Differential Equations reference sheet

This sheet follows the thirteen units of the course in order. Each unit starts with the big idea in a few sentences, then gives the definitions, formulas, theorems and methods you will look up most often, with the conditions they need. When you want the full explanation, a derivation or more examples, follow the lesson links under each topic.

Unless a topic says otherwise, the independent variable is $t$ and a prime is a derivative with respect to it, so $y' = dy/dt$. Topics about space or power series use $x$ and say so. A scalar unknown is $y$, a state vector is a plain capital letter such as $X = (x_1, \ldots, x_n)^T$, and matrices are plain capitals such as $A$, $I$ and $\Phi$. Partial derivatives are written $u_t = \partial u/\partial t$ and $u_{xx} = \partial^2 u/\partial x^2$, and the only dot is in $\dot V$, the rate of change of a function $V$ along solutions. The order of an equation is its highest derivative, which is a different thing from the degree of a polynomial. Logarithms are natural logarithms, and the conventions for the Laplace transform, Fourier series and the Fourier transform are stated where they are used. A formula holds wherever both sides are defined unless a condition says otherwise.

Language, Solutions and Modeling

A differential equation relates an unknown function to its derivatives. A solution is a function that satisfies it at every point of an interval, and initial or boundary data pick one solution out of a family. Before using any method, identify the kind of equation, the data and the interval, and check whether a solution is guaranteed to exist and to be unique.

Classifying an equation

An ordinary differential equation (ODE) has one independent variable, and a partial differential equation (PDE) has several and involves partial derivatives. The order is the highest derivative that appears, whatever power it is raised to. An equation of order $n$ is linear when it can be written as

with coefficients $a_k$ and forcing $g$ that depend on $t$ only. It is homogeneous when $g = 0$ and nonhomogeneous otherwise. Powers such as $y^2$ or $(y')^2$, products such as $y\,y'$ and functions such as $\sin y$ make an equation nonlinear. A first-order equation $y' = F(y/x)$ is also called homogeneous, in a second sense that is about scaling, not about forcing.

Lessons Differential equations and their language

Solutions and side conditions

A solution of an equation of order $n$ on an open interval $I$ is a function with $n$ derivatives on $I$ that satisfies the equation at every point of $I$. It is explicit when it is given as a formula for $y$ and implicit when a relation $G(t, y) = 0$ defines $y$ as a differentiable function on $I$. A general solution is a family with arbitrary constants, usually $n$ of them, and a particular solution is one member of the family.

• Initial value problem. The values $y(t_0), y'(t_0), \ldots, y^{(n-1)}(t_0)$ are given at one point.
• Boundary value problem. Conditions are given at two or more points. A linear one has exactly one solution, none or infinitely many.

In the variable $x$, the equation $y'' + y = 0$ has the general solution $C_1\cos x + C_2\sin x$. The initial values $y(0) = 2$ and $y'(0) = -1$ select $2\cos x - \sin x$. The boundary conditions $y(0) = y(\pi) = 0$ allow every $C\sin x$, while $y(0) = 0$ and $y(\pi) = 1$ allow none. The relation $x^2 + y^2 = 25$ gives the explicit solution $y = \sqrt{25 - x^2}$ of $y' = -x/y$ only on $-5 < x < 5$, since $y = 0$ at the endpoints. To verify a solution, substitute it into the equation and the data and state the interval.

Lessons Solutions and conditions

Calculus and complex tools

The product rule $(uv)' = u'v + uv'$ and the chain rule $\frac{d}{dt}f(y(t)) = f'(y(t))\,y'(t)$ are used to check every solution. If $f$ is continuous on an interval containing $t_0$, the solutions of $y' = f(t)$ there are

and an indefinite integral carries one arbitrary constant on each interval. In particular $\int \frac{dt}{t} = \ln\lvert t \rvert + C$ on each interval that does not contain $0$.

