Unit 1: Ordinary Differential Equations of First Order

1. Mathematical Foundations and Definitions

An Ordinary Differential Equation (ODE) is a mathematical equation establishing an algebraic relationship between an independent variable, a single dependent variable, and one or more ordinary derivatives of the dependent variable with respect to the independent variable.

$\Phi\left(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \dots, \frac{d^ny}{dx^n}\right) = 0$

Where:

• $x \in \mathbb{R}$ is the independent variable.
• $y = y(x)$ is the unknown dependent function.
• $\frac{d^ky}{dx^k}$ denotes the $k$-th order ordinary derivative of $y$ with respect to $x$.

CODECopy Code
                              Differential Equations
                                        |
                   +--------------------+--------------------+
                   |                                         |
       Ordinary Differential Equations            Partial Differential Equations
                   (ODEs)                                    (PDEs)
         (Single independent variable)            (Two or more independent variables)

1.1 Order and Degree of a Differential Equation

• Order: The order of a differential equation is the order of the highest derivative appearing in the equation.
• Degree: The degree of a differential equation is the power (positive integer exponent) of the highest derivative, after the equation has been expressed in a form free of fractional exponents and radicals involving derivatives.

Illustrative Examples:

1. $\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^5 + y = \sin x$
   • Highest derivative is $\frac{d^2y}{dx^2} \implies \text{Order} = 2$.
   • Power of highest derivative is $3 \implies \text{Degree} = 3$.
2. $\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = k \frac{d^2y}{dx^2}$
   • Squaring both sides to eliminate the fractional exponent:
     $\left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 = k^2 \left(\frac{d^2y}{dx^2}\right)^2$
   • Highest derivative is $\frac{d^2y}{dx^2} \implies \text{Order} = 2$.
   • Power of highest derivative is $2 \implies \text{Degree} = 2$.
3. $\frac{dy}{dx} = e^{dy/dx}$ or $\sin\left(\frac{dy}{dx}\right) + y = 0$
   • $\text{Order} = 1$; however, the equation cannot be expanded as a polynomial in $\frac{dy}{dx} \implies \text{Degree is undefined}$.

1.2 Linearity of Differential Equations

A differential equation is defined as linear if it satisfies two strict criteria:

1. The dependent variable $y$ and all its derivatives $\frac{dy}{dx}, \dots, \frac{d^ny}{dx^n}$ occur strictly to the first power ($1$).
2. There are no product terms involving the dependent variable and any of its derivatives (e.g., no $y \frac{dy}{dx}$ or $\left(\frac{dy}{dx}\right)^2$) and no transcendental functions of $y$ (e.g., $\sin y, e^y, \ln y$).

A first-order linear differential equation always conforms to the standard canonical structure:

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

1.3 Nature of Solutions: General, Particular, and Singular

• General Solution (Complete Integral): A solution containing a number of arbitrary, independent constants equal to the order of the differential equation. For a first-order ODE, the general solution contains exactly one arbitrary constant $C$:
  $F(x, y, C) = 0$
• Particular Solution: A solution obtained from the general solution by assigning specific numerical values to the arbitrary constants, typically dictated by initial conditions (e.g., $y(x_0) = y_0$) or boundary conditions.
• Singular Solution: A solution that satisfies the differential equation but cannot be obtained from the general solution for any choice of the arbitrary constant. Geometrically, a singular solution represents the envelope of the one-parameter family of curves described by the general solution.

1.4 Existence and Uniqueness: The Picard–Lindelöf Theorem

Consider the first-order Initial Value Problem (IVP):

$\frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0$

Let $R = \{(x, y) \in \mathbb{R}^2 : |x - x_0| \le a, |y - y_0| \le b\}$ be a closed rectangle centered at $(x_0, y_0)$.

