First order ordinary differential equations are fundamental in modeling dynamic systems in automotive mechatronics, such as engine temperature control, sensor signal processing, and vehicle suspension behavior. Understanding the types of these equations is essential for automotive technicians and engineers to analyze and predict system responses accurately. This chapter focuses on classifying first order differential equations and forming them from practical automotive scenarios to equip professionals with the skills to solve real-world problems.

7.1 Types of first order differential equations

The classification of first order differential equations is critical in automotive mechatronics because different types require specific solution methods. Recognizing the form of an equation derived from automotive system behavior enables the correct analytical approach, improving diagnostic and design accuracy. Common types include separable, linear, homogeneous, and exact equations, each with distinct characteristics and solution strategies.

7.1.1 Separable differential equations

Separable differential equations can be expressed as a product of a function of \(x\) and a function of \(y\), allowing the variables to be separated on opposite sides of the equation for integration.

The general form is:

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

which can be rearranged as

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

Worked Examples

Example 1: In an automotive cooling system, the temperature \(T\) of the coolant changes at a rate proportional to the difference between the coolant temperature and ambient temperature \(T_a = 25^\circ C\). The rate of change is given by:

$$\frac{dT}{dt} = -k (T - 25)$$

where \(k = 0.1\) min\(^{-1}\). Solve for \(T\) given initial temperature \(T_0 = 90^\circ C\) at \(t=0\).

Given: \(k = 0.1\), \(T_a = 25\), \(T_0 = 90\), \(t=0\)

Rearranged:

$$\frac{dT}{dt} = -0.1 (T - 25)$$

Separate variables:

$$\frac{1}{T - 25} dT = -0.1 dt$$

Integrate both sides:

$$\int \frac{1}{T - 25} dT = \int -0.1 dt$$

$$\ln |T - 25| = -0.1 t + C$$

Apply initial condition \(t=0, T=90\):

$$\ln |90 - 25| = C$$

$$\ln 65 = C$$

So,

$$\ln |T - 25| = -0.1 t + \ln 65$$

Exponentiate both sides:

$$|T - 25| = 65 e^{-0.1 t}$$

Since \(T > 25\), drop absolute value:

$$T = 25 + 65 e^{-0.1 t}$$

Answer:

$$T(t) = 25 + 65 e^{-0.1 t} \quad (^\circ C)$$

Example 2: The voltage \(V\) across a capacitor in an automotive electronic circuit discharges at a rate proportional to the voltage itself:

$$\frac{dV}{dt} = -5 V$$

Given initial voltage \(V_0 = 12 V\) at \(t=0\), find \(V\) at time \(t\).

Separate variables:

$$\frac{1}{V} dV = -5 dt$$

Integrate:

$$\int \frac{1}{V} dV = \int -5 dt$$

$$\ln |V| = -5 t + C$$

Initial condition:

$$\ln 12 = C$$

So,

$$\ln |V| = -5 t + \ln 12$$

Exponentiate:

$$V = 12 e^{-5 t}$$

Answer:

$$V(t) = 12 e^{-5 t} \quad (V)$$

Example 3: The concentration \(C\) of a pollutant in air in a garage changes according to:

$$\frac{dC}{dt} = k (C_0 - C)$$

where \(k = 0.3 \, \text{min}^{-1}\), \(C_0 = 50 \, \text{ppm}\), and initial concentration \(C(0) = 0\). Find \(C(t)\).

Rewrite:

$$\frac{dC}{dt} = 0.3 (50 - C)$$

Separate variables:

$$\frac{1}{50 - C} dC = 0.3 dt$$

Integrate:

$$-\ln |50 - C| = 0.3 t + C_1$$

Multiply both sides by -1:

$$\ln |50 - C| = -0.3 t + C_2$$

Apply initial condition:

$$\ln 50 = C_2$$

So,

$$\ln |50 - C| = -0.3 t + \ln 50$$

Exponentiate:

$$50 - C = 50 e^{-0.3 t}$$

Solve for \(C\):

$$C = 50 (1 - e^{-0.3 t})$$

Answer:

$$C(t) = 50 (1 - e^{-0.3 t}) \quad \text{ppm}$$

7.1.2 Linear first order differential equations

Linear first order differential equations have the standard form:

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

where \(P(x)\) and \(Q(x)\) are functions of \(x\) only. The integrating factor method is used to solve them.

