Calculus II

An ordinary differential equation (ODE) is an equation whose unknown is not a number but a function $y(x)$, and which involves its derivatives. "Ordinary" means $y$ depends on a single variable (unlike partial differential equations). The order is that of the highest derivative appearing: here we treat the first order, where at most $y'$ occurs.

The most general form is $F(x,y,y')=0$, but one can almost always solve for the derivative, obtaining the normal form:

How to read it. Think of $y(x)$ as the position of a point and $x$ as time. The equation $y'=f(x,y)$ gives the slope of the trajectory at every point of the plane: at each $(x,y)$ a tiny arrow is drawn with slope $f(x,y)$. Solving the ODE means finding the curves that, instant by instant, follow those arrows. This picture is the direction field.

General and particular solution. Integrating always introduces an arbitrary constant: the general solution of a first-order equation is a family of curves depending on a constant $C$. To pick exactly one we impose an initial condition $y(x_0)=y_0$ — geometrically, "make the curve pass through $(x_0,y_0)$". Equation plus initial condition is the initial value problem (Cauchy problem).

Checking a solution is always easy and worth doing: differentiate the candidate and substitute it into the equation. If both sides agree, it is a solution — no matter how it was found.

This is the simplest type and the first to try. An equation is separable when the right-hand side factors as a function of $x$ alone times a function of $y$ alone:

Idea. Write $y'=\dfrac{dy}{dx}$ and treat $dy$ and $dx$ as quantities to move ("separate") to opposite sides: everything with $y$ on the left, everything with $x$ on the right. Then integrate each side separately.

The result is generally an implicit relation between $x$ and $y$; solve for $y$ when possible.

Mind the constant solutions. Dividing by $h(y)$ implicitly assumed $h(y)\neq 0$. But every value $y_0$ with $h(y_0)=0$ gives the constant solution $y\equiv y_0$: these must be added by hand because separation "loses" them. They are often exactly the equilibrium states of the system.

The fundamental model: $y'=ky$. Separating, $\int dy/y=\int k\,dx$ gives $\ln|y|=kx+C$, i.e. $y=y_0e^{k(x-x_0)}$. With $k>0$ it is exponential growth (populations, compound interest); with $k<0$ exponential decay (radioactive substances, capacitor discharge, Newton's cooling). The constant $1/|k|$ is the characteristic time in which the quantity multiplies/divides by $e$.

An equation is first-order linear if $y$ and $y'$ appear only to the first power, with no products or nonlinear functions:

$Q(x)$ is the forcing term: if $Q\equiv 0$ the equation is homogeneous (and also separable), otherwise complete.

The integrating-factor trick. We want the left side to be the derivative of a product. Multiply everything by an unknown function $\mu(x)$: $\mu y' + \mu P y = \mu Q$. For the left side to equal $(\mu y)' = \mu y' + \mu' y$, we need $\mu' = \mu P$, i.e. (separable!) $\mu(x)=e^{\int P(x)\,dx}$. This $\mu$ is the integrating factor.

With such $\mu$ the equation becomes $(\mu y)' = \mu Q$, integrated directly:

Structure of the solution. It always reads $y = y_{\text{hom}} + y_{\text{part}}$: a part $y_{\text{hom}}=C/\mu$ solving the homogeneous equation (the "memory" of the initial conditions, usually a vanishing transient) plus a particular solution forced by the source term (the "steady state"). The same structure recurs in second-order ODEs and electrical circuits.

Physical example: RC circuit. The charge $q$ on a capacitor in series with resistance $R$ and source $V$ obeys $R\,q' + q/C = V$, i.e. $q' + q/(RC)=V/R$: first-order linear. The solution $q(t)=CV(1-e^{-t/RC})$ shows the charge tending to the steady value $CV$ with time constant $\tau=RC$.

The Bernoulli equation has the form

It is nonlinear because of the $y^n$ term, but a substitution turns it linear. (The cases $n=0$ and $n=1$ are already linear or separable, hence excluded.)