A complex number $z = a + ib$ with $i^2 = -1$ has real part $a$, imaginary part $b$, conjugate $\overline{z} = a - ib$ and modulus $\lvert z \rvert = \sqrt{a^2 + b^2}$. The nonreal roots of a polynomial with real coefficients come in conjugate pairs. For real $\theta$, $a$ and $b$, Euler’s formula and the complex exponential are

For every complex constant $r$, $\frac{d}{dt}e^{rt} = re^{rt}$. If a complex function solves a linear equation with real coefficients, its real and imaginary parts are real solutions. For example $(-1 + 2i)^2 + 2(-1 + 2i) + 5 = 0$, so the real part $e^{-t}\cos 2t$ of $e^{(-1 + 2i)t}$ solves $y'' + 2y' + 5y = 0$.

Lessons Calculus and complex numbers

Direction fields

For $y' = f(x, y)$ in the variable $x$, the direction field draws at each point $(x, y)$ a short segment of slope $f(x, y)$. A solution curve is tangent to the field at every one of its points. An isocline $f(x, y) = m$ is a curve along which all segments have the same slope $m$, and in general it is not a solution curve. For $y' = x - y$ the isoclines are the lines $y = x - m$ and the solutions are $y = x - 1 + Ce^{-x}$. The isocline $m = 1$, the line $y = x - 1$, happens to be the solution with $C = 0$, and every other solution approaches it as $x \to \infty$.

Lessons Direction fields

Existence and uniqueness

Consider $y' = f(t, y)$ with $y(t_0) = y_0$.

• Existence. If $f$ is continuous on an open rectangle containing $(t_0, y_0)$, at least one solution exists on some open interval containing $t_0$. The interval may be shorter than the rectangle.
• Uniqueness. If moreover $f$ is locally Lipschitz in $y$, meaning that near each point some constant $L$ gives $\lvert f(t, y_1) - f(t, y_2) \rvert \le L\,\lvert y_1 - y_2 \rvert$, the solution is unique. A continuous $\partial f/\partial y$ is enough for this but not necessary, since $f = \lvert y \rvert$ is Lipschitz.
• Maximal interval. Under both hypotheses on an open set $D$, the solution extends to a largest open interval of existence $(\alpha, \omega)$. If $\omega < \infty$, the point $(t, y(t))$ leaves every closed and bounded subset of $D$ as $t \to \omega^-$, for example by blowing up, and the same holds at $\alpha$.

The theorems are local. The function $f(t, y) = y^2$ is smooth everywhere, yet $y' = y^2$ with $y(0) = 1$ has the solution $y = \frac{1}{1 - t}$ only on $(-\infty, 1)$, which is finite-time blowup. The equation $y' = 2\sqrt{\lvert y \rvert}$ with $y(0) = 0$ has a continuous right side that is not Lipschitz at $y = 0$, and it has infinitely many solutions, the zero solution and, for each $a \ge 0$,

Lessons Existence and uniqueness

Models, units and scaling

A balance law says that the rate of change of an amount equals what flows in minus what flows out, plus what is produced minus what is used up. Every term has the same units, those of the amount per unit time, and every exponent or argument of a function such as $e^{-kt}$ is dimensionless. The state is the quantity the equation evolves, and the parameters are fixed numbers that specify the model. To nondimensionalize, divide the state and the time by characteristic scales, $u = y/Y$ and $\tau = t/T$, chosen so that fewer parameters remain.

For a tank holding $V = 10$ liters, with inflow and outflow of $2$ liters per minute, inflow concentration $3$ grams per liter and $M(0) = 10$ grams of salt, the amount $M$ in grams after $t$ minutes satisfies

The scales $u = M/30$ and $\tau = t/5$ give $\frac{du}{d\tau} = 1 - u$ with $u(0) = \frac{1}{3}$, an equation with no parameters left.

Lessons Modeling and units

First-Order Solution Methods

Many first-order equations can be solved exactly once their form is recognized. Each method below changes the equation into one that can be integrated, and each can lose solutions or hide the interval where a solution is valid, so those are checked after solving.

