Differential Equations › Section A · Basic concepts › Topic A.1
Definitions and classification
Before you can solve a differential equation you need to know what kind it is. A handful of labels, order, ordinary or partial, linear or nonlinear, tell you which methods can work, and a few more say exactly what counts as an answer.
Equations about rates
A differential equation is any equation that involves derivatives of an unknown function. You have met one already: Newton's second law, force equals mass times acceleration. Since acceleration is the derivative of velocity, the law for an object of mass \(m\) reads
where the force may depend on time and on the velocity itself, as air resistance does. The unknown is not a number but a whole function \(v(t)\). Equations like this describe any system whose future is set by its present state and the laws acting on it: populations, circuits, chemical reactions, springs and orbits.
Order
The order of a differential equation is the order of the highest derivative that appears. The equation \(y' = ty\) is first order; \(y'' + 9y = 0\) is second order; \(y''' - y' = t\) is third order. Powers do not change the order: \((y')^4 = y + 1\) is still first order, because the highest derivative is \(y'\), however it is raised. Order matters because it predicts how much information a solution needs. A first-order equation usually has a one-parameter family of solutions, a second-order equation a two-parameter family, and so on.
Ordinary or partial
If the unknown function depends on a single variable, the derivatives are ordinary derivatives and the equation is an ordinary differential equation, or ODE. If the unknown depends on several variables, the equation contains partial derivatives and is a partial differential equation, or PDE. The temperature \(u(x, t)\) along a metal rod depends on position and time and satisfies the heat equation
which is a second-order PDE. Almost all of this course concerns ODEs; the final section returns to PDEs, using ODE methods to solve them.
Linear or nonlinear
An ODE is linear if it can be written as
Read the form carefully: \(y\) and its derivatives appear only to the first power, never multiplied together and never inside another function such as \(\sin y\) or \(e^y\). The coefficients \(a_k(t)\) and \(g(t)\) can be any functions of \(t\), however complicated. Anything else is nonlinear. The pendulum equation \(\theta'' + \sin\theta = 0\) is nonlinear because of \(\sin\theta\). Linear equations have a well-developed general theory; nonlinear ones are often solvable only numerically.
Worked example: classifying four equations
Classify each equation by type, order and linearity. 1. \(t^2y'' + (\sin t)y = e^t\): an ODE of second order, and linear, because \(t^2\), \(\sin t\) and \(e^t\) involve only \(t\). 2. \(y' = y^2 + t\): a first-order ODE, nonlinear because of \(y^2\). 3. \(y''' + yy' = 0\): a third-order ODE, nonlinear because \(y\) multiplies its own derivative. 4. \(u_t = 3u_{xx}\): a second-order PDE, linear. Check: in each case the decision rested only on how the unknown and its derivatives appear, never on how complicated the functions of \(t\) are.
What counts as a solution
A solution of a differential equation on an interval is a function that, when it and its derivatives are substituted, makes the equation true at every point of that interval. The interval is part of the answer. A formula can satisfy an equation symbolically yet fail to exist, or fail to be differentiable, at some points, and a solution cannot cross such a point. A useful rule: an equation with real coefficients should have real-valued solutions, so values of \(t\) that would force square roots of negative numbers are excluded too. Checking a solution is always mechanical: differentiate, substitute, simplify.
Worked example: checking a solution
Show that \(y = \sqrt{t}\) solves \(4t^2y'' + y = 0\), and find a suitable interval. 1. Write \(y = t^{1/2}\), so \(y' = \frac{1}{2}t^{-1/2}\) and \(y'' = -\frac{1}{4}t^{-3/2}\). 2. Substitute:
3. Interval: \(\sqrt{t}\) is not real for \(t < 0\), and \(y'\) does not exist at \(t = 0\), so the solution holds on \(t > 0\). Check: \(y = \sqrt{t}\ln t\) also passes the same substitution, so this equation has more than one solution. That is typical, and it leads to the next idea.
General solutions and initial value problems
The general solution is a formula, with arbitrary constants, that captures every solution, at least on a given interval. For an \(n\)th-order equation it usually carries \(n\) constants. To pick out one member you add initial conditions: values of the solution and its derivatives at a single point \(t_0\), such as \(y(t_0) = y_0\) and \(y'(t_0) = y_1\). An equation together with the right number of initial conditions, one per unit of order, is an initial value problem, or IVP. The member that meets the conditions is the particular solution, which some books call the actual solution.
Worked example: from general to particular
The general solution of \(ty' - y = t^2\) is \(y = t^2 + Ct\). Verify it, then solve the IVP with \(y(1) = 4\). 1. Differentiate: \(y' = 2t + C\). 2. Substitute:
as required, for every \(C\). 3. Apply the condition: \(y(1) = 1 + C = 4\), so \(C = 3\). 4. The particular solution is \(y = t^2 + 3t\). Check: \(y(1) = 1 + 3 = 4\). The order of work matters: find the general solution first, then use the condition to fix the constant.
Explicit and implicit solutions
An explicit solution gives \(y\) directly as a function, \(y = y(t)\). An implicit solution is an equation linking the variables that defines \(y\) without isolating it. For instance, \(x^2 + xy + y^2 = 7\) is an implicit solution of
as implicit differentiation of the curve shows, and the point \((1, 2)\) lies on it. Solving for \(y\) gives two branches, from the quadratic formula, and the initial condition \(y(1) = 2\) selects the one with the plus sign. Implicit answers are complete answers; many equations have no explicit solution at all.
Common mistakes to avoid
First, confusing powers with order: \((y')^3 = t\) is first order, and the \(t^3\) in \(t^3y'' = y\) does not make that equation third order or nonlinear. Second, judging linearity by the coefficients: \(e^{t^2}y' = y\) is linear, while \(y' = \cos y\) is not, because only the unknown's appearance counts. Third, applying an initial condition before finding the general solution. Fourth, giving a solution formula with no interval, or an interval that crosses a point where the formula breaks down. Fifth, keeping both branches of an explicit solution when the initial condition allows only one.
Key takeaways
A differential equation involves derivatives of an unknown function. Its order is the highest derivative present. It is an ODE if the unknown depends on one variable and a PDE if it depends on several. It is linear if the unknown and its derivatives appear only to the first power, with coefficients depending on the independent variable alone. A solution satisfies the equation on an interval. The general solution carries one constant per order, and an IVP adds that many conditions at one point to give a particular solution, which may be explicit or implicit.
Flashcards
Flashcard
An equation involving derivatives of an unknown function
Test your understanding
Question 1 of 5
How is t³y′′ − 4y′ + y = cos t classified?
Question 2 of 5
Which of these equations is nonlinear?
Question 3 of 5
Which function is a solution of y′′ + 4y = 0?
Question 4 of 5
The general solution of y′ + 3y = 6 is y = Ce^(−3t) + 2. Which function satisfies y(0) = 7?
Question 5 of 5
What does an initial value problem for a third-order equation need?