Differential Equations

1. Introduction and Classification

1.1 Basic Definitions

A differential equation (DE) is an equation involving an unknown function and its derivatives. An ordinary differential equation (ODE) involves a function of one variable and its ordinary Derivatives. A partial differential equation (PDE) involves a function of several variables and Its partial derivatives.

1.2 Classification of ODEs

An ODE is:

• Ordinary vs. partial: depends on whether partial derivatives appear.
• Order: the highest derivative that appears.
• Linear vs. nonlinear: linear if the unknown function and its derivatives appear linearly.
• Homogeneous vs. nonhomogeneous: for linear ODEs, homogeneous if the forcing term is zero.

1.3 Initial and Boundary Value Problems

An initial value problem (IVP) specifies the value of the function (and possibly its Derivatives) at a single point. A boundary value problem (BVP) specifies conditions at two or More points.

1.4 Examples from Physics and Biology

Differential equations arise throughout the natural sciences. A few canonical examples:

1. Newton”s law of cooling. The temperature $T(t)$ of a body in a medium at temperature $T_m$ satisfies $\frac{dT}{dt} = -k(T - T_m)$A first-order linear ODE.

2. Harmonic oscillator. A mass on a spring with damping obeys $m\frac{d^2 x}{dt^2} + c\frac{dx}{dt} + kx = F(t)$A second-order linear ODE.

3. Logistic population growth. $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$A first-order nonlinear (Bernoulli) ODE.

4. Lotka-Volterra predator-prey model. $\frac{dx}{dt} = x(\alpha - \beta y)$, $\frac{dy}{dt} = y(-\gamma + \delta x)$A coupled nonlinear system.

5. RC circuit. The charge $q(t)$ on a capacitor satisfies $R\frac{dq}{dt} + \frac{q}{C} = V(t)$ a first-order linear ODE.

6. Heat equation. The temperature $u(x, t)$ in a rod satisfies $u_t = \alpha^2 u_{xx}$A second-order linear PDE.

7. Wave equation. The displacement $u(x, t)$ of a string satisfies $u_{tt} = c^2 u_{xx}$A second-order linear PDE.

8. Laplace’s equation. The steady-state temperature satisfies $u_{xx} + u_{yy} = 0$A second-order linear PDE.

1.5 Classification Tree

Differential Equations
├── ODE (one independent variable)
│   ├── By order
│   │   ├── First-order: y' = f(x, y)
│   │   ├── Second-order: y'' = f(x, y, y')
│   │   └── n-th order: y^(n) = f(x, y, ..., y^(n-1))
│   ├── By linearity
│   │   ├── Linear: a_n(x)y^(n) + ... + a_0(x)y = g(x)
│   │   │   ├── Homogeneous (g = 0)
│   │   │   └── Nonhomogeneous (g ≠ 0)
│   │   └── Nonlinear (y or derivatives appear nonlinearly)
│   └── By coefficients
│       ├── Constant coefficient
│       └── Variable coefficient
└── PDE (multiple independent variables)
    ├── Elliptic: B² - 4AC < 0  (e.g., Laplace)
    ├── Parabolic: B² - 4AC = 0 (e.g., Heat)
    └── Hyperbolic: B² - 4AC > 0 (e.g., Wave)

1.6 Worked Example: Classifying ODEs

Problem. Classify each equation by order, linearity, and homogeneity (if linear).

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

(b) $(y')^2 + y = 0$

(c) $x^2 y'' + xy' + (x^2 - 1)y = 0$

(d) $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$

2. First-Order ODEs

2.1 Separable Equations

A first-order ODE $\frac{dy}{dx} = f(x, y)$ is separable if $f(x, y) = g(x)h(y)$.

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

Integrating both sides: $\int \frac{dy}{h(y)} = \int g(x)\, dx + C$.

Example. Solve $\frac{dy}{dx} = xy$.

Separating: $\frac{dy}{y} = x\, dx$. Integrating: $\ln|y| = \frac{x^2}{2} + C$. Thus $y = Ce^{x^2/2}$ where $C \neq 0$Plus the trivial solution $y = 0$.

2.2 Linear First-Order Equations

A linear first-order ODE has the form

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

Theorem 2.1 (Integrating Factor). The solution is

$y(x) = e^{-\int P(x)\, dx}\left(\int Q(x) e^{\int P(x)\, dx}\, dx + C\right)$

Proof. Multiply both sides by $\mu(x) = e^{\int P(x)\, dx}$:

$\frac{d}{dx}(\mu y) = \mu \frac{dy}{dx} + \mu P y = \mu \frac{dy}{dx} + \mu' y = \mu\left(\frac{dy}{dx} + Py\right) = \mu Q$

Integrating: $\mu y = \int \mu Q\, dx + C$. Solving for $y$ gives the result. $\blacksquare$

2.3 Worked Example: Linear Equation

Problem. Solve $y' + \frac{2}{x}y = x^2$ for $x > 0$.

Solution. $P(x) = 2/x$, $Q(x) = x^2$.

$\mu(x) = e^{\int 2/x\, dx} = e^{2\ln x} = x^2$.

$y = x^{-2}\left(\int x^2 \cdot x^2\, dx + C\right) = x^{-2}\left(\frac{x^5}{5} + C\right) = \frac{x^3}{5} + \frac{C}{x^2}$. $\blacksquare$

2.4 Exact Equations

The ODE $M(x, y)\, dx + N(x, y)\, dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

When exact, there exists $\Psi(x, y)$ such that $\frac{\partial \Psi}{\partial x} = M$ and $\frac{\partial \Psi}{\partial y} = N$And the solution is $\Psi(x, y) = C$.

Theorem 2.2. If $M$ and $N$ have continuous partial derivatives on a connected domain $D$Then $M\, dx + N\, dy = 0$ is exact if and only if $M_y = N_x$.

Proof. If exact, $M = \Psi_x$ and $N = \Psi_y$So $M_y = \Psi_{xy} = \Psi_{yx} = N_x$ by Clairaut. Conversely, if $M_y = N_x$Define $\Psi(x, y) = \int_{x_0}^x M(t, y)\, dt + \int_{y_0}^y N(x_0, s)\, ds$. Then $\Psi_x = M(x, y)$ and $\Psi_y = \int_{x_0}^x M_y(t, y)\, dt + N(x_0, y) = \int_{x_0}^x N_x(t, y)\, dt + N(x_0, y) = N(x, y) - N(x_0, y) + N(x_0, y) = N(x, y)$. $\blacksquare$

2.5 Worked Example: Exact Equation

Problem. Solve $(2xy + 3)\, dx + (x^2 - 1)\, dy = 0$.

Solution. $M = 2xy + 3$, $N = x^2 - 1$. Check: $M_y = 2x = N_x$. Exact.

$\Psi_x = 2xy + 3 \implies \Psi = x^2 y + 3x + h(y)$.

$\Psi_y = x^2 + h'(y) = x^2 - 1 \implies h'(y) = -1 \implies h(y) = -y$.

Solution: $x^2 y + 3x - y = C$. $\blacksquare$

2.6 Integrating Factors for Non-Exact Equations

If $M_y \neq N_x$One can sometimes find an integrating factor $\mu(x, y)$ such that $(\mu M)_y = (\mu N)_x$.

Case 1: If $\frac{M_y - N_x}{N}$ depends only on $x$Then $\mu(x) = e^{\int \frac{M_y - N_x}{N}\, dx}$.

Case 2: If $\frac{N_x - M_y}{M}$ depends only on $y$Then $\mu(y) = e^{\int \frac{N_x - M_y}{M}\, dy}$.

2.7 Bernoulli Equations

A Bernoulli equation has the form

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

Where $n \neq 0, 1$. The substitution $v = y^{1-n}$ transforms it into a linear equation:

$\frac{dv}{dx} + (1 - n)P(x)v = (1 - n)Q(x)$

Example. Solve $y' + y = y^2 e^x$.

Here $n = 2$So set $v = y^{-1}$. Then $v' = -y^{-2}y'$And the equation becomes $v' - v = -e^x$. Integrating factor: $e^{-x}$. So $(v e^{-x})' = -1$Giving $ve^{-x} = -x + C$, $v = -xe^x + Ce^x$ And $y = 1/(C - x)e^x$.

2.8 Existence and Uniqueness

Theorem 2.3 (Picard-Lindelöf). If $f$ and $\partial f/\partial y$ are continuous on a rectangle Containing $(x_0, y_0)$Then the IVP $y' = f(x, y)$, $y(x_0) = y_0$ has a unique solution in some Neighbourhood of $x_0$.

2.9 Substitution Methods

Several substitutions reduce specific equations to separable or linear form:

1. Homogeneous equations: $y' = f(y/x)$. Set $v = y/x$.
2. Equations of the form $y' = f(ax + by + c)$: set $v = ax + by + c$.

2.10 Homogeneous Equations

An ODE of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ is called homogeneous (not to be Confused with the linearity sense). The substitution $v = y/x$I.e., $y = vx$Gives $y' = v + xv'$ So the equation becomes:

$v + x\frac{dv}{dx} = F(v) \implies x\frac{dv}{dx} = F(v) - v$

This is separable: $\frac{dv}{F(v) - v} = \frac{dx}{x}$.

2.11 Worked Example: Homogeneous Equation

Problem. Solve $y' = \frac{x^2 + y^2}{xy}$.

2.12 Riccati Equations

A Riccati equation has the form

$\frac{dy}{dx} = q_0(x) + q_1(x)y + q_2(x)y^2$

If a particular solution $y_1(x)$ is known, the substitution $y = y_1 + \frac{1}{v}$ reduces the Riccati equation to a linear first-order equation in $v$:

$\frac{dv}{dx} = -(q_1 + 2q_2 y_1)v - q_2$

Example. Solve $y' = 1 + x^2 - 2xy + y^2$ given that $y_1 = x$ is a particular solution.

Substituting $y = x + 1/v$: $y' = 1 - v'/v^2$. The equation becomes

$1 - v'/v^2 = 1 + x^2 - 2x(x + 1/v) + (x + 1/v)^2$

$1 - v'/v^2 = 1 + x^2 - 2x^2 - 2x/v + x^2 + 2x/v + 1/v^2$

$1 - v'/v^2 = 1 + 1/v^2$

$-v'/v^2 = 1/v^2 \implies v' = -1$

So $v = -x + C$And $y = x + \frac{1}{C - x}$.

2.13 Worked Example: Newton’s Law of Cooling

Problem. A body at $90\degree\mathrm{C}$ is placed in a room at $20\degree\mathrm{C}$. After 10 Minutes, its temperature is $60\degree\mathrm{C}$. When will it reach $30\degree\mathrm{C}$?

2.14 Worked Example: Mixing Problem

Problem. A tank contains 100 L of brine with 20 kg of salt. Fresh water enters at 3 L/min and The mixture leaves at 3 L/min. Find the amount of salt after 30 minutes.

2.15 Worked Example: Integrating Factor

Problem. Solve $(3xy + 2y^2)\, dx + (x^2 + 2xy)\, dy = 0$.

2.16 Orthogonal Trajectories

Given a one-parameter family of curves $F(x, y, C) = 0$The orthogonal trajectories are curves That intersect every member of the family at right angles. To find them:

1. Find the differential equation $\frac{dy}{dx} = f(x, y)$ of the given family.
2. Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ (equivalently, negate the slope).
3. Solve the new ODE.

Example. Find the orthogonal trajectories of $y = Cx^2$.

$\frac{dy}{dx} = 2Cx = \frac{2y}{x}$.

Orthogonal trajectories satisfy $\frac{dy}{dx} = -\frac{x}{2y}$.

Separating: $2y\, dy = -x\, dx$. Integrating: $y^2 = -\frac{x^2}{2} + C$Or $\frac{x^2}{2} + y^2 = C$. These are ellipses.

2.17 Common Pitfalls for First-Order ODEs

:::caution Common Pitfall When separating variables, dividing by $h(y)$ can lose solutions where $h(y) = 0$. Always check whether $h(y) = 0$ yields valid solutions before dividing. :::

:::caution Common Pitfall Not every first-order ODE falls into a standard category. Equations like $y' = e^{x^2} + \sin(y^2)$ cannot be solved by elementary methods and require numerical Techniques. :::

3. Second-Order Linear ODEs

3.1 General Theory

A second-order linear ODE has the form

$y'' + p(x)y' + q(x)y = g(x)$

Theorem 3.1. If $y_1$ and $y_2$ are solutions of the homogeneous equation $y'' + py' + qy = 0$ Then $c_1 y_1 + c_2 y_2$ is also a solution (superposition principle).

Theorem 3.2 (Wronskian Criterion). Two solutions $y_1, y_2$ of the homogeneous equation form a fundamental set (i.e., span all solutions) if and only if their Wronskian is non-zero:

$W(y_1, y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \neq 0$

Abel’s identity states that $W(x) = W(x_0) e^{-\int_{x_0}^x p(t)\, dt}$.

Theorem 3.3. The general solution of $y'' + py' + qy = g$ is $y = y_h + y_p$Where $y_h$ is the General solution of the homogeneous equation and $y_p$ is any particular solution.

3.2 Homogeneous Equations with Constant Coefficients

For $y'' + ay' + by = 0$ with $a, b$ constants, try $y = e^{rx}$:

$r^2 + ar + b = 0$

Case 1: Two distinct real roots $r_1 \neq r_2$. $y_h = c_1 e^{r_1 x} + c_2 e^{r_2 x}$.

Case 2: Repeated root $r$. $y_h = c_1 e^{rx} + c_2 x e^{rx}$.

Case 3: Complex conjugate roots $r = \alpha \pm i\beta$. $y_h = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))$.

3.3 Worked Example: Homogeneous Equation

Problem. Solve $y'' - 5y' + 6y = 0$ with $y(0) = 1$, $y'(0) = 0$.

Solution. Characteristic equation: $r^2 - 5r + 6 = (r-2)(r-3) = 0$. Roots: $r = 2, 3$.

$y_h = c_1 e^{2x} + c_2 e^{3x}$.

