4.5 First-order Linear Equations

Standard Form

Consider the differential equation

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of $y^{\prime }$ be equal to $1.$ To make this happen, we divide both sides by $3x^2-4.$

This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to Equation 4.14, we can divide both sides of the equation by $a\left(x\right).$ This leads to the equation

Now define $p\left(x\right)=\frac{b\left(x\right)}{a\left(x\right)}$ and $q\left(x\right)=\frac{c\left(x\right)}{a\left(x\right)}.$ Then Equation 4.14 becomes

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Integrating Factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

The first term on the left-hand side of Equation 4.15 is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the product rule from the Differentiation Rules section. If we multiply Equation 4.16 by a yet-to-be-determined function $\mu \left(x\right),$ then the equation becomes

The left-hand side Equation 4.18 can be matched perfectly to the product rule:

Matching term by term gives $y=f\left(x\right),g\left(x\right)=\mu \left(x\right),$ and $g^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right).$ Taking the derivative of $g\left(x\right)=\mu \left(x\right)$ and setting it equal to the right-hand side of $g^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right)$ leads to

This is a first-order, separable differential equation for $\mu \left(x\right).$ We know $p(x)$ because it appears in the differential equation we are solving. Separating variables and integrating yields

Here $C_2$ can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of Equation 4.16 by the integrating factor $\mu \left(x\right).$ This gives

The left-hand side of Equation 4.20 can be rewritten as $\frac{d}{dx}\left(\mu \left(x\right)y\right).$

Next integrate both sides of Equation 4.21 with respect to $x.$

Divide both sides of Equation 4.22 by $\mu \left(x\right)\text{:}$

Since $\mu \left(x\right)$ was previously calculated, we are now finished. An important note about the integrating constant $C\text{:}$ It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving $p\left(x\right)$ is necessary in order to find an integrating factor for Equation 4.15. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for $C$ for this integral. We chose $C=0.$ When calculating the integral inside the brackets in Equation 4.21, it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to Equation 4.15. This integrating factor guarantees just that.

Now we use the same strategy to find the solution to an initial-value problem.

Applications of First-order Linear Differential Equations

We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.

Electrical Circuits

A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff’s Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:

1. The voltage drop across a resistor is given by
   
   $$E_R=Ri,$$
   
   where $R$ is a constant of proportionality called the resistance, and $i$ is the current.
2. The voltage drop across an inductor is given by
   
   $$E_L=L{i}^{\prime },$$
   
   where $L$ is a constant of proportionality called the inductance, and $i$ again denotes the current.
3. The voltage drop across a capacitor is given by
   
   $$E_C=\frac{1}{C}q,$$

where $C$ is a constant of proportionality called the capacitance, and $q$ is the instantaneous charge on the capacitor. The relationship between $i$ and $q$ is $i=q^{\prime }.$

We use units of volts $\left(\text{V}\right)$ to measure voltage $E,$ amperes $\left(\text{A}\right)$ to measure current $i,$ coulombs $\left(\text{C}\right)$ to measure charge $q,$ ohms $\left(\text{Ω}\right)$ to measure resistance $R,$ henrys $\left(\text{H}\right)$ to measure inductance $L,$ and farads $\left(\text{F}\right)$ to measure capacitance $C.$ Consider the circuit in Figure 4.25.

Figure 4.25 A typical electric circuit, containing a voltage generator $\left(V_S\right),$ capacitor $\left(C\right),$ inductor $\left(L\right),$ and resistor $\left(R\right).$

Applying Kirchhoff’s Loop Rule to this circuit, we let $E$ denote the electromotive force supplied by the voltage generator. Then

$$E_L+E_R+E_C=E.$$

Substituting the expressions for $E_L,E_R,$ and $E_C$ into this equation, we obtain

$$L{i}^{\prime }+Ri+\frac{1}{C}q=E.$$

(4.24)

If there is no capacitor in the circuit, then the equation becomes

$$L{i}^{\prime }+Ri=E.$$

(4.25)

This is a first-order differential equation in $i.$ The circuit is referred to as an $LR$ circuit.

Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so $L=0,R\ne 0,$ and $C\ne 0.$ Then Equation 4.23 can be rewritten as

$$R{q}^{\prime }+\frac{1}{C}q=E,$$

(4.26)

which is a first-order linear differential equation. This is referred to as an RC circuit. In either case, we can set up and solve an initial-value problem.

Example 4.19

Finding Current in an RL Electric Circuit

A circuit has in series an electromotive force given by $E=50\text{sin}20t\text{V},$ a resistor of $5\text{Ω},$ and an inductor of $0.4\text{H}\text{.}$ If the initial current is $0,$ find the current at time $t>0.$

Solution

We have a resistor and an inductor in the circuit, so we use Equation 4.24. The voltage drop across the resistor is given by $E_R=Ri=5i.$ The voltage drop across the inductor is given by $E_L=L{i}^{\prime }=0.4i^{\prime }.$ The electromotive force becomes the right-hand side of Equation 4.24. Therefore Equation 4.24 becomes

$$0.4i^{\prime }+5i=50\text{sin}20t.$$

Dividing both sides by $0.4$ gives the equation

$$i^{\prime }+12.5i=125\text{sin}20t.$$

Since the initial current is 0, this result gives an initial condition of $i\left(0\right)=0.$ We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.

