Why Differential Equation Notation Matters
A derivative describes how a quantity changes as its input changes. A differential equation uses one or more derivatives to state a relationship that an unknown function must satisfy. Reading the notation correctly tells us what is changing, which variable it changes with respect to, and what slope the equation prescribes.
In this unit, a central form is \(y'=f(x,y)\). The equation can be read geometrically: at a point with coordinates \((x,y)\), the value \(f(x,y)\) specifies a slope. This interpretation is the starting point for studying slope fields, but first it is important to distinguish the variables and the equation’s order.
Independent and Dependent Variables
For a function \(y=y(x)\), the independent variable is the input \(x\). The dependent variable is the output \(y\), because its value depends on the chosen input. The notation \(y'\) abbreviates the derivative \(y'(x)=\frac{dy}{dx}\): the rate at which \(y\) changes with respect to \(x\).
The letters are not fixed by the mathematics. In an equation involving time, for example, \(t\) is often the independent variable and \(y(t)\) is the dependent variable. Then \(\frac{dy}{dt}\) describes the rate of change of \(y\) with respect to time. The variable in the denominator of Leibniz notation identifies what is being varied.
In \(y'=f(x,y)\), the occurrence of \(y\) on the right means the derivative can depend on the current value of the unknown function as well as on \(x\). More fully, the equation means
The shorter notation suppresses the repeated \(x\) inside \(y(x)\). This suppression is convenient, but it does not make \(x\) and \(y\) interchangeable: \(x\) is the input, while \(y\) is the function’s value at that input.
What “First-Order” Means
The order of a differential equation is the highest derivative of the unknown function that appears in the equation. Thus an equation involving \(y'\) but no higher derivative is first-order. An equation involving \(y''\) is second-order, even if it also contains \(y\) and \(y'\).
For example, \(y'=x+y\) is first-order. So is \(3y'-2y=x\), because its highest derivative is still \(y'\). By contrast, \(y''+3y'=t\) is second-order because \(y''\) is the highest derivative. The order is not determined by how many terms appear or by whether the equation is written in a particular arrangement.
An equation may be written explicitly as \(y'=f(x,y)\), with the first derivative isolated, or implicitly, with derivative terms mixed into another relation. An implicit first-order equation is still first-order if it involves no derivative higher than \(y'\). But it may not determine a single value of \(y'\) at every point.
From a Formula to a Slope at a Point
Suppose \(f\) is defined at a point \((a,b)\). For an equation \(y'=f(x,y)\), substituting \(x=a\) and \(y=b\) gives the prescribed derivative value at that point. If a solution curve passes through \((a,b)\), then \(y(a)=b\), and its derivative there must satisfy
This is a pointwise rule: the coordinates are inputs to \(f\), and the output is a slope. The equation describes a constraint on any solution, not a list of solution values. In particular, evaluating \(f(a,b)\) tells us what slope a solution through the point must have; by itself, it does not prove that a solution exists or that it is unique.
Worked Example: Evaluating a Prescribed Slope
Consider \(y'=2x+y\). Find the slope prescribed at the point \((1,3)\).
Here, \(x=1\) and \(y=3\). Substitute both coordinates into the right-hand side:
The equation prescribes slope \(5\) at \((1,3)\). If a differentiable solution passes through this point, then \(y(1)=3\) and \(y'(1)=5\).
Check: Substituting \(x=1\), \(y=3\), and \(y'=5\) into the original equation gives \(5=2(1)+3=5\), as required.
Why the Prescribed Number Is a Geometric Slope
The derivative at a point gives the slope of the tangent to a differentiable graph there. This follows from the definition of derivative. If the graph is \(y=y(x)\), its secant slope from \(x=a\) to \(x=a+h\), for \(h\ne 0\), is
As \(h\) approaches \(0\), the secant line approaches the tangent line, and the secant slope approaches \(y'(a)\), provided the derivative exists. Equivalently, the graph can be represented near the point by the position vector \((x,y(x))\). Its derivative with respect to \(x\) is \((1,y'(x))\), so at \(x=a\) the tangent direction is \((1,y'(a))\). Since the horizontal component is \(1\), the tangent’s slope is its vertical component, \(y'(a)\).
Now suppose \(y'=f(x,y)\) and the graph passes through \((a,b)\). By the equation, its derivative there is \(f(a,b)\). By the derivative interpretation just established, its tangent slope is therefore \(f(a,b)\). This is why the formula can be read as assigning a slope to each point where \(f\) is defined. It assigns a prescribed slope; it does not itself draw a graph or guarantee a solution through every point.
Reading Notation in Different Variables
Worked Example: Time as the Independent Variable
Consider \(\frac{dy}{dt}=t-y\) with the condition \(y(0)=2\). Identify the independent and dependent variables, determine the order, and find the prescribed slope at the point specified by the condition.