The substitution. Divide by $y^n$: $y^{-n}y' + P\,y^{1-n}=Q$. Set $z = y^{1-n}$; then $z' = (1-n)y^{-n}y'$, i.e. $y^{-n}y' = z'/(1-n)$. Substituting gives a linear equation in $z$:

Solve it with the integrating factor (§3), then return to $y$ via $y = z^{1/(1-n)}$ (remembering the possible solution $y\equiv 0$ lost in the division).

Where it appears. Logistic growth $y'=ry(1-y/K)$ is a Bernoulli with $n=2$; it also shows up in fluid and chemical-reaction models. Bernoulli's value is the general method: "if you see $y^n$ on the right and linear on the left, substitute $z=y^{1-n}$".

An equation $y'=f(x,y)$ is homogeneous (of degree zero) if the right-hand side depends only on the ratio $y/x$, i.e. $y' = F\!\left(\dfrac{y}{x}\right)$. This typically happens when $f$ is a quotient of polynomials of the same degree.

The substitution. Set $v = y/x$, so $y = vx$ and $y' = v + x\,v'$. The equation becomes $v + x v' = F(v)$, i.e.

which is separable in $v$ and $x$:

Once solved for $v$, return to $y = vx$. Values $v_0$ with $F(v_0)=v_0$ give the straight-line solutions $y=v_0 x$ (lines through the origin), often geometrically meaningful.

Write the equation in symmetric differential form:

It is exact if the left side is the total differential $dF$ of a function $F(x,y)$, i.e. if there is an $F$ with $F_x = M$ and $F_y = N$. Then $dF=0$ along solutions, so the general solution is simply $F(x,y)=C$ (level curves of $F$).

Exactness criterion. Since mixed second partials agree ($F_{xy}=F_{yx}$), a necessary condition — and, on a simply connected domain, sufficient — is:

How to find $F$. Integrate $F_x=M$ in $x$ to get $F=\int M\,dx + \varphi(y)$ (the "constant" may depend on $y$); then impose $F_y=N$ to determine $\varphi(y)$. This is the same procedure as the potential of a conservative field, here in two dimensions.

If not exact. Sometimes an integrating factor $\mu$ makes $\mu M\,dx+\mu N\,dy=0$ exact. Two practical cases: if $(M_y-N_x)/N$ depends only on $x$, then $\mu(x)=e^{\int (M_y-N_x)/N\,dx}$; if $(N_x-M_y)/M$ depends only on $y$, then $\mu(y)=e^{\int (N_x-M_y)/M\,dy}$.

Before solving it is worth asking: does a solution exist? Is it unique? The existence and uniqueness theorems answer without computing the solution.

Peano's theorem (existence). If $f(x,y)$ is continuous near $(x_0,y_0)$, the Cauchy problem $y'=f(x,y)$, $y(x_0)=y_0$ has at least one local solution. Continuity guarantees existence, but not uniqueness.

Picard–Lindelöf theorem (existence and uniqueness). If in addition $f$ is Lipschitz in $y$, i.e. there is $L$ with

then the local solution is unique. In practice it suffices that $\partial f/\partial y$ be continuous (bounded) near the point: the Lipschitz condition follows.

Why it matters. Uniqueness has a strong physical meaning: if a deterministic system starts in a precise state, its future is determined. When uniqueness fails, several solutions emerge from one initial datum — the system "branches". Classic example: $y'=\sqrt{|y|}$, $y(0)=0$ has both $y\equiv 0$ and $y=x^2/4$ as solutions, because $\sqrt{|y|}$ is not Lipschitz at $y=0$ (the slope blows up).

Global existence. The hypotheses guarantee only a local solution: it may "blow up" in finite time (e.g. $y'=y^2$, $y(0)=1$ gives $y=1/(1-x)$, diverging at $x=1$). The maximal interval of existence is generally smaller than all of $\mathbb{R}$.