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A differential equation is an equation involving an unknown function and its derivatives. They are the natural language of physics, engineering, biology, and economics — anywhere a system evolves over time according to local rules.

A first-order differential equation is one in which only the first derivative appears. They are the simplest case but appear constantly in applications. This article covers what an ODE is and is not, separable equations, linear first-order equations and the integrating factor method, exact equations, substitution methods, and applications in physics, biology, and economics.

What an ODE is (and isn't)

An ordinary differential equation (ODE) is an equation involving a function of one variable and its derivatives. Notation: $y(x)$ is the unknown function, $y^{\prime }(x),y^{\prime \prime }(x),\dots$ are its derivatives.

Example: $\frac{dy}{dx}=2x$, which we'd write as $y^{\prime }=2x$. The solution is $y=x^2+C$ for any constant $C$.

A partial differential equation (PDE) involves a function of multiple variables and partial derivatives. These are harder and beyond the scope of this article. Examples: heat equation, wave equation, Maxwell's equations.

Order of an ODE: the highest derivative that appears. First-order: only $y^{\prime }$. Second-order: includes $y^{\prime \prime }$. Etc.

Linear vs nonlinear: An ODE is linear if it can be written in the form $a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots +a_0(x)y=f(x)$. Otherwise nonlinear. Linear equations are usually much easier than nonlinear ones.

Initial value problem (IVP): an ODE together with a condition $y(x_0)=y_0$. Most physical problems are IVPs — you know the state at some time and want to know it at later times.

Separable equations

The simplest case. An ODE is separable if it can be written as

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

Solve by separating variables and integrating:

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

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

Example 1: Solve $y^{\prime }=xy$.

Separate: $\frac{dy}{y}=xdx$. Integrate: $\ln |y|=\frac{x^2}{2}+C_1$. Exponentiate: $y=C{e}^{x^2/2}$ (where $C=\pm e^{C_1}$).

Example 2: A radioactive substance decays at a rate proportional to its amount. If $N(t)$ is the amount at time $t$, then $N^{\prime }=-kN$ for some $k>0$.

Separate and integrate: $N(t)=N_0{e}^{-kt}$, where $N_0=N(0)$.

This is the famous exponential decay formula. Half-life: solve $N(t)=N_0/2$ for $t$: $t_{1/2}=\frac{\ln 2}{k}$.

Linear first-order equations: the integrating factor

A linear first-order ODE has the form

$y^{\prime }+p(x)y=q(x)$

The technique: multiply both sides by an integrating factor $\mu (x)$ to make the left side a derivative of a product.

The integrating factor is

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

Why this works: $(\mu y{)}^{\prime }=\mu y^{\prime }+{\mu }^{\prime }y=\mu (y^{\prime }+py)=\mu q$. So $\mu y=\int \mu qdx+C$.

Example 3: Solve $y^{\prime }+2xy=x$.

Integrating factor: $\mu =e^{\int 2xdx}=e^{x^2}$. Multiply both sides:

$e^{x^2}{y}^{\prime }+2x{e}^{x^2}y=x{e}^{x^2}$

The left side is $(e^{x^2}y{)}^{\prime }$. Integrate both sides:

$e^{x^2}y=\int x{e}^{x^2}dx=\frac{1}{2}{e}^{x^2}+C$

So $y=\frac{1}{2}+C{e}^{-x^2}$.

Example 4 (with initial condition): Solve $y^{\prime }+y=e^{-x}$, $y(0)=1$.

Integrating factor: $\mu =e^x$. Multiply: $e^x{y}^{\prime }+e^{x}y=1$. So $(e^{x}y{)}^{\prime }=1$. Integrate: $e^{x}y=x+C$. So $y=(x+C)e^{-x}$.

Apply initial condition: $1=(0+C)\cdot 1$, so $C=1$. Solution: $y=(x+1)e^{-x}$.

Exact equations

An ODE in the form $M(x,y)dx+N(x,y)dy=0$ is exact if there exists a function $F(x,y)$ such that $\frac{\partial F}{\partial x}=M$ and $\frac{\partial F}{\partial y}=N$.

Test for exactness: $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

If exact, find $F$ by integrating $M$ with respect to $x$ (treating $y$ as constant) and using the second condition to determine the constant. The solution is $F(x,y)=C$.

Example 5: Solve $(2x+y)dx+(x+2y)dy=0$.

Check: $\frac{\partial M}{\partial y}=1$ and $\frac{\partial N}{\partial x}=1$. Exact.