$y(0) = c_1 + c_2 = 1$. $y'(0) = 2c_1 + 3c_2 = 0$. Solving: $c_1 = 3$, $c_2 = -2$.

$y = 3e^{2x} - 2e^{3x}$. $\blacksquare$

3.4 Worked Example: Complex Roots

Problem. Solve $y'' + 2y' + 5y = 0$ with $y(0) = 1$, $y'(0) = 0$.

3.5 Worked Example: Repeated Roots

Problem. Solve $y'' - 4y' + 4y = 0$ with $y(0) = 1$, $y'(0) = 3$.

3.6 Nonhomogeneous Equations: Undetermined Coefficients

For equations $y'' + ay' + by = g(x)$ where $g(x)$ is a polynomial, exponential, sine, cosine, or Products of these, guess the form of $y_p$ and solve for coefficients.

Rule. If any term of the guess is a solution of the homogeneous equation, multiply by $x$ (or $x^2$ if already multiplied by $x$).

3.7 Worked Example: Undetermined Coefficients

Problem. Solve $y'' - y = 2e^x$.

Solution. Homogeneous: $r^2 - 1 = 0$Roots $\pm 1$. $y_h = c_1 e^x + c_2 e^{-x}$.

Since $e^x$ is a homogeneous solution, guess $y_p = Axe^x$. $y_p' = Ae^x + Axe^x$ $y_p'' = 2Ae^x + Axe^x$. $y_p'' - y_p = (2Ae^x + Axe^x) - Axe^x = 2Ae^x = 2e^x$So $A = 1$.

$y = c_1 e^x + c_2 e^{-x} + xe^x$. $\blacksquare$

3.8 Worked Example: Undetermined Coefficients with Polynomial

Problem. Solve $y'' + 3y' + 2y = x^2 + 1$.

3.8b Worked Example: Undetermined Coefficients with Product

Problem. Solve $y'' + 2y' + y = 3e^{-x}\sin x$.

Consider the forced harmonic oscillator

$y'' + \omega_0^2 y = F_0 \cos(\omega t)$

Case 1: $\omega \neq \omega_0$ (Non-resonant). The particular solution is $y_p = \frac{F_0}{\omega_0^2 - \omega^2}\cos(\omega t)$With bounded amplitude.

Case 2: $\omega = \omega_0$ (Resonant). Since $\cos(\omega_0 t)$ is a homogeneous solution, Guess $y_p = At\sin(\omega_0 t)$. Substituting:

$y_p'' + \omega_0^2 y_p = 2A\omega_0 \cos(\omega_0 t) - A\omega_0^2 t\sin(\omega_0 t) + A\omega_0^2 t\sin(\omega_0 t) = 2A\omega_0 \cos(\omega_0 t)$

Setting equal to $F_0 \cos(\omega_0 t)$: $A = \frac{F_0}{2\omega_0}$.

$y_p = \frac{F_0}{2\omega_0} t \sin(\omega_0 t)$

The amplitude grows linearly with $t$ --- this is resonance. Physically, the system absorbs energy From the periodic forcing at its natural frequency, causing unbounded oscillations.

Worked Example. Solve $y'' + 9y = 6\cos(3t)$, $y(0) = 0$, $y'(0) = 0$.

3.10 Variation of Parameters

Theorem 3.4 (Variation of Parameters). For $y'' + p(x)y' + q(x)y = g(x)$Let $y_1, y_2$ be a Fundamental set of solutions of the homogeneous equation. Then a particular solution is

$y_p = -y_1 \int \frac{y_2 g}{W}\, dx + y_2 \int \frac{y_1 g}{W}\, dx$

Where $W = W(y_1, y_2) = y_1 y_2' - y_2 y_1'$.

Proof. Seek $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$. Impose the constraint $u_1'y_1 + u_2'y_2 = 0$. Then $y_p' = u_1 y_1' + u_2 y_2'$ and $y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''$. Substituting into the ODE: $(u_1'y_1' + u_2'y_2') + u_1(y_1'' + py_1' + qy_1) + u_2(y_2'' + py_2' + qy_2) = g$. Since $y_1, y_2$ satisfy the homogeneous equation, this reduces to $u_1'y_1' + u_2'y_2' = g$. Together With $u_1'y_1 + u_2'y_2 = 0$Solving gives the formulas above. $\blacksquare$

3.11 Worked Example: Variation of Parameters

Problem. Solve $y'' + y = \tan x$ using variation of parameters.

Solution. $y_1 = \cos x$, $y_2 = \sin x$. $W = \cos x \cdot \cos x - \sin x \cdot (-\sin x) = 1$.

$u_1' = -\frac{y_2 g}{W} = -\sin x \tan x = -\frac{\sin^2 x}{\cos x} = -(1 - \cos^2 x)/\cos x = -\sec x + \cos x$.