The integrating factor (I.F.) is:

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

The solution is:

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

Worked Examples

Example 1: An automotive sensor output voltage \(y\) satisfies:

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

Find \(y(x)\) given initial condition \(y(0) = 1\).

Given: \(P(x) = 2\), \(Q(x) = 4\), \(y(0) = 1\)

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:

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

Integrate RHS:

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

So,

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

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

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

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

$$1 = 2 + C \Rightarrow C = -1$$

Answer:

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

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

$$\frac{dI}{dt} + 3 I = 6 t$$

with initial condition \(I(0) = 0\). Find \(I(t)\).

Given: \(P(t) = 3\), \(Q(t) = 6 t\), \(I(0) = 0\)

Integrating factor:

$$\mu(t) = e^{\int 3 dt} = e^{3 t}$$

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

$$e^{3 t} \frac{dI}{dt} + 3 e^{3 t} I = 6 t e^{3 t}$$

LHS is derivative:

$$\frac{d}{dt} (I e^{3 t}) = 6 t e^{3 t}$$

Integrate both sides:

$$I e^{3 t} = \int 6 t e^{3 t} dt + C$$

Use integration by parts:

Let \(u = t\), \(dv = 6 e^{3 t} dt\)

Then \(du = dt\), \(v = 2 e^{3 t}\)

So,

$$\int 6 t e^{3 t} dt = 6 \int t e^{3 t} dt = 6 (t \times \frac{e^{3 t}}{3} - \int \frac{e^{3 t}}{3} dt) = 6 (\frac{t e^{3 t}}{3} - \frac{e^{3 t}}{9}) = 2 t e^{3 t} - \frac{2}{3} e^{3 t}$$

Therefore:

$$I e^{3 t} = 2 t e^{3 t} - \frac{2}{3} e^{3 t} + C$$

Divide by \(e^{3 t}\):

$$I = 2 t - \frac{2}{3} + C e^{-3 t}$$

Apply initial condition \(t=0, I=0\):

$$0 = 0 - \frac{2}{3} + C \Rightarrow C = \frac{2}{3}$$

Answer:

$$I(t) = 2 t - \frac{2}{3} + \frac{2}{3} e^{-3 t}$$

Example 3: In an automotive fuel injection system, the rate of change of fuel concentration \(c\) is described by:

$$\frac{dc}{dt} + 4 c = 8 e^{-2 t}$$

with \(c(0) = 1\). Find \(c(t)\).

Given: \(P(t) = 4\), \(Q(t) = 8 e^{-2 t}\), \(c(0) = 1\)

Integrating factor:

$$\mu(t) = e^{\int 4 dt} = e^{4 t}$$

Multiply both sides:

$$e^{4 t} \frac{dc}{dt} + 4 e^{4 t} c = 8 e^{4 t} e^{-2 t} = 8 e^{2 t}$$

LHS is derivative:

$$\frac{d}{dt} (c e^{4 t}) = 8 e^{2 t}$$

Integrate both sides:

$$c e^{4 t} = \int 8 e^{2 t} dt + C$$

Integrate RHS:

$$\int 8 e^{2 t} dt = 8 \times \frac{e^{2 t}}{2} = 4 e^{2 t}$$

So,

$$c e^{4 t} = 4 e^{2 t} + C$$

Divide by \(e^{4 t}\):

$$c = 4 e^{-2 t} + C e^{-4 t}$$

Apply initial condition \(t=0, c=1\):

$$1 = 4 + C \Rightarrow C = -3$$

Answer:

$$c(t) = 4 e^{-2 t} - 3 e^{-4 t}$$

7.1.3 Homogeneous first order differential equations

A first order differential equation is homogeneous if it can be written as:

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

This means the right side is a function of the ratio \( \frac{y}{x} \) only.

The substitution \(v = \frac{y}{x}\) is used, so \(y = vx\) and

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

The equation becomes separable in \(v\) and \(x\).

Worked Examples

Example 1: Solve

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

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

Rewrite:

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

Substitute \(y = v x\):

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

So:

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

Subtract \(v\) from both sides:

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

Separate variables:

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

Integrate both sides:

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

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

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

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

Multiply both sides by \(x\):

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

Apply initial condition \(x=1, y=2\):

$$2 = 1 \times \ln 1 + C \times 1 = 0 + C \Rightarrow C = 2$$

Answer:

$$y = x \ln |x| + 2 x$$

Example 2: Solve

$$\frac{dy}{dx} = \frac{y}{x} - (\frac{y}{x})^3$$

Initial condition: \(y(1) = 1\).

Rewrite:

$$\frac{dy}{dx} = v, v^3$$

Substitute \(y = v x\), so:

$$\frac{dy}{dx} = v + x \frac{dv}{dx} = v, v^3$$

$$v + x \frac{dv}{dx} = v, v^3$$

Subtract \(v\):

$$x \frac{dv}{dx} = -v^3$$

Separate variables:

$$\frac{dv}{v^3} = -\frac{dx}{x}$$

Rewrite:

$$v^{-3} dv = - \frac{dx}{x}$$

Integrate both sides:

$$\int v^{-3} dv = - \int \frac{1}{x} dx$$

Integral of \(v^{-3}\):

$$\int v^{-3} dv = \int v^{-3} dv = \frac{v^{-2}}{-2} + C = -\frac{1}{2 v^{2}} + C$$

Integral of RHS:

$$-\ln |x| + C$$

So,

$$-\frac{1}{2 v^{2}} = -\ln |x| + K$$

Multiply both sides by -1:

$$\frac{1}{2 v^{2}} = \ln |x| - K$$

Multiply by 2:

$$\frac{1}{v^{2}} = 2 \ln |x| - 2 K$$

Let \(C_1 = 2K\):

$$\frac{1}{v^{2}} = 2 \ln |x| - C_1$$

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

$$\frac{1}{(y/x)^2} = 2 \ln |x| - C_1$$

Simplify:

$$\frac{x^2}{y^2} = 2 \ln |x| - C_1$$

Invert:

$$\frac{y^2}{x^2} = \frac{1}{2 \ln |x| - C_1}$$

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

$$y^2 = \frac{x^2}{2 \ln |x| - C_1}$$

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

$$1^2 = \frac{1^2}{2 \ln 1 - C_1} = \frac{1}{0 - C_1} = -\frac{1}{C_1}$$

So,

$$- \frac{1}{C_1} = 1 \Rightarrow C_1 = -1$$

Final solution:

$$y^2 = \frac{x^2}{2 \ln |x| + 1}$$

Answer:

$$y = \frac{x}{\sqrt{2 \ln |x| + 1}}$$

Example 3: Solve

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

with initial condition \(y(2) = 3\).

Rewrite:

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

Substitute \(y = v x\):

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

So,

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

Subtract \(v\):

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

Separate variables:

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

Integrate:

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

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

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

Multiply both sides by \(x\):

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

Apply initial condition \(x=2, y=3\):

$$3 = 2 \ln 2 + 2 C$$

Calculate:

$$\ln 2 \approx 0.693$$

So,

$$3 = 2 \times 0.693 + 2 C = 1.386 + 2 C$$

Subtract:

$$3 - 1.386 = 2 C$$

$$1.614 = 2 C$$

$$C = 0.807$$

Answer:

$$y = x \ln |x| + 0.807 x$$

7.1.4 Exact differential equations

An equation of the form

$$M(x,y) + N(x,y) \frac{dy}{dx} = 0$$

is exact if

$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

The solution is found by integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\).

Worked Examples

Example 1: Solve the exact equation:

$$(2x + 3y) + (3x + 4y) \frac{dy}{dx} = 0$$

Check exactness:

$$M = 2x + 3y, \quad N = 3x + 4y$$

Calculate partial derivatives:

$$\frac{\partial M}{\partial y} = 3$$

$$\frac{\partial N}{\partial x} = 3$$

Since equal, equation is exact.

Integrate \(M\) with respect to \(x\):

$$\int (2x + 3y) dx = x^2 + 3 y x + h(y)$$

Differentiate with respect to \(y\):

$$\frac{d}{dy} (x^2 + 3 y x + h(y)) = 3 x + h'(y)$$

Set equal to \(N\):

$$3 x + h'(y) = 3 x + 4 y$$

Hence,

$$h'(y) = 4 y$$

Integrate:

$$h(y) = 2 y^2 + C$$

Solution:

$$x^2 + 3 x y + 2 y^2 = C$$

Answer:

$$x^2 + 3 x y + 2 y^2 = C$$

Example 2: Solve

$$(y \cos x + 2) + (\sin x + 3 y^2) \frac{dy}{dx} = 0$$

Check exactness:

$$M = y \cos x + 2, \quad N = \sin x + 3 y^2$$

Calculate partial derivatives:

$$\frac{\partial M}{\partial y} = \cos x$$

$$\frac{\partial N}{\partial x} = \cos x$$

Equal, so exact.

Integrate \(M\) with respect to \(x\):

$$\int (y \cos x + 2) dx = y \sin x + 2 x + h(y)$$

Differentiate with respect to \(y\):

$$\frac{d}{dy} (y \sin x + 2 x + h(y)) = \sin x + h'(y)$$

Set equal to \(N\):

$$\sin x + h'(y) = \sin x + 3 y^2$$

So,

$$h'(y) = 3 y^2$$

Integrate:

$$h(y) = y^3 + C$$

Solution:

$$y \sin x + 2 x + y^3 = C$$

Answer:

$$y \sin x + 2 x + y^3 = C$$

Example 3: Solve

$$(2 x y + y^2) + (x^2 + 2 x y) \frac{dy}{dx} = 0$$

Check exactness:

$$M = 2 x y + y^2, \quad N = x^2 + 2 x y$$

Calculate partial derivatives:

$$\frac{\partial M}{\partial y} = 2 x + 2 y$$

$$\frac{\partial N}{\partial x} = 2 x + 2 y$$

Equal, so exact.

Integrate \(M\) with respect to \(x\):

$$\int (2 x y + y^2) dx = y x^2 + y^2 x + h(y)$$

Differentiate with respect to \(y\):

$$\frac{d}{dy} (y x^2 + y^2 x + h(y)) = x^2 + 2 y x + h'(y)$$

Set equal to \(N\):

$$x^2 + 2 y x + h'(y) = x^2 + 2 y x$$

So,

$$h'(y) = 0 \Rightarrow h(y) = C$$

Solution:

$$y x^2 + y^2 x = C$$

Answer:

$$x^2 y + x y^2 = C$$

Chapter Summary

This chapter introduced the various types of first order differential equations, providing a clear understanding of their characteristics and differences. It then explained how to form first order differential equations from given problems, establishing a foundation for their solution. The chapter explored methods to solve first order differential equations, including separation of variables, linear equations, and homogeneous equations, with detailed worked examples. Applications of first order differential equations in engineering contexts were demonstrated to show their practical relevance. Moving to second order differential equations, the chapter covered their formation based on different system behaviors and conditions. It then focused on solving these second order equations using the method of undetermined coefficients, illustrating step-by-step procedures. Finally, the chapter highlighted practical applications of second order differential equations in engineering problems, reinforcing their importance in modelling and analysis.