Integrate $M$ with respect to $x$: $F=x^2+xy+g(y)$. Differentiate with respect to $y$: $\frac{\partial F}{\partial y}=x+g^{\prime }(y)$. Set equal to $N=x+2y$: $g^{\prime }(y)=2y$, so $g(y)=y^2$.

Solution: $x^2+xy+y^2=C$.

Substitution methods

For some equations, a clever substitution converts the problem to a solvable form.

Bernoulli equation: $y^{\prime }+p(x)y=q(x)y^n$. Substitute $v=y^{1-n}$. The equation becomes linear in $v$.

Homogeneous equation (in the ODE sense, not the linear-algebra sense): $y^{\prime }=F(y/x)$. Substitute $v=y/x$. The equation becomes separable in $v$.

Example 6: Solve $y^{\prime }+y=y^2$.

This is Bernoulli with $n=2$. Substitute $v=y^{-1}$, so $y=1/v$ and $y^{\prime }=-v^{\prime }/v^2$.

Original: $-v^{\prime }/v^2+1/v=1/v^2$. Multiply by $-v^2$: $v^{\prime }-v=-1$. Linear in $v$, integrating factor $e^{-x}$: $(e^{-x}v{)}^{\prime }=-e^{-x}$, so $e^{-x}v=e^{-x}+C$, so $v=1+C{e}^x$.

Substitute back: $y=\frac{1}{1+C{e}^x}$.

Applications

Population dynamics

The simplest model: population grows at a rate proportional to its size. $P^{\prime }=rP$, where $r$ is the growth rate. Solution: $P(t)=P_0{e}^{rt}$. Exponential growth.

A better model: the logistic equation $P^{\prime }=rP(1-P/K)$, where $K$ is the carrying capacity. The growth rate slows as $P$ approaches $K$. Solution: $P(t)=\frac{K}{1+A{e}^{-rt}}$ for some constant $A$ determined by initial conditions.

Newton's law of cooling

The rate of cooling of an object is proportional to the difference between its temperature and the surroundings:

$T^{\prime }=-k(T-T_{\text{env}})$

This is linear first-order. Solution: $T(t)=T_{\text{env}}+(T_0-T_{\text{env}})e^{-kt}$.

The temperature approaches the environment temperature exponentially, with time constant $1/k$.

Compound interest

Continuously compounded interest at rate $r$ gives $A^{\prime }=rA$, with solution $A(t)=A_0{e}^{rt}$. This is why "continuously compounded" gives slightly higher returns than discrete compounding: the equation is solved exactly by the exponential.

Chemical reactions

First-order reactions follow $C^{\prime }=-kC$, where $C$ is the concentration of the reactant. Solution: $C(t)=C_0{e}^{-kt}$. The half-life $t_{1/2}=\ln 2/k$ is constant — a defining feature of first-order kinetics.

Second-order reactions follow $C^{\prime }=-k{C}^2$, which is separable. Solution: $\frac{1}{C}=\frac{1}{C_0}+kt$. The half-life for second-order reactions depends on initial concentration.

A note on numerical methods

Most real-world ODEs cannot be solved in closed form. When the analytical methods above fail, numerical methods compute approximate solutions.

The simplest is Euler's method: $y_{n+1}=y_n+h\cdot f(x_n,y_n)$, where $h$ is the step size. This is first-order accurate.

Better methods include Runge-Kutta (the standard 4th-order RK4 is the workhorse) and adaptive step-size methods. Modern numerical packages (scipy.integrate.solve_ivp in Python, ode45 in MATLAB) implement these efficiently.

Common pitfalls

Pitfall 1 — Forgetting the constant of integration. Every indefinite integral introduces a constant. For an IVP, the constant is determined by the initial condition.

Pitfall 2 — Missing solutions. When you separate $\frac{dy}{h(y)}=g(x)dx$, you implicitly assume $h(y)\ne 0$. The solutions where $h(y)=0$ (called equilibrium solutions) must be checked separately. Example: in $y^{\prime }=y(1-y)$, the solutions $y=0$ and $y=1$ are equilibria you'd miss by separation alone.

Pitfall 3 — Confusing order and degree. Order is the highest derivative; degree is the highest power of the highest derivative. $y^{\prime \prime }+(y^{\prime }{)}^3=0$ is order 2, degree 1 (in $y^{\prime \prime }$).

Pitfall 4 — Applying the integrating factor to nonlinear equations. The integrating factor method only works for linear equations. For nonlinear, use other techniques.