Differential Equations

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.

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.

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.

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.

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)

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

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 $y = 0$.

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. $\blacksquare$

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

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

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

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

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 $\Psi_y = \int_{x_0}^x M_y(t, y)\, dt + N(x_0, y) = \int_{x_0}^x N_x(t, y)\, dt + N(x_0, y) = N(x, y) - N(x_0, y) + N(x_0, y) = N(x, y)$. $\blacksquare$

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

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 $\mu(y) = e^{\int \frac{N_x - M_y}{M}\, dy}$.

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 $y = 1/(C - x)e^x$.

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 $x_0$.

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 $v = ax + by + c$.

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

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

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

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 $30\degree\mathrm{C}$?

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.

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

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.

:::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. :::

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.

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 $r = \alpha \pm i\beta$. $y_h = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))$.

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: $c_1 = 3$, $c_2 = -2$.

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

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

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

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 $x$).

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 $A = 1$.

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

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

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

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. $\blacksquare$