_2=2i{v}_1$$

Choosing $v_1=1$, we get:

$${\mathbf{v}}_1=\left[\begin{array}{c}1\\ 2i\end{array}\right]$$

Note: we have flexibility in choosing $v_1$ because eigenvectors are not unique - they can be scaled by any non-zero constant and still remain eigenvectors.

For $\lambda =-2i$, the eigenvector is:

$${\mathbf{v}}_2=\left[\begin{array}{c}1\\ -2i\end{array}\right]$$

Constructing the General Homogeneous Solution:

Using Euler’s formula $e^{2it}=\cos (2t)+i\sin (2t)$, we get:

$${\mathbf{q}}_h(t)=c_1\left[\begin{array}{c}\cos (2t)\\ -2\sin (2t)\end{array}\right]+c_2\left[\begin{array}{c}\sin (2t)\\ 2\cos (2t)\end{array}\right]$$

where $c_1,c_2$ are real constants.

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Step 2: Find a Particular Solution

Since the inhomogeneous term is constant, assume a constant solution:

$${\mathbf{q}}_p=\left[\begin{array}{c}a\\ b\end{array}\right]$$

Substituting into the equation:

$$\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]+\left[\begin{array}{c}0\\ 1\end{array}\right]=0$$

Note: If $\mathbf{q}$ is constant ${\mathbf{q}}^{\prime }$ is $0$

This gives:

$$\{\begin{array}{l}a+0=0\Rightarrow a=0\\ -4a+0+1=0\Rightarrow b=-\frac{1}{4}\end{array}$$

Thus, the particular solution is:

$${\mathbf{q}}_p=\left[\begin{array}{c}0\\ -\frac{1}{4}\end{array}\right]$$

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Step 3: General Solution

$$\mathbf{q}(t)=c_1\left[\begin{array}{c}\cos (2t)\\ -2\sin (2t)\end{array}\right]+c_2\left[\begin{array}{c}\sin (2t)\\ 2\cos (2t)\end{array}\right]+\left[\begin{array}{c}0\\ -\frac{1}{4}\end{array}\right]$$

where $c_1,c_2$ are arbitrary real constants.

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