Step 1. Rewrite the differential equation as $i^{\prime }+12.5i=125\text{sin}20t.$ This gives $p\left(t\right)=12.5$ and $q\left(t\right)=125\text{sin}20t.$

Step 2. The integrating factor is $\mu \left(t\right)=e^{{\displaystyle \int 12.5dt}}=e^{12.5t}.$

Step 3. Multiply the differential equation by $\mu \left(t\right)\text{:}$

$$\begin{array}{ccc}\hfill e^{12.5t}{i}^{\prime }+12.5e^{12.5t}i& =\hfill & 125e^{12.5t}\text{sin}20t\hfill \\ \hfill \frac{d}{dt}\left[i{e}^{12.5t}\right]& =\hfill & 125e^{12.5t}\text{sin}20t.\hfill \end{array}$$

Step 4. Integrate both sides:

$$\begin{array}{ccc}\hfill {\displaystyle \int \frac{d}{dt}\left[i{e}^{12.5t}\right]}dt& =\hfill & {\displaystyle \int 125e^{12.5t}\text{sin}20tdt}\hfill \\ \hfill i{e}^{12.5t}& =\hfill & \left(\frac{250\text{sin}20t-400\text{cos}20t}{89}\right)e^{12.5t}+C\hfill \\ \hfill i\left(t\right)& =\hfill & \frac{250\text{sin}20t-400\text{cos}20t}{89}+C{e}^{\mathrm{-12.5}t}.\hfill \end{array}$$

Step 5. Solve for $C$ using the initial condition $v\left(0\right)=2\text{:}$

$$\begin{array}{ccc}\hfill i\left(t\right)& =\hfill & \frac{250\text{sin}20t-400\text{cos}20t}{89}+C{e}^{\mathrm{-12.5}t}\hfill \\ \hfill i\left(0\right)& =\hfill & \frac{250\text{sin}20\left(0\right)-400\text{cos}20\left(0\right)}{89}+C{e}^{\mathrm{-12.5}\left(0\right)}\hfill \\ \hfill 0& =\hfill & -\frac{400}{89}+C\hfill \\ \hfill C& =\hfill & \frac{400}{89}.\hfill \end{array}$$

Therefore the solution to the initial-value problem is $i\left(t\right)=\frac{250\text{sin}20t-400\text{cos}20t+400e^{\mathrm{-12.5}t}}{89}=\frac{250\text{sin}20t-400\text{cos}20t}{89}+\frac{400e^{\mathrm{-12.5}t}}{89}.$

The first term can be rewritten as a single cosine function. First, multiply and divide by $\sqrt{{250}^2+{400}^2}=50\sqrt{89}\text{:}$

$$\begin{array}{cc}\hfill \frac{250\text{sin}20t-400\text{cos}20t}{89}& =\frac{50\sqrt{89}}{89}\left(\frac{250\text{sin}20t-400\text{cos}20t}{50\sqrt{89}}\right)\hfill \\ & =-\frac{50\sqrt{89}}{89}\left(\frac{8\text{cos}20t}{\sqrt{89}}-\frac{5\text{sin}20t}{\sqrt{89}}\right).\hfill \end{array}$$

Next, define $\phi$ to be an acute angle such that $\text{cos}\phi =\frac{8}{\sqrt{89}}.$ Then $\text{sin}\phi =\frac{5}{\sqrt{89}}$ and

$$\begin{array}{cc}\hfill -\frac{50\sqrt{89}}{89}\left(\frac{8\text{cos}20t}{\sqrt{89}}-\frac{5\text{sin}20t}{\sqrt{89}}\right)& =-\frac{50\sqrt{89}}{89}\left(\text{cos}\phi \text{cos}20t-\text{sin}\phi \text{sin}20t\right)\hfill \\ & =-\frac{50\sqrt{89}}{89}\text{cos}\left(20t+\phi \right).\hfill \end{array}$$

Therefore the solution can be written as

$$i\left(t\right)=-\frac{50\sqrt{89}}{89}\text{cos}\left(20t+\phi \right)+\frac{400e^{\mathrm{-12.5}t}}{89}.$$

The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is given by $\phi ,$ and the amplitude of the steady-state current is given by $\frac{50\sqrt{89}}{89}.$ The graph of this solution appears in Figure 4.26:

Figure 4.26

Checkpoint 4.19

A circuit has in series an electromotive force given by $E=20\text{sin}5t$ V, a capacitor with capacitance $0.02\text{F},$ and a resistor of $8\text{Ω}.$ If the initial charge is $4\text{C},$ find the charge at time $t>0.$