Separable equations

An equation $y' = a(t)\,b(y)$ is separable. Take $a$ continuous and $b$ continuously differentiable, so that solutions are unique.

1. Find the constant solutions $y = c$ with $b(c) = 0$ first, because dividing by $b(y)$ loses them.
2. Where $b(y) \ne 0$, integrate both sides of $\frac{dy}{b(y)} = a(t)\,dt$.
3. Solve for $y$ if possible, apply the initial value and state the interval where the solution is defined.

By uniqueness a nonconstant solution never reaches a zero of $b$, so the constant solutions and the separated ones do not mix. For $y' = 2ty$ the constant solution is $y = 0$, and elsewhere $\ln\lvert y \rvert = t^2 + C_0$. Together they give $y = Ce^{t^2}$ for every real $C$, and $y(0) = 3$ selects $y = 3e^{t^2}$. An implicit answer is often the natural one. In the variable $x$, the equation $y\,y' = -x$ gives $x^2 + y^2 = C$, and with $y(0) = 2$ the solution is $y = \sqrt{4 - x^2}$ on $-2 < x < 2$.

Lessons Separable equations

Linear equations and integrating factors

Divide $a(t)\,y' + b(t)\,y = g(t)$ by $a(t)$ on an interval where $a(t) \ne 0$ to get the standard form $y' + p(t)\,y = q(t)$. The integrating factor $\mu = e^{\int p\,dt}$ satisfies $\mu' = p\mu$, which turns the left side into a product derivative, $(\mu y)' = \mu q$.

If $p$ and $q$ are continuous on an open interval $I$ containing $t_0$, the problem with $y(t_0) = y_0$ has exactly one solution, it exists on all of $I$, and

For $y' + 2y = 4t$ with $y(0) = 3$, the factor $e^{2t}$ gives $(e^{2t}y)' = 4te^{2t}$ and $y = 2t - 1 + 4e^{-2t}$. Points where the leading coefficient vanishes split the line. In the variable $x$, the equation $x\,y' + y = 1$ has the solutions $y = 1 + C/x$, each defined on $x > 0$ or on $x < 0$ but never across $x = 0$.

Lessons Linear first-order equations

Exact equations and integrating factors

The equation $M(x, y)\,dx + N(x, y)\,dy = 0$, with $M$ and $N$ continuously differentiable on an open set $D$, is exact on $D$ when it has a potential function $F$ there, with $F_x = M$ and $F_y = N$. Its solutions lie on the level curves $F(x, y) = C$, and where $N \ne 0$ they solve $y' = -M/N$.

• Exactness on $D$ implies $M_y = N_x$ on $D$.
• If $D$ is simply connected, such as a disk or a rectangle, then $M_y = N_x$ on $D$ implies exactness. So the test always works locally.
• On a region with a hole it can fail. The form $\frac{-y\,dx + x\,dy}{x^2 + y^2}$ satisfies $M_y = N_x$ on the plane without the origin, but its integral around the unit circle is $2\pi$, so it has no potential there.

To find a potential, integrate $M$ in $x$ to get $F = \int M\,dx + h(y)$, then choose $h$ so that $F_y = N$. For $(2xy + 3x^2)\,dx + (x^2 + 4y)\,dy = 0$ the test gives $M_y = N_x = 2x$, and the potential is $F = x^2y + x^3 + 2y^2$.

A nonexact equation can become exact after multiplication by an integrating factor $\mu$. Two tests look for one that depends on a single variable.

The new equation has the same solutions only where $\mu$ is defined and nonzero, so state that region. For $y\,dx + 2x\,dy = 0$ on $x > 0$, the first quotient is $-\frac{1}{2x}$, so $\mu = x^{-1/2}$, the potential is $F = 2y\sqrt{x}$ and the solutions are $y\sqrt{x} = C$.

Lessons Exact equations, Integrating factors

Substitutions

Each substitution misses some solutions, which have to be added back.

