Solution of First-Order Linear Ordinary Differential Equations

First-Order Linear Ordinary Differential Equation

If a first-order ordinary differential equation can be algebraically brought to the form

\[y'+p(x)y=r(x) \tag{1}\]

it is called linear, otherwise it is called nonlinear.

The form in equation (1) is called the standard form of a first-order linear ordinary differential equation. If the first term of a given first-order linear ordinary differential equation is $f(x)y’$, we can obtain the standard form by dividing both sides of the equation by $f(x)$.

In engineering, $r(x)$ is often called the input, and $y(x)$ is called the output or the response to the input (and initial conditions).

Homogeneous Linear Ordinary Differential Equation

Let’s say $J$ is some interval $a<x<b$ where we want to solve equation (1). If $r(x)\equiv 0$ for the interval $J$ in equation (1), then

\[y'+p(x)y=0 \tag{2}\]

and this is called homogeneous. In this case, we can use the method of separation of variables.

\[\frac{dy}{y} = -p(x)dx\] \[\log |y| = -\int p(x)dx + c^*\] \[y(x) = ce^{-\int p(x)dx} \tag{3}\]

When $c=0$, we get the trivial solution $y(x)=0$.

Nonhomogeneous Linear Ordinary Differential Equation

If $r(x)\not\equiv 0$ in the interval $J$, it is called nonhomogeneous. It is known that the nonhomogeneous linear ordinary differential equation (1) has an integrating factor that depends only on $x$. This integrating factor $F(x)$ can be found using equation (11) from the method of finding integrating factors, or it can be found directly as follows.

Multiplying equation (1) by $F(x)$, we get

\[Fy'+pFy=rF \tag{1*}\]

If

\[pF=F'\]

then the left side of equation (1*) becomes the derivative $(Fy)’=F’y+Fy’$. Separating variables in $pF=F’$, we get $dF/F=p\ dx$, and integrating with $h=\int p\ dx$, we have

\[\log |F|=h=\int p\ dx\] \[F = e^h\]

Substituting this into equation (1*), we get

\[e^hy'+h'e^hy=e^hy'+(e^h)'=(e^hy)'=re^h\]

Integrating, we get

\(e^hy=\int e^hr\ dx + c\) and dividing by $e^h$, we obtain the desired solution formula.

\[y(x)=e^{-h}\left(\int e^hr\ dx + c\right),\qquad h=\int p(x)\ dx \tag{4}\]

Here, the integration constant in $h$ does not matter.

In equation (4), the only value that depends on the given initial condition is $c$, so if we write equation (4) as the sum of two terms

\[y(x)=e^{-h}\int e^hr\ dx + ce^{-h} \tag{4*}\]

we can see that:

\[\text{Total output}=\text{Response to input }r+\text{Response to initial condition} \tag{5}\]

Example: RL Circuit

Suppose an RL circuit consists of a battery with electromotive force $E=48\textrm{V}$, a resistor with $R=11\mathrm{\Omega}$, and an inductor with $L=0.1\text{H}$, and the initial current is 0. Construct the model of this RL circuit and solve the resulting ordinary differential equation for the current $I(t)$.

Ohm’s law
The current $I$ in the circuit causes a voltage drop $RI$ across the resistor.

Faraday’s law of electromagnetic induction
The current $I$ in the circuit causes a voltage drop $LI’=L\ dI/dt$ across the inductor.

Kirchhoff’s Voltage Law (KVL)
The electromotive force applied to a closed circuit is equal to the sum of the voltage drops across all other elements in the circuit.

Solution

According to the above laws, the model of the RL circuit is $LI’+RI=E(t)$, and in standard form:

\[I'+\frac{R}{L}I=\frac{E(t)}{L} \tag{6}\]

We can solve this linear ordinary differential equation using equation (4) with $x=t, y=I, p=R/L, h=(R/L)t$.

\[I=e^{-(R/L)t}\left(\int e^{(R/L)t} \frac{E(t)}{L}dt+c\right)\] \[I=e^{-(R/L)t}\left(\frac{E}{L}\frac{e^{(R/L)t}}{R/L}+c\right)=\frac{E}{R}+ce^{-(R/L)t} \tag{7}\]

Here, $R/L=11/0.1=110$ and $E(t)=48$, so

\[I=\frac{48}{11}+ce^{-110t}\]

From the initial condition $I(0)=0$, we get $I(0)=E/R+c=0$, $c=-E/R$. From this, we can find the following particular solution:

\[I=\frac{E}{R}(1-e^{-(R/L)t}) \tag{8}\] \[\therefore I=\frac{48}{11}(1-e^{-110t})\]