Engineering electronics systems often involve dynamic processes described by differential equations, where the rate of change of currents, voltages, or charge storage is fundamental. Understanding how to form and solve ordinary differential equations (ODEs) equips electronics engineers with tools to analyze circuits involving capacitors, inductors, and signal responses. This chapter focuses on the formation and solution of first and second order ODEs commonly encountered in electronics engineering applications, such as RC, RL, and RLC circuits.

3.1 Formation and Solution of 1st and 2nd Order Differential Equations

In electronics engineering practice in Kenya, differential equations model transient responses in circuits and control systems. Mastery of solving these ODEs enables engineers to predict system behavior under various inputs, ensure stability, and design filters and oscillators. We begin with first order ODEs, progressing through variable separable, homogeneous, and linear types, fundamental in analyzing circuits with single energy storage elements.

3.1.1 1st Order Variable Separable Differential Equations

A first order variable separable differential equation can be expressed as

$$\frac{dy}{dx} = g(x)h(y)$$

where the variables \(x\) and \(y\) can be separated on opposite sides of the equation. This form is common in simple RC circuit charging or discharging problems where voltage or current changes with time.

The general solution involves rewriting as

$$\frac{1}{h(y)} dy = g(x) dx$$

and integrating both sides.

Worked Examples

Example 1: An RC circuit discharges through a resistor with voltage across the capacitor \(V\) governed by

$$\frac{dV}{dt} = -\frac{1}{RC} V$$

Given \(R = 2\,k\Omega\), \(C = 10\,\mu F\), and initial voltage \(V_0 = 12\,V\), find \(V\) at time \(t\).

Given: \(R=2000\,\Omega\), \(C=10 \times 10^{-6} F\), \(V_0=12\,V\), find \(V(t)\).

Separate variables:

$$\frac{dV}{V} = -\frac{1}{RC} dt$$

Substitute values:

$$\frac{dV}{V} = -\frac{1}{2000 \times 10 \times 10^{-6}} dt = -50 dt$$

Integrate both sides:

$$\int \frac{1}{V} dV = -50 \int dt$$

$$\ln |V| = -50 t + C$$

Solve for \(V\):

$$V = e^{-50 t + C} = Ae^{-50 t}$$

Apply initial condition \(V(0) = 12\):

$$12 = A e^{0} \Rightarrow A = 12$$

Final solution:

$$V(t) = 12 e^{-50 t}$$

Answer: \(V(t) = 12 e^{-50 t}\) volts

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Example 2: The current \(I\) in an RL circuit satisfies

$$\frac{dI}{dt} = -\frac{R}{L} I$$

Given \(R=5\,\Omega\), \(L=0.1\,H\), and initial current \(I_0=3\,A\), find \(I\) at \(t=0.05\,s\).

Separate variables:

$$\frac{dI}{I} = -\frac{R}{L} dt$$

Substitute values:

$$\frac{dI}{I} = -\frac{5}{0.1} dt = -50 dt$$

Integrate:

$$\ln |I| = -50 t + C$$

Solve for \(I\):

$$I = Ae^{-50 t}$$

Apply initial condition:

$$3 = A e^{0} \Rightarrow A=3$$

Calculate \(I\) at \(t=0.05\):

$$I = 3 e^{-50 \times 0.05} = 3 e^{-2.5}$$

Calculate \(e^{-2.5} \approx 0.0821\):

$$I = 3 \times 0.0821 = 0.2463\,A$$

Answer: \(I = 0.2463\,A\) at \(t=0.05\,s\)

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Example 3: A capacitor voltage \(V\) in an RC circuit is governed by

$$\frac{dV}{dt} = \frac{1}{RC} (V_{in} - V)$$

Assuming \(V_{in} = 5\,V\), \(R=1\,k\Omega\), \(C=20\,\mu F\), and initial voltage \(V(0)=0\), find \(V\) at \(t=0.01\,s\).

