One Slope Rule, Many Curves
A differential equation such as \(y'=2x\) gives a rule for the slope of a solution curve at each point. At \(x=1\), for instance, any solution curve must have slope \(2\). The equation does not, by itself, say which height the curve has at \(x=1\). Curves can follow the same slope rule while passing through different points.
An initial condition supplies that missing information. It specifies the value of the dependent variable at a particular value of the independent variable. The equation and the initial condition together can identify one curve from a family of possible solutions.
Solution Families and Particular Solutions
A solution of a differential equation on an interval is a differentiable function that satisfies the equation at every \(x\) in that interval. For example, to be a solution of \(y'=2x\), a function \(y=f(x)\) must satisfy \(f'(x)=2x\) throughout its interval.
For this equation, every function of the form \(y=x^2+C\), where \(C\) is a constant, is a solution: its derivative is \(2x\), whatever value \(C\) has. Thus the equation admits a family of solution curves. Each value of \(C\) gives a different vertical position, while leaving the slope rule unchanged.
An initial condition is a specified value \(y(x_0)=y_0\). It says that at the input \(x_0\), the output must be \(y_0\). Geometrically, the graph must pass through \((x_0,y_0)\). A solution that also satisfies the initial condition is called a particular solution of the initial-value problem.
The distinction between these pieces of information matters. The differential equation specifies the derivative; the initial condition specifies a function value. In general, the number \(y_0\) is not the slope at \(x_0\). The slope there is determined by the differential equation and the point, while the initial condition determines the curve's required height.
Why a Point Selects a Parameter
Suppose a solution family is written using a parameter \(C\). To find the member passing through \((x_0,y_0)\), substitute \(x_0\) for \(x\) and \(y_0\) for \(y\) in the family. This gives an equation for \(C\). Solve it, then place the resulting value back into the family.
Keep the arbitrary constant visible so the possible members are clear.
Set the family’s value at \(x_0\) equal to \(y_0\).
Use the equation from the initial condition to determine \(C\).
Differentiate the proposed function and substitute \(x_0\) to verify the initial value.
Deriving the Family for a Simple Slope Rule
For \(y'=2x\), the family \(y=x^2+C\) is easy to check by differentiation. We can also justify that it includes every differentiable solution on an interval, not just some solutions.
This argument explains why the arbitrary constant appears: the slope rule determines how the solution changes, but does not determine its vertical position. The initial condition supplies that position by fixing the constant.
Worked Examples
Worked Example: Select a Curve from \(y'=2x\)
Find the particular solution of \(y'=2x\) satisfying \(y(1)=5\).
The solution family is \(y=x^2+C\). Apply the initial condition by substituting \(x=1\) and setting the output equal to \(5\):
Subtract \(1\) from both sides to obtain \(C=4\). The particular solution is therefore
Check the differential equation: \(\frac{dy}{dx}=2x\), as required. Check the initial condition: \(y(1)=1^2+4=5\). Both requirements hold, so \(y=x^2+4\) is the required particular solution.
Worked Example: Apply an Initial Condition to a Constant Slope
Find the particular solution of \(y'=3\) satisfying \(y(-2)=1\).
A function with derivative \(3\) has the solution family \(y=3x+C\), since differentiating \(3x+C\) gives \(3\). Substitute the initial condition:
Add \(6\) to both sides, giving \(C=7\). The selected solution is
Check the equation: \(\frac{dy}{dx}=3\). Check the initial value: \(y(-2)=3(-2)+7=-6+7=1\). The function satisfies both the slope rule and the specified point.
Worked Example: Use a Point with a Nonzero Independent Variable
Find the particular solution of \(y'=2x+1\) satisfying \(y(2)=7\).
The family \(y=x^2+x+C\) has the required derivative because
Substitute \(x=2\) into the family and set its value equal to \(7\):
Thus \(C=1\), and the particular solution is
Check the differential equation: \(\frac{dy}{dx}=2x+1\). Check the initial condition: \(y(2)=2^2+2+1=4+2+1=7\). Both checks succeed.
What the Initial Condition Does—and Does Not—Say
The notation \(y(1)=5\) gives the height of the curve when \(x=1\); it does not assert that the slope there is \(5\). In the first worked example, the equation says the slope at \(x=1\) is \(2(1)=2\), while the initial condition says the height there is \(5\). These statements describe different features of the same point on the curve.
In the simple families above, each initial condition determines exactly one value of \(C\). It is important not to turn that pattern into an unconditional claim about every differential equation. Whether an initial-value problem has a solution, and whether that solution is unique, depends on the equation and its domain. Here, the point selects a single member because the displayed families have one adjustable constant and the initial condition determines it uniquely.
The initial condition is useful because a slope rule alone can leave the position of a curve undetermined. In an application, the differential equation can describe how a quantity changes, while a measured value at a known time or position locates the particular solution relevant to the situation. For the examples here, that idea is visible directly in the constant \(C\).
Common Mistake: Confusing the Initial Value with the Slope
Suppose \(y'=2x\) and \(y(1)=5\). A common incorrect step is to use the differential equation at \(x=1\), obtaining \(y'(1)=2\), and then claim that the initial condition must be \(y(1)=2\). This confuses the derivative with the function value: \(y'(1)\) is a slope, whereas \(y(1)\) is a height.
The correct calculation uses the solution family \(y=x^2+C\) and the specified function value:
The resulting function \(y=x^2+4\) has \(y(1)=5\), while its slope at that point is \(y'(1)=2\). There is no conflict: the two numbers answer different questions.
Check Your Understanding
For each question, distinguish the slope rule from the point condition, and verify your proposed answer in both.
- For \(y'=2x\), find the particular solution satisfying \(y(0)=3\).
- For \(y'=4\), find the particular solution satisfying \(y(2)=1\).
- In the condition \(y(-1)=6\), what point must the graph pass through?
- For \(y'=2x+1\), what slope does the equation require at \(x=3\)? Does that give the value of \(y(3)\)?
- Why must a proposed particular solution be checked in both the differential equation and the initial condition?