The derivative is with respect to \(t\), so \(t\) is the independent variable and \(y=y(t)\) is the dependent variable. The highest derivative is \(\frac{dy}{dt}\), so the equation is first-order.
The condition \(y(0)=2\) specifies the point \((t,y)=(0,2)\). Substituting its coordinates into the right-hand side gives
The equation prescribes slope \(-2\) at that point. Here slope means change in \(y\) per unit change in \(t\).
Check: With \(t=0\), \(y=2\), and \(\frac{dy}{dt}=-2\), the original equation reads \(-2=0-2=-2\).
A condition such as \(y(0)=2\) is called an initial condition when it specifies the value of the dependent variable at a given value of the independent variable. Together, a differential equation and an initial condition describe a problem with both a rate rule and a specified point. The condition identifies the point; the equation specifies the slope there.
Identifying Order in Implicit Equations
Worked Example: Finding the Order Before Rearranging
Consider \(x y''+(y')^2=1\). Identify the independent variable, dependent variable, and order. Is this a first-order equation?
The derivatives are written with respect to \(x\), so \(x\) is independent and \(y=y(x)\) is dependent. The equation contains \(y''\), the second derivative, and no derivative higher than \(y''\). Therefore its order is \(2\).
The term \((y')^2\) does not change the order: it is a power of the first derivative, not a second derivative. The presence of \(y''\) makes the equation second-order, so it is not first-order.
Check: The highest derivative displayed in the original equation is \(y''\); hence the order is \(2\), in agreement with the classification.
When an Implicit Equation Does Not Specify One Slope
It is important not to assume that every first-order equation can be rearranged into \(y'=f(x,y)\) at every point. Some implicit equations determine one derivative only under a condition, and some fail to determine one derivative at particular points.
Worked Example: A Slope Rule That Breaks Down at Some Points
Consider \(y\,y'=x\). This equation involves only the first derivative, so it is first-order. When \(y\ne 0\), divide both sides by \(y\) to obtain
At \((2,1)\), the equation therefore prescribes
Check at \((2,1)\): Substitution into the implicit equation gives \(1\cdot 2=2\), which matches \(x=2\).
At \((2,0)\), however, the original equation would require \(0\cdot y'=2\), which is impossible for any finite value of \(y'\). So no differentiable solution satisfying the equation can pass through that point.
At \((0,0)\), the equation reads \(0\cdot y'=0\). This equality holds for every value of \(y'\) at that point, so the equation there does not determine a unique slope. The rearranged formula \(x/y\) cannot be used when \(y=0\), because division by zero is undefined.
Check: The original equation gives different conclusions at \((2,0)\) and \((0,0)\); treating \(x/y\) as valid at either point would ignore its restriction \(y\ne 0\).
A Common Notation Mistake
A frequent error is to substitute only the independent-variable coordinate into \(f\), leaving out the dependent-variable coordinate. For \(y'=2x+y\), a mistaken calculation at \((1,3)\) might say, “The slope is \(2(1)=2\).” This is wrong because the right-hand side includes \(y\) as well as \(x\). The correct calculation is \(2(1)+3=5\).
Checking a Proposed Solution
A function is a solution of a differential equation on an interval if it is differentiable there and, after substitution, satisfies the equation at every point of that interval. A single correct slope calculation at one point is not enough to verify a function as a solution on an interval.
Worked Example: Verifying a Function
Check whether \(y=e^x\) is a solution of \(y'=y\) on \(\mathbb{R}\).
Differentiate the proposed function. Using \(\frac{d}{dx}e^x=e^x\), we get \(y'=e^x\). The right-hand side of the equation is \(y=e^x\). Thus, for every real \(x\),
Therefore \(y=e^x\) satisfies the equation on \(\mathbb{R}\).
Check: Substitution of \(y=e^x\) and \(y'=e^x\) into \(y'=y\) gives \(e^x=e^x\) for every \(x\), so the equation holds throughout the stated interval.
Summary
In \(y'=f(x,y)\), \(x\) is the independent variable, \(y=y(x)\) is the dependent variable, and \(y'\) is shorthand for the derivative of \(y\) with respect to \(x\). The equation is first-order because its highest derivative is \(y'\). At a point \((a,b)\) in the domain of \(f\), it prescribes slope \(f(a,b)\) for any solution passing through that point. Implicit equations require care: rearrangement may involve division, and the equation may fail to determine a slope at some points.
Check Your Understanding
For each question, pay attention to the variables and to the conditions under which a slope is defined.
- In \(\frac{dy}{dt}=t+2y\), identify the independent variable, dependent variable, and order.
- For \(y'=x-3y\), what slope is prescribed at \((2,1)\)?
- What is the order of \(y''+4y'=x\), and which derivative determines that order?
- For \(y\,y'=x\), at which points is it valid to rearrange to \(y'=x/y\)?
- What two checks are needed to verify that a proposed differentiable function solves a differential equation on an interval?