• Homogeneous. Every root $v_0$ of $F(v) = v$ gives a line solution $y = v_0x$ that is lost when dividing by $F(v) - v$, and $x = 0$ is excluded. With $x > 0$, the problem $y' = 1 + y/x$, $y(1) = 0$ gives $xv' = 1$ and $y = x\ln x$.
• Bernoulli. For $n > 0$ the zero function is a solution that the new variable misses, and for fractional $n$ the solution must stay where $y^n$ is defined. For $y' + y = y^2$ with $y(0) = \frac{1}{2}$, the variable $z = 1/y$ satisfies $z' - z = -1$, so $z = 1 + e^t$ and $y = \frac{1}{1 + e^t}$.
• Riccati. The known solution $y_p$ itself corresponds to no finite $u$, and $y$ has a pole wherever $u = 0$. For $y' = y^2 - 1$ with $y_p = 1$ and $y(0) = 0$, the correction satisfies $u' + 2u = -1$ with $u(0) = -1$, so $u = -\frac{1}{2}(1 + e^{-2t})$ and $y = 1 + \frac{1}{u} = -\tanh t$.

Lessons Homogeneous equations, Bernoulli equations, Riccati equations

First-Order Models and Qualitative Behavior

An autonomous equation can be understood from the sign of its right side, without a formula for the solution. The models in this unit, growth and decay, cooling, logistic growth with harvesting, mixing and falling with drag, are first-order equations whose parameters carry units and whose solutions make sense only on a physical domain.

Equilibria and the phase line

For an autonomous equation $y' = f(y)$ with $f$ continuously differentiable, an equilibrium is a number $y_*$ with $f(y_*) = 0$, and $y = y_*$ is then a constant solution. Between two equilibria $f$ keeps one sign, so a solution there increases where $f > 0$, decreases where $f < 0$ and never reaches an equilibrium in finite time. The phase line marks the equilibria on the $y$ axis and draws arrows in these directions.

• Stable. For every $\varepsilon > 0$ there is $\delta > 0$ such that $\lvert y(0) - y_* \rvert < \delta$ gives $\lvert y(t) - y_* \rvert < \varepsilon$ for all $t \ge 0$.
• Asymptotically stable. Stable, and every solution that starts close enough tends to $y_*$.
• Semistable. Solutions approach $y_*$ from one side and move away on the other, so the equilibrium is unstable.

For $y' = y(2 - y)$ the equilibrium $0$ is unstable and $2$ is asymptotically stable, and $y(0) = 1$ gives $y = \frac{2}{1 + e^{-2t}}$, which rises toward $2$. The equation $y' = (y - 1)^2$ is semistable at $1$. From $y(0) = 0$ the solution $1 - \frac{1}{1 + t}$ approaches $1$, while from $y(0) = 2$ the solution $1 + \frac{1}{1 - t}$ blows up at $t = 1$.

Lessons Phase-line stability

Growth, decay and cooling

After one time constant a decaying amount is $Q_0/e$, about 37 percent of its starting value. A half-life of $4$ hours gives $\lambda = \frac{\ln 2}{4}$ per hour, so $80$ grams decays to $40$ grams after $4$ hours and to $20$ grams after $8$ hours. A cup at $90$ degrees in a room at $A = 20$ degrees with $k = \frac{1}{5}$ per minute follows $T = 20 + 70e^{-t/5}$. If the surrounding temperature changes with time, the difference $D = T - A$ obeys $D' = -kD - A'(t)$, which is no longer pure decay.

Lessons Growth, decay and cooling

Logistic growth and harvesting

With growth rate $r > 0$ and carrying capacity $K > 0$, the logistic model and its solution for $P(0) = P_0 > 0$ are

The equilibrium $0$ is unstable and $K$ is asymptotically stable. The growth rate is largest at $P = K/2$, where it equals $rK/4$. With a constant harvesting rate $H \ge 0$ the model becomes $P' = rP(1 - P/K) - H$.

A population that starts below the threshold, or below $K/2$ when $H = rK/4$, or at any level when $H > rK/4$, reaches $0$ in finite time, where the model stops applying. With $r = \frac{1}{5}$ and $K = 100$ the critical rate is $5$, and $H = \frac{16}{5}$ gives the threshold $20$ and the stable level $80$.