Rearranged:

$$\frac{dV}{dt} + \frac{1}{RC} V = \frac{V_{in}}{RC}$$

This is not separable directly, but if we consider the complementary solution:

Separate variables for the homogeneous part:

$$\frac{dV}{dt} = -\frac{1}{RC} V$$

Solution:

$$V_c = Ae^{-\frac{t}{RC}}$$

The particular solution is \(V_p = V_{in} = 5\,V\).

General solution:

$$V = V_p + V_c = 5 + Ae^{-\frac{t}{RC}}$$

Apply initial condition:

$$0 = 5 + A e^{-\frac{0}{RC}} = 5 + A \Rightarrow A = -5$$

Calculate at \(t=0.01\):

$$V = 5 - 5 e^{-\frac{0.01}{1000 \times 20 \times 10^{-6}}} = 5 - 5 e^{-0.5}$$

Calculate \(e^{-0.5} \approx 0.6065\):

$$V = 5 - 5 \times 0.6065 = 5 - 3.0325 = 1.9675\,V$$

Answer: \(V = 1.9675\,V\) at \(t=0.01\,s\)

3.1.2 1st Order Homogeneous Differential Equations

A first order homogeneous differential equation has the form

$$\frac{dy}{dx} = F(\frac{y}{x})$$

which implies the function depends on the ratio \(y/x\). This form appears in electronics when variables scale proportionally, such as in transistor characteristic curves or normalized circuit parameters.

The substitution \(v = \frac{y}{x}\) transforms the equation into a separable one.

Worked Examples

Example 1: Solve

$$\frac{dy}{dx} = \frac{y}{x}$$

Given initial condition \(y(1)=2\).

Substitute \(v = \frac{y}{x}\), then \(y = vx\), so

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

Substitute into original equation:

$$v + x \frac{dv}{dx} = v$$

Simplify:

$$x \frac{dv}{dx} = 0$$

$$\frac{dv}{dx} = 0$$

Integrate:

$$v = C$$

Recall \(v = \frac{y}{x}\), so

$$\frac{y}{x} = C \Rightarrow y = C x$$

Apply initial condition:

$$2 = C \times 1 \Rightarrow C=2$$

Answer: \(y = 2x\)

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Example 2: Solve

$$\frac{dy}{dx} = \frac{x + y}{x}$$

Given \(y(1) = 1\).

Rewrite:

$$\frac{dy}{dx} = 1 + \frac{y}{x}$$

Let \(v = \frac{y}{x}\), \(y = v x\), so

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

Substitute into equation:

$$v + x \frac{dv}{dx} = 1 + v$$

Simplify:

$$x \frac{dv}{dx} = 1$$

Separate variables:

$$\frac{dv}{dx} = \frac{1}{x}$$

Integrate:

$$v = \ln |x| + C$$

Recall \(v = \frac{y}{x}\):

$$\frac{y}{x} = \ln |x| + C \Rightarrow y = x \ln |x| + C x$$

Apply initial condition \(y(1) = 1\):

$$1 = 1 \times \ln 1 + C \times 1 = 0 + C \Rightarrow C=1$$

Answer: \(y = x \ln |x| + x\)

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Example 3: Solve

$$\frac{dy}{dx} = \frac{y, x}{y + x}$$

Given \(y(1) = 0\).

Rewrite the right side:

$$\frac{dy}{dx} = \frac{y, x}{y + x}$$

Let \(v = \frac{y}{x}\), so \(y = v x\).

Calculate \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).

Substitute:

$$v + x \frac{dv}{dx} = \frac{v x, x}{v x + x} = \frac{v - 1}{v + 1}$$

Rearrange:

$$x \frac{dv}{dx} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v (v + 1)}{v + 1} = \frac{v - 1 - v^2 - v}{v + 1} = \frac{-v^2 - 1}{v + 1}$$

Simplify numerator:

$$-v^2 - 1 = -(v^2 + 1)$$

So:

$$x \frac{dv}{dx} = -\frac{v^2 + 1}{v + 1}$$

Separate variables:

$$\frac{v + 1}{v^2 + 1} dv = -\frac{dx}{x}$$

Integrate left side:

$$\int \frac{v + 1}{v^2 + 1} dv = \int \frac{v}{v^2 + 1} dv + \int \frac{1}{v^2 + 1} dv$$

