From a Family of Antiderivatives to One Function
Knowing the derivative of a function does not usually tell you exactly which function you started with. If \(F'(x)=q(x)\), then every function \(F(x)+C\), where \(C\) is a constant, also has derivative \(q(x)\). The derivative records how a function changes, but it does not record its vertical position.
A function value supplies that missing information. If we know both the derivative and the value of the function at one input, we can determine the constant and recover a particular function. In an initial-value problem, the derivative rule describes the change and the stated function value is called the initial condition.
The Constant Is Determined by the Function Value
Suppose \(F'(x)=q(x)\). The family of antiderivatives is \(F(x)+C\). To make one member of this family satisfy \(y(a)=b\), substitute \(x=a\) and set the result equal to \(b\):
Solving the second equation gives \(C=b-F(a)\). Thus the particular function is
The order matters: evaluate the antiderivative at the specified input first, then compare that value with the required function value. The constant adjusts the difference between them.
Write a general family \(y(x)=F(x)+C\) for the given derivative.
Substitute the specified input into the family and set the result equal to the specified function value.
Substitute that constant back into the family to state the particular function.
Differentiate the result and evaluate it at the initial input.
Why One Initial Condition Is Enough
The method works because antiderivatives of the same function on an interval can differ only by a constant. The derivative determines the shape of the function’s changes; one value determines its vertical position. Here is a justification that also explains why the answer is unique.
Suppose \(y_1'(x)=q(x)\) and \(y_2'(x)=q(x)\) throughout an interval. Define \(h(x)=y_1(x)-y_2(x)\). Then \(h'(x)=y_1'(x)-y_2'(x)=0\).
For any two points \(u\) and \(v\) in the interval, the Mean Value Theorem applies to \(h\) on the segment between them. For some point \(c\) between \(u\) and \(v\), \(h(u)-h(v)=h'(c)(u-v)=0\). Therefore \(h(u)=h(v)\).
Because \(h(u)=h(v)\) for any two points, \(h\) is constant on the interval. Thus two antiderivatives of \(q\) differ by a constant there.
If both candidate functions also satisfy \(y_1(a)=b\) and \(y_2(a)=b\), then \(h(a)=y_1(a)-y_2(a)=0\). Since \(h\) is constant, it is zero everywhere. Therefore \(y_1=y_2\) throughout the interval.
This proves uniqueness: on an interval, two functions with the same derivative and the same value at one point must be the same function. The calculation \(C=b-F(a)\) also constructs that function, so the condition determines exactly one member of the antiderivative family.
Worked Examples
Worked Example: A Polