# Lineair Differential Equations

$$\require{AMSsymbols} \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\lfzraised#1{\raise{10mu}{#1}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \def\crw{\lfz{}} \require{cancel} \newcommand\ccancel[black] {\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[black] {\color{#1}{\cancelto{#2}{\color{black}{#3}}}}$$For a lineair non-homogeneous differential equation with constant coefficients $$a_1\ldots a_n$$ in the form
\begin{align} \frac{\text{d}^nf(t)}{\text{d}t^n}+ a_1\frac{\text{d}^{n-1}f(t)}{\text{d}t^{n-1}}+ \cdots+ a_{n-1}\frac{\text{d}f(t)}{\text{d}t}+ a_nf(t)&=g(t)\nonumber\\[10mu] \overset{abbrev}{\Rightarrow}\quad f^{(n)}(t)+a_1f^{n-1}(t)+\cdots+a_{n-1}f'(t)+a_nf(t)&=g(t)\nonumber \end{align}\label{eq:bDV} the solution is the superposition of the natural response and the forced response of the system. In math speak, these are called the homogeneous solution $$f_h(t)$$ and the particular solution $$f_p(t)$$
$$\shaded{f(t)=f_h(t)+f_p(t)}$$

To solve the lineair non-homogeneous differential equation, we

1. set the force $$g(t)=0$$ and solve the natural response $$f_h(t)$$,
2. set the initial conditions $$f(0)=f^\prime(0)=f^{\prime\prime}(0)=\ldots=0$$ and solve the forced response $$f_p(t)$$,
3. sum the forced response to the natural response to get the total response,
4. use the initial conditions to resolve any constants.

## Natural (homogeneous solution)

The natural response, $$f_h(t)$$, is the behavior of a circuit due to initial conditions, but without input force. We suppress the input force $$g(t)=0$$ and solve just the circuit itself. This makes the non-homogeneous differential equation $$\eqref{eq:bDV}$$ into a homegeneous differential equation.
$${f_h}^{(n)}(t)+a_1{f_h}^{(n-1)}(t)+\cdots+a_{n-1}{f_h}^\prime(t)+a_nf_h(t)=0\label{eq:bDVh}$$

Leonhard Euler, in E10 (1728) and E62 (1739), realized that general homogeneous solutions have the form
$$\shaded{f_{h,i}=e^{pt}}$$ where $$p\in \mathbb{C}$$. Substituting $$f_h(t)=\mathrm{e}^{pt}$$ in homegeneous differential equation $$\eqref{eq:bDVh}$$ gives the so-called characteristic equation
\begin{align} p^n\mathrm{e}^{pt}+ a_1 p^{n-1}\mathrm{e}^{pt} + \cdots + a_n\mathrm{e}^{pt}&=0,&\div \mathrm{e}^{pt}\nonumber\\ \Rightarrow\quad p^n+ a_1 p^{n-1} + \cdots + a_n&=0\\ \end{align}

Substituting any of the roots of the polynomial $$p_1,p_2,\ldots p_n$$ in $$\mathrm{e}^{pt}$$ results in a solution base $$f_i(t)=\mathrm{e}^{p_it}$$.

Given that homogeneous linear differential equations obey the superposition principle, any linear combination of these functions also satisfies the differential equation. Therefore, combining the $$n$$ lineair independent solutions $$f_1(t), f_2(t),\ldots,f_n(t)$$, leads to the homogeneous solution with real arbitrary constants $$c_1,c_2,\ldots,c_n$$
$$\shaded{f_h(t)=c_1f_1(t)+c_2f_2(t)+\cdots+c_nf_n(t)}$$

Two notes:

1. For double roots, or in general when a root $$p_i$$ has multiplicity $$m$$, the solution base is $$f(t)=t^{k-1}\,\mathrm{e}^{p_it}$$ where $$k\in {0,1,\ldots,m-1}$$.
2. If the differential equation $$\eqref{eq:bDV}$$ has real coefficients $$a_i$$, complex solutions for $$p$$ will only occur in complex conjugate pairs. The real-valued solutions are obtained by replacing each pair with their real-valued linear combinations $$\Re(f_1)$$ and $$\Im(f_2)$$, as in
\begin{align} \Re(f_1)=\frac{f_1+f_2}{2}\\ \Im(f_1)=\frac{f_1-f_2}{2j} \end{align} Applying Euler’s trigonometry identities, these will solutions will turn into $$\cos$$ and $$\sin$$ terms.

The example below shows how an homogeneous lineair differential equation is solved.

### Example

Assume an homogeneous lineair differential equation
$$\DeclareMathOperator*{\dprime}{\prime\prime} \DeclareMathOperator*{\tprime}{\prime\prime\prime} \DeclareMathOperator*{\qprime}{\prime\prime\prime\prime} f^{\qprime}(t)-2f^{\tprime}(t)+2f^{\dprime}(t)-2f^{\prime}(t)+f(t)=0$$

The characteristic equation and its factorized form follow as
\begin{align} p^4-2p^3+2p^2-2p+1&=0\nonumber\\[6mu] \Rightarrow\quad (p-j)\,(p+j)\,(p-1)^2&=0 \end{align} the solution basis becomes \begin{align} f_1&=\mathrm{e}^{jt}&\text{based on }p_1=j\nonumber\\ f_2&=\mathrm{e}^{-jt}&\text{based on }p_2=-j\nonumber\\ f_3&=\mathrm{e}^{t}&\text{based on }p_3=p_4=1\nonumber\\ f_4&=t\mathrm{e}^{t}&\text{based on }p_3=p_4=1\nonumber \end{align}

Using Euler’s trigonometry identities
\begin{align} \Re(f_1)&=\frac{f_1+f_2}{2}=\frac{\mathrm{e}^{jt}+\mathrm{e}^{-jt}}{2}=\cos(t)\nonumber\\ \Im(f_1)&=\frac{f_1-f_2}{2j}=\frac{\mathrm{e}^{jt}-\mathrm{e}^{-jt}}{2j}=\sin(t)\nonumber\\ \end{align} simplifies the solution basis to
\begin{align} f_1&=cos(t)&\text{based on }p_1={p_2}^*=j\nonumber\\ f_2&=sin(t)&\text{based on }p_1={p_2}^*=j\nonumber\\ f_3&=\mathrm{e}^{t}&\text{based on }p_3=p_4=1\nonumber\\ f_4&=t\,\mathrm{e}^{t}&\text{based on }p_3=p_4=1\nonumber \end{align}

The homogeneous solution follows as a linear combination
$$f_h(t)=c_1\cos (t)+c_2\sin(t)+c_3\,\mathrm{e}^{t}+c_4\,t\,\mathrm{e}^{t}$$ where the constants $$c_{1,\ldots,4}$$ follow from the initial conditions.

## Forced (particular solution)

The forced response $$f_p(t)$$, is the part of the response caused directly by the input force assuming all initial conditions are zero.
$${f_p}^{(n)}(t)+a_1{f_p}^{(n-1)}(t)+\cdots+a_{n-1}{f_p}^\prime(t)+a_nf_p(t)=g(t)\label{eq:bDVp}$$

The solution for the forced response is usually a scaled version of the input. In the examples below we will show two methods of finding the particular solution. As you will learn the using a complex forcing function is the easiest way of obtaining the particular solution. In all other cases, we find $$f_p(t)$$ by either the method of undetermined coefficients or the variation of parameters method. [link]

The particular solution is typically found using trigonometry identities, as shown in the examples in RC Low-pass Filter Appendix B, and RC Low-pass Filter Appendix B. Even for these linear first order linear systems this is a fairly painstaking process. Here we explain a less involved method to find the response to a sinusoid forcing function.

### Complex superposition

The superposition property:

When two signals are added together and forced on a linear system, the system response is the same as if one had forced each signal through the system separately and then added the responses.

The linearity of the system implies that if we use an input of the form $$=\hat{u}\cos(\omega t)$$ then the output will have the same frequency but with a different phase and amplitude. As shown in the table below, if we scale the input by a factor $$k$$ then the output will scaled by the same factor. This applies even when that factor is the imaginary number $$j$$.

Linear system
input output
$$\hat{u}\cos(\omega t+\theta)\nonumber$$ $$\longrightarrow\nonumber$$ $$A\hat{u}\cos(\omega t+\phi)\nonumber$$
$$\color{olive}{k}\hat{u}\cos(\omega t+\theta)\nonumber$$ $$\longrightarrow\nonumber$$ $$\color{olive}{k}A\hat{u}\cos(\omega t+\phi)\nonumber$$
$$\color{blue}{j}\hat{u}\cos(\omega t+\theta)\nonumber$$ $$\longrightarrow\nonumber$$ $$\color{blue}{j}A\hat{u}\cos(\omega t+\phi)\nonumber$$

By the superposition principle of linear systems, a forcing function of a summed $$\cos$$ and $$j\sin$$, will produce a scaled response of $$\cos$$ and $$j\sin$$
$$\hat{u}\cos(\omega t)+\color{green}{j}\,\hat{u}\sin(\omega t) \longrightarrow A\cos(\omega t+\phi)+\color{green}{j}A\sin(\omega t+\phi)$$

By applying Euler’s formula

$$\mathrm{e}^{j\varphi}=\cos\varphi+j\sin\varphi\nonumber$$

the complex input $$\underline{u}(t)$$ and output $$\underline{f}(t)$$ can be expressed as
$$\underline{u}(t)=\hat{u}\,\mathrm{e}^{j\omega t} \longrightarrow \underline{f}(t)=A\,\mathrm{e}^{j(\omega t+\phi)}$$

Even if the forcing function is only the real-part of $$\underline{u}$$, to derive the system response we may assume that it is the mathematically more convenient $$\underline{u}(t)$$ even though that also includes an imaginary part, for as long as we ignore the imaginary part of the response.

In other words: if the forcing function is a $$\hat{u}\cos(\omega t)$$, we may pretend that the forcing function is $$\underline{u}(t)=\hat{u}\cos(\omega t)+\color{green}{j}\,\hat{u}\sin(\omega t)=\hat{u}\,\mathrm{e}^{j\omega t}$$, derive the response and then consider only the real part of the complex solution.

### Example

An example can be found under the heading “complex arithmetic method” in the examples in RC Low-pass Filter Appendix B. The main section of that article describes the use of an even more convenient method using a Laplace Transform.

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