Calculate each integral:

1. \(\int \frac{v}{v^2 + 1} dv = \frac{1}{2} \ln (v^2 + 1)\)

2. \(\int \frac{1}{v^2 + 1} dv = \tan^{-1} v\)

Therefore:

$$\int \frac{v + 1}{v^2 + 1} dv = \frac{1}{2} \ln (v^2 + 1) + \tan^{-1} v + C$$

Integrate right side:

$$\int -\frac{dx}{x} = -\ln |x| + C$$

Combine:

$$\frac{1}{2} \ln (v^2 + 1) + \tan^{-1} v = -\ln |x| + C$$

Apply initial condition \(y(1) = 0\), so \(v = \frac{0}{1} = 0\):

$$\frac{1}{2} \ln (0^2 + 1) + \tan^{-1} 0 = -\ln 1 + C$$

$$0 + 0 = 0 + C \Rightarrow C=0$$

Implicit solution:

$$\frac{1}{2} \ln (v^2 + 1) + \tan^{-1} v = -\ln |x|$$

Recall \(v = \frac{y}{x}\):

$$\frac{1}{2} \ln (\frac{y^2}{x^2} + 1 ) + \tan^{-1} (\frac{y}{x}) = -\ln |x|$$

3.1.3 1st Order Linear Differential Equations

A first order linear differential equation has the standard form

$$\frac{dy}{dx} + P(x) y = Q(x)$$

which appears frequently in electronics for circuits with forcing functions, such as RC circuits with input voltage sources.

The integrating factor method solves this equation using

$$\mu(x) = e^{\int P(x) dx}$$

and the solution formula

$$y \times \mu(x) = \int Q(x) \times \mu(x) dx + C$$

Worked Examples

Example 1: Solve

$$\frac{dy}{dx} + 2 y = 4$$

with initial condition \(y(0) = 1\).

Calculate integrating factor:

$$\mu(x) = e^{\int 2 dx} = e^{2x}$$

Multiply both sides by \(\mu(x)\):

$$e^{2x} \frac{dy}{dx} + 2 e^{2x} y = 4 e^{2x}$$

Left side is derivative:

$$\frac{d}{dx} (y e^{2x}) = 4 e^{2x}$$

Integrate both sides:

$$y e^{2x} = \int 4 e^{2x} dx + C$$

Calculate integral:

$$\int 4 e^{2x} dx = 4 \times \frac{e^{2x}}{2} = 2 e^{2x}$$

So:

$$y e^{2x} = 2 e^{2x} + C$$

Divide both sides by \(e^{2x}\):

$$y = 2 + C e^{-2x}$$

Apply initial condition \(y(0) = 1\):

$$1 = 2 + C e^{0} = 2 + C \Rightarrow C = -1$$

Final solution:

$$y = 2 - e^{-2x}$$

Answer: \(y = 2 - e^{-2x}\)

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Example 2: An RC circuit voltage \(V\) satisfies

$$\frac{dV}{dt} + \frac{1}{RC} V = \frac{V_{in}}{RC}$$

Given \(R = 1\,k\Omega\), \(C=10\,\mu F\), \(V_{in} = 10\,V\), and \(V(0) = 0\), find \(V(t)\).

Calculate \(\mu(t)\):

$$\mu(t) = e^{\int \frac{1}{RC} dt} = e^{\frac{t}{RC}}$$

Substitute \(RC = 1000 \times 10 \times 10^{-6} = 0.01\,s\):

$$\mu(t) = e^{\frac{t}{0.01}} = e^{100 t}$$

Multiply equation by \(\mu(t)\):

$$e^{100 t} \frac{dV}{dt} + \frac{1}{RC} e^{100 t} V = \frac{V_{in}}{RC} e^{100 t}$$

Left side is derivative:

$$\frac{d}{dt} (V e^{100 t}) = \frac{V_{in}}{RC} e^{100 t}$$

Integrate both sides:

$$V e^{100 t} = \int \frac{V_{in}}{RC} e^{100 t} dt + C = \frac{V_{in}}{RC} \int e^{100 t} dt + C$$

Calculate integral:

$$\int e^{100 t} dt = \frac{e^{100 t}}{100}$$

Therefore:

$$V e^{100 t} = \frac{V_{in}}{RC} \times \frac{e^{100 t}}{100} + C = \frac{10}{0.01} \times \frac{e^{100 t}}{100} + C = 1000 \times \frac{e^{100 t}}{100} + C = 10 e^{100 t} + C$$

Divide both sides by \(e^{100 t}\):

$$V = 10 + C e^{-100 t}$$

Apply initial condition \(V(0) = 0\):

$$0 = 10 + C e^{0} \Rightarrow C = -10$$

Final solution:

$$V = 10 - 10 e^{-100 t}$$

Answer: \(V(t) = 10 - 10 e^{-100 t}\) volts

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Example 3: Solve

$$\frac{dy}{dx} + \frac{2}{x} y = \sin x$$

with \(y(\pi) = 0\).

Calculate integrating factor:

$$\mu(x) = e^{\int \frac{2}{x} dx} = e^{2 \ln |x|} = x^{2}$$

Multiply both sides by \(x^{2}\):

$$x^{2} \frac{dy}{dx} + 2 x y = x^{2} \sin x$$

Left side is derivative:

$$\frac{d}{dx} (x^{2} y) = x^{2} \sin x$$

Integrate both sides:

$$x^{2} y = \int x^{2} \sin x dx + C$$

Use integration by parts to evaluate \(\int x^{2} \sin x dx\):

Let

\(u = x^{2}\), \(dv = \sin x dx\)

Then

\(du = 2x dx\), \(v = -\cos x\)

Apply integration by parts:

\(\int x^{2} \sin x dx = -x^{2} \cos x + \int 2x \cos x dx\)

Evaluate \(\int 2x \cos x dx\) by parts again:

Let

\(u = 2x\), \(dv = \cos x dx\)

Then

\(du = 2 dx\), \(v = \sin x\)

So

\(\int 2x \cos x dx = 2x \sin x - \int 2 \sin x dx = 2x \sin x + 2 \cos x + C\)

Therefore:

\(\int x^{2} \sin x dx = -x^{2} \cos x + 2x \sin x + 2 \cos x + C\)

So:

$$x^{2} y = -x^{2} \cos x + 2x \sin x + 2 \cos x + C$$

Divide by \(x^{2}\):

$$y = -\cos x + \frac{2 \sin x}{x} + \frac{2 \cos x}{x^{2}} + \frac{C}{x^{2}}$$

Apply initial condition \(y(\pi) = 0\):

Calculate each term at \(x = \pi\):

\(-\cos \pi = -(-1) = 1\)

\(\frac{2 \sin \pi}{\pi} = 0\)

\(\frac{2 \cos \pi}{\pi^{2}} = \frac{2 \times -1}{\pi^{2}} = -\frac{2}{\pi^{2}}\)

So:

$$0 = 1 + 0 - \frac{2}{\pi^{2}} + \frac{C}{\pi^{2}}$$

Solve for \(C\):

$$\frac{C}{\pi^{2}} = -1 + \frac{2}{\pi^{2}}$$

Multiply both sides by \(\pi^{2}\):

$$C = -\pi^{2} + 2$$

Final solution:

$$y = -\cos x + \frac{2 \sin x}{x} + \frac{2 \cos x}{x^{2}} + \frac{-\pi^{2} + 2}{x^{2}}$$

Chapter Summary

This chapter introduced the formation and solution of first and second order ordinary differential equations, beginning with first order equations that are variable separable, homogeneous, and linear. It then progressed to solving second order differential equations by determining unknown coefficients using characteristic equations. The D-operator method was presented as a powerful technique to simplify solving linear differential equations, including its extension to simultaneous differential equations and solutions through the series method. Throughout, emphasis was placed on step-by-step procedures to obtain general and particular solutions. The chapter concluded by exploring practical applications of differential equations in engineering contexts, demonstrating their relevance in modeling physical phenomena and solving real-world problems. Mastery of these methods equips students to analyze and solve a wide range of engineering problems involving rates of change and dynamic systems.