Lessons Logistic growth and harvesting

Mixing and compartments

For a perfectly mixed tank holding an amount $M(t)$ of a substance in volume $V(t)$, with inflow rate $q_{\text{in}}$ at concentration $c_{\text{in}}$ and outflow rate $q_{\text{out}}$, all constant,

valid while the tank neither empties nor overflows. Solve for the amount, which is linear in $M$, then compute the concentration $c = M/V$. For $V = 100 + 2t$ liters with $M(0) = 100$ grams, an inflow of $3$ liters per minute at $2$ grams per liter and an outflow of $1$ liter per minute, the equation $M' = 6 - M/V$ has the integrating factor $\sqrt{V}$ and

For several compartments, write one balance per compartment, and each exchange appears with opposite signs in the two compartments it connects. The system $M_1' = -(M_1 - M_2)/10$, $M_2' = (M_1 - M_2)/10$ keeps $M_1 + M_2$ constant, while the difference decays at rate $\frac{1}{5}$. With $M_1(0) = 20$ and $M_2(0) = 10$ this gives $M_1 = 15 + 5e^{-t/5}$ and $M_2 = 15 - 5e^{-t/5}$.

Lessons Mixing and compartments

Drag and terminal velocity

Take velocity $v$ positive downward. A mass $m$ falling under gravity $g$ obeys $m v' = mg + F_{\text{drag}}$, where the drag force always opposes the motion.

The terminal velocity is the equilibrium of the velocity equation, where gravity and drag balance. For quadratic drag the simpler form $-cv^2$ is valid only while $v \ge 0$, because an object moving upward has drag pointing down. With $m = 2$, $g = 10$ and $b = 4$ in SI units, $v_T = 5$ meters per second and, from rest, $v = 5(1 - e^{-2t})$. Quadratic drag with $c = \frac{4}{5}$ has the same terminal velocity but approaches it as $v = 5\tanh 2t$.

Lessons Drag and terminal velocity

Second-Order and Higher-Order Equations

Linear equations have a clear solution structure. The solutions of a homogeneous equation of order $n$ form a family with exactly $n$ free constants, and every solution of a forced equation is one particular solution plus a homogeneous one. With constant coefficients the homogeneous solutions come from the roots of a polynomial.

Linear structure and the Wronskian

Write a linear equation of order $n$ in standard form, with leading coefficient $1$,

A regular interval is an open interval $I$ on which every $p_k$ and $g$ is continuous. For every $t_0$ in $I$ and any $n$ values $y(t_0), \ldots, y^{(n-1)}(t_0)$ there is exactly one solution, and it exists on all of $I$.

• Superposition. Because $L$ is linear, sums and constant multiples of solutions of $L[y] = 0$ are again solutions. This fails for nonlinear equations and for $L[y] = g$ with $g \ne 0$.
• General solution. If $y_1, \ldots, y_n$ is a fundamental set of solutions of $L[y] = 0$ on $I$ and $y_p$ is one solution of $L[y] = g$, every solution is $c_1y_1 + \cdots + c_ny_n + y_p$.
• Wronskian. For two functions $W = y_1y_2' - y_1'y_2$, and for $n$ functions it is the determinant of the matrix whose rows are the functions and their derivatives up to order $n - 1$.

For solutions of $y'' + p(t)\,y' + q(t)\,y = 0$ on $I$, Abel’s identity

shows that $W$ is zero everywhere on $I$ or nowhere, and order $n$ works the same way with $p_{n-1}$ in place of $p$. So $n$ solutions form a fundamental set exactly when $W \ne 0$ at one point of $I$. The test is about solutions. For arbitrary functions $W = 0$ everywhere does not imply dependence, as $t^2$ and $t\lvert t \rvert$ show on the whole line. For $y'' - 3y' + 2y = 0$ the solutions $e^t$ and $e^{2t}$ have $W = e^{3t}$, and $y(0) = 4$, $y'(0) = 5$ select $y = 3e^t + e^{2t}$.

Lessons Superposition and Wronskians

Constant coefficients

For $y'' + ay' + by = 0$ with real constants $a$ and $b$, the function $e^{rt}$ is a solution exactly when $r^2 + ar + b = 0$.

For order $n$, with characteristic polynomial $a_nr^n + \cdots + a_1r + a_0$ and $a_n \ne 0$, each real root $r$ of multiplicity $m$ contributes $e^{rt}, te^{rt}, \ldots, t^{m-1}e^{rt}$, and each pair $\alpha \pm i\beta$ of multiplicity $m$ contributes $t^ke^{\alpha t}\cos\beta t$ and $t^ke^{\alpha t}\sin\beta t$ for $k = 0, \ldots, m - 1$. The multiplicities add up to $n$, so this gives a fundamental set of $n$ functions, and the $n$ initial values fix the constants.

• The problem $y'' + 2y' + 5y = 0$, $y(0) = 3$, $y'(0) = 1$ has roots $-1 \pm 2i$ and the solution $e^{-t}(3\cos 2t + 2\sin 2t)$.
• The polynomial $(r^2 + 1)^2$ belongs to $y^{(4)} + 2y'' + y = 0$, with the solutions $\cos t$, $\sin t$, $t\cos t$ and $t\sin t$.
• The polynomial $(r - 1)^3$ with $y(0) = 1$ and $y'(0) = y''(0) = 0$ gives $y = e^t(1 - t + \frac{1}{2}t^2)$.

Lessons Second-order equations, Higher-order equations

Undetermined coefficients

The solution of $L[y] = g$ is $y = y_c + y_p$, a complementary solution $y_c$ of $L[y] = 0$ plus one particular solution $y_p$. When $L$ has constant coefficients and $g$ is a sum of terms $P_m(t)\,e^{\alpha t}\cos\beta t$ or $P_m(t)\,e^{\alpha t}\sin\beta t$, with $P_m$ a polynomial of degree $m$ and possibly $\beta = 0$, try for each term

Here $A$ and $B$ are polynomials of degree $m$ with unknown coefficients, both needed even when $g$ has only a cosine or only a sine, and $s$ is the multiplicity of $\alpha + i\beta$ as a root of the characteristic polynomial, with $s = 0$ when it is not a root.

The factor $t^s$ handles overlap with the complementary solutions. For $y'' - 2y' + y = e^t$ the root $1$ is double, so $s = 2$, the trial is $At^2e^t$ and $A = \frac{1}{2}$. The annihilator method explains the rule. If a constant-coefficient operator $Q(D)$, with $D = d/dt$, satisfies $Q(D)\,g = 0$, then every solution of $L[y] = g$ solves $Q(D)L[y] = 0$, and the trial is a combination of the basis solutions of that larger equation that do not solve $L[y] = 0$. Here $(D - 1)\,e^t = 0$, so $(D - 1)^3y = 0$ and the new term is $t^2e^t$. For other forcings or variable coefficients, use variation of parameters.

Lessons Undetermined coefficients

Variation of parameters

Take $y'' + p(t)\,y' + q(t)\,y = g(t)$ in standard form, with $p$, $q$ and $g$ continuous on $I$ and a fundamental set $y_1$, $y_2$ with Wronskian $W$. Then

Taking $u_1$ and $u_2$ as integrals from $t_0$ to $t$ gives the particular solution with $y_p(t_0) = y_p'(t_0) = 0$. The leading coefficient must be $1$, so for $P\,y'' + Q\,y' + R\,y = G$ use $g = G/P$ on an interval where $P \ne 0$. For $y'' + y = \sec t$ on $-\frac{\pi}{2} < t < \frac{\pi}{2}$, with $y_1 = \cos t$, $y_2 = \sin t$ and $W = 1$, the derivatives are $u_1' = -\tan t$ and $u_2' = 1$, and

Lessons Variation of parameters