• Existence Condition: If $f(x, y)$ is continuous on $R$, then there exists at least one solution $y(x)$ in an interval $|x - x_0| \le h$.
• Uniqueness Condition: If, in addition, the partial derivative $\frac{\partial f}{\partial y}$ is continuous on $R$ (satisfying the Lipschitz condition $|f(x, y_1) - f(x, y_2)| \le L|y_1 - y_2|$), then the solution $y(x)$ is unique.

2. Classical First-Order ODE Solution Techniques

2.1 Separation of Variables

An equation of the form:

$M(x)dx + N(y)dy = 0 \quad \text{or} \quad \frac{dy}{dx} = g(x)h(y)$

is solved by isolating terms containing $x$ with $dx$ and terms containing $y$ with $dy$:

$\frac{1}{h(y)}dy = g(x)dx \implies \int \frac{1}{h(y)}dy = \int g(x)dx + C$

Equations Reducible to Separable Form:

Equations of the form $\frac{dy}{dx} = f(ax + by + c)$ are made separable by the linear transformation:

$u = ax + by + c \implies \frac{du}{dx} = a + b\frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{b}\left(\frac{du}{dx} - a\right)$

Substituting into the ODE yields:

$\frac{du}{dx} = a + bf(u) \implies \frac{du}{a + bf(u)} = dx$

2.2 Homogeneous Differential Equations

A function $f(x, y)$ is homogeneous of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$. A first-order ODE:

$\frac{dy}{dx} = \frac{f_1(x, y)}{f_2(x, y)}$

is homogeneous if $f_1$ and $f_2$ are homogeneous functions of the exact same degree. In this case, the equation can be expressed as:

$\frac{dy}{dx} = \phi\left(\frac{y}{x}\right)$

Solution Strategy:

1. Substitute $y = vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
2. Substitute into the differential equation:
   $v + x\frac{dv}{dx} = \phi(v) \implies x\frac{dv}{dx} = \phi(v) - v$
3. Separate variables:
   $\frac{dv}{\phi(v) - v} = \frac{dx}{x}$
4. Integrate both sides and substitute back $v = \frac{y}{x}$.

2.3 Non-Homogeneous Equations Reducible to Homogeneous Form

Equations having linear fractional arguments:

$\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}$

• Case 1: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Intersecting Lines):
  Shift the origin to the intersection point $(h, k)$ by setting $x = X + h$ and $y = Y + k$, where $(h, k)$ satisfies:
  $a_1h + b_1k + c_1 = 0 \quad \text{and} \quad a_2h + b_2k + c_2 = 0$
  Since $dx = dX$ and $dy = dY$, the equation transforms into the homogeneous form:
  $\frac{dY}{dX} = \frac{a_1X + b_1Y}{a_2X + b_2Y}$
  Now solve by setting $Y = vX$.

• Case 2: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = m$ (Parallel Lines):
  Here $a_2x + b_2y = m(a_1x + b_1y)$. Substitute $u = a_1x + b_1y$, reducing the equation directly to variable separable form.

3. Exact Differential Equations

A first-order differential equation expressed in Pfaffian differential form:

$M(x, y)dx + N(x, y)dy = 0$

is called Exact if there exists a scalar potential function $u(x, y)$ such that its total differential equals the left-hand side:

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

Since $du = 0$, the general solution is simply:

$u(x, y) = C$

3.1 Euler's Condition for Exactness

The necessary and sufficient condition for $M(x, y)dx + N(x, y)dy = 0$ to be an exact differential equation (assuming $M, N$ and their first partials are continuous on a simply connected domain) is:

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

Proof Outline: Since $M = \frac{\partial u}{\partial x}$ and $N = \frac{\partial u}{\partial y}$, Clairaut's theorem on the equality of mixed second partial derivatives guarantees:
$\frac{\partial M}{\partial y} = \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial N}{\partial x}$

3.2 Standard Integration Formula for Exact Equations

Once exactness is verified ($\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$), the general solution is evaluated directly using the standard formula:

$\int_{y = \text{constant}} M(x, y)\,dx + \int (\text{terms of } N \text{ free from } x)\,dy = C$

4. Integrating Factors (Non-Exact Differential Equations)

If $M(x, y)dx + N(x, y)dy = 0$ is not exact ($\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$), it can often be made exact by multiplying the entire equation by a non-zero integrating factor $\mu(x, y)$:

$\mu(x, y)M(x, y)dx + \mu(x, y)N(x, y)dy = 0$

The condition for exactness of the new equation requires:

$\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N) \implies \mu \frac{\partial M}{\partial y} + M \frac{\partial \mu}{\partial y} = \mu \frac{\partial N}{\partial x} + N \frac{\partial \mu}{\partial x}$

From this governing PDE, five standardized rules for finding Integrating Factors (IF) are derived:

CODECopy Code
                       Rules for Integrating Factors (IF)
                                       |
     +-------------+-------------+-----+-------------+-------------+
     |             |             |                   |             |
  Rule 1        Rule 2        Rule 3              Rule 4        Rule 5
Inspection   Homogeneous   f₁(xy)y dx +        Function of   Function of
               Equations   f₂(xy)x dy = 0       x alone       y alone

4.1 Rule 1: Method of Grouping (Inspection)

Recognize standard exact total differential pairs:

• $x\,dy + y\,dx = d(xy)$
• $\frac{x\,dy - y\,dx}{x^2} = d\left(\frac{y}{x}\right)$
• $\frac{y\,dx - x\,dy}{y^2} = d\left(\frac{x}{y}\right)$
• $\frac{x\,dy - y\,dx}{xy} = d\left(\ln\frac{y}{x}\right)$
• $\frac{x\,dy - y\,dx}{x^2 + y^2} = d\left(\arctan\frac{y}{x}\right)$
• $\frac{x\,dx + y\,dy}{x^2 + y^2} = d\left(\frac{1}{2}\ln(x^2 + y^2)\right)$

4.2 Rule 2: Homogeneous Non-Exact Equations

If $M(x, y)dx + N(x, y)dy = 0$ is a homogeneous equation of degree $n$ and $Mx + Ny \neq 0$:

$\text{IF} = \frac{1}{Mx + Ny}$

4.3 Rule 3: Equations of the Form $f_1(xy)y\,dx + f_2(xy)x\,dy = 0$

If the equation can be written as $f_1(xy)y\,dx + f_2(xy)x\,dy = 0$ and $Mx - Ny \neq 0$:

$\text{IF} = \frac{1}{Mx - Ny}$

4.4 Rule 4: Function of $x$ Alone

If the quantity:

$\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = f(x)$

is a function of $x$ only (or constant), then:

$\text{IF} = e^{\int f(x)\,dx}$

4.5 Rule 5: Function of $y$ Alone

If the quantity:

$\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = g(y)$

is a function of $y$ only (or constant), then:

$\text{IF} = e^{\int g(y)\,dy}$

5. Linear First-Order Equations (Leibniz's Equation)

5.1 Standard Form and Solution

The canonical form of a first-order linear differential equation is:

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

Where $P(x)$ and $Q(x)$ are continuous functions of $x$ alone.

Derivation of the Integrating Factor:

Multiplying by an integrating factor $\mu(x)$:

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

The left-hand side is identical to the product rule expansion $\frac{d}{dx}(\mu y) = \mu \frac{dy}{dx} + \frac{d\mu}{dx}y$ if:

$\frac{d\mu}{dx} = \mu P(x) \implies \frac{d\mu}{\mu} = P(x)dx \implies \ln \mu = \int P(x)dx \implies \mu(x) = e^{\int P(x)\,dx}$

Hence, the Integrating Factor is:

$\text{IF} = e^{\int P(x)\,dx}$

Integrating both sides yields the general solution:

$y \cdot e^{\int P(x)\,dx} = \int Q(x) e^{\int P(x)\,dx}\,dx + C \quad \implies \quad y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF})\,dx + C$

5.2 Equations Linear in $x$

When an equation cannot be expressed linearly in $y$, it may be linear when viewed with $x$ as the dependent variable:

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

• $\text{IF} = e^{\int P(y)\,dy}$
• General solution: $x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF})\,dy + C$

6. Non-Linear Equations Reducible to Linear Form

6.1 Bernoulli's Differential Equation

A first-order non-linear differential equation of the form:

$\frac{dy}{dx} + P(x)y = Q(x)y^n \quad (n \neq 0, n \neq 1)$

is known as Bernoulli's Equation.

Transformation to Linear Form:

1. Divide both sides by $y^n$:
   $y^{-n}\frac{dy}{dx} + P(x)y^{1-n} = Q(x)$
2. Introduce the substitution:
   $v = y^{1-n} \implies \frac{dv}{dx} = (1 - n)y^{-n}\frac{dy}{dx} \implies y^{-n}\frac{dy}{dx} = \frac{1}{1-n}\frac{dv}{dx}$
3. Substitute into the equation:
   $\frac{1}{1-n}\frac{dv}{dx} + P(x)v = Q(x) \implies \frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$
   This is a standard first-order linear ODE in $v$, solvable via integrating factor:
   $\text{IF} = e^{(1-n)\int P(x)\,dx}$

6.2 Clairaut's Equation

An equation of the form:

$y = px + f(p) \quad \text{where } p = \frac{dy}{dx}$

Differentiating both sides with respect to $x$:

$\frac{dy}{dx} = p = p + x\frac{dp}{dx} + f'(p)\frac{dp}{dx} \implies \frac{dp}{dx}\left[x + f'(p)\right] = 0$

This yields two distinct branches of solutions:

1. General Solution: $\frac{dp}{dx} = 0 \implies p = C$. Substituting into the original equation gives:
   $y = Cx + f(C)$
   (A one-parameter family of straight lines)
2. Singular Solution: $x + f'(p) = 0$. Eliminating $p$ between $y = px + f(p)$ and $x + f'(p) = 0$ yields the singular solution, which forms the envelope of the general solution lines.

7. Engineering and Physical Applications

First-order ordinary differential equations serve as mathematical models for a wide variety of dynamic engineering phenomena.

CODECopy Code
                         First-Order ODE Applications
                                      |
         +--------------------+-------+--------------------+
         |                    |                            |
  Thermal Modeling       RL / RC Circuits         Kinematics & Geometry
(Newton's Law of        (Kirchhoff's Laws:        (Orthogonal Trajectories,
    Cooling)            L di/dt + Ri = E(t))        Pursuit curves, Decay)

7.1 Orthogonal Trajectories

Given a one-parameter family of plane curves $F(x, y, C) = 0$, an Orthogonal Trajectory is a curve that intersects every member of the family at right angles ($90^\circ$).

Algorithm for Cartesian Coordinates $F(x, y, C) = 0$:

1. Differentiate $F(x, y, C) = 0$ with respect to $x$ to obtain an expression involving $\frac{dy}{dx}$.
2. Eliminate the parameter $C$ between the original equation and the derivative to find the governing ODE:
   $f\left(x, y, \frac{dy}{dx}\right) = 0$
3. Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ (since the product of orthogonal slopes $m_1 m_2 = -1$).
4. Solve the resulting differential equation:
   $f\left(x, y, -\frac{dx}{dy}\right) = 0$

Algorithm for Polar Coordinates $F(r, \theta, C) = 0$:

1. Form the differential equation by eliminating $C$, yielding $f\left(r, \theta, \frac{dr}{d\theta}\right) = 0$.
2. Replace $\frac{dr}{d\theta}$ with $-r^2 \frac{d\theta}{dr}$.
3. Solve the resulting orthogonal trajectory ODE.

7.2 Newton's Law of Cooling

The rate of change of temperature $T(t)$ of a body is directly proportional to the difference between its temperature and the ambient medium temperature $T_m$:

$\frac{dT}{dt} = -k(T - T_m), \quad k > 0$

Separating variables:

$\frac{dT}{T - T_m} = -k\,dt \implies \ln|T - T_m| = -kt + C_1 \implies T(t) = T_m + (T_0 - T_m)e^{-kt}$

Where $T_0 = T(0)$ is the initial temperature at $t = 0$.

7.3 Transient Electrical Circuits

Series RL Circuit:

Applying Kirchhoff's Voltage Law (KVL) to a resistor $R$ and inductor $L$ connected to voltage source $E(t)$:

$L\frac{di}{dt} + Ri = E(t) \implies \frac{di}{dt} + \frac{R}{L}i = \frac{E(t)}{L}$

• This is a linear first-order ODE with $\text{IF} = e^{(R/L)t}$.
• If $E(t) = E_0$ (constant DC voltage) and $i(0) = 0$:
  $i(t) = \frac{E_0}{R}\left(1 - e^{-(R/L)t}\right)$
• The time constant of the circuit is $\tau = \frac{L}{R}$.

Series RC Circuit:

Applying KVL to a resistor $R$ and capacitor $C$ with charge $q(t)$ (where current $i = \frac{dq}{dt}$):

$R\frac{dq}{dt} + \frac{q}{C} = E(t) \implies \frac{dq}{dt} + \frac{1}{RC}q = \frac{E(t)}{R}$

• This is a linear first-order ODE with $\text{IF} = e^{t/(RC)}$.
• For constant voltage $E_0$ and uncharged capacitor $q(0) = 0$:
  $q(t) = CE_0\left(1 - e^{-t/(RC)}\right)$
• The capacitive time constant is $\tau = RC$.

7.4 Radioactive Decay and Population Growth

• Radioactive Decay: The rate of disintegration is proportional to the active mass present:
  $\frac{dN}{dt} = -\lambda N \implies N(t) = N_0 e^{-\lambda t}$
  The half-life $t_{1/2}$ is the time when $N = N_0/2$:
  $t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$
• Malthusian Population Growth:
  $\frac{dP}{dt} = kP \implies P(t) = P_0 e^{kt}$

8. Summary Checklist for Solving First-Order ODEs

Before selecting an integration strategy for a given first-order ODE, follow this diagnostic checklist:

1. Can variables be separated directly? Check if $\frac{dy}{dx} = g(x)h(y)$ or if it can be transformed via $u = ax + by + c$.
2. Is it homogeneous? Check if replacing $(x, y)$ with $(\lambda x, \lambda y)$ leaves the equation invariant; if so, substitute $y = vx$.
3. Is it exact? Test Euler's condition: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. If exact, integrate using $\int_{y=\text{const}} M\,dx + \int (\text{terms of } N \text{ free from } x)\,dy = C$.
4. If not exact, does it admit an Integrating Factor?
   • Check if $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = f(x) \implies \text{IF} = e^{\int f(x)\,dx}$.
   • Check if $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = g(y) \implies \text{IF} = e^{\int g(y)\,dy}$.
   • Check if homogeneous with $Mx + Ny \neq 0 \implies \text{IF} = \frac{1}{Mx + Ny}$.
5. Is it linear in $y$? If $\frac{dy}{dx} + P(x)y = Q(x)$, use $\text{IF} = e^{\int P\,dx}$.
6. Is it linear in $x$? If $\frac{dx}{dy} + P(y)x = Q(y)$, use $\text{IF} = e^{\int P\,dy}$.
7. Is it a Bernoulli equation? If $\frac{dy}{dx} + Py = Qy^n$, substitute $v = y^{1-n}$.
8. Is it Clairaut's form? If $y = px + f(p)$, the general solution is $y = cx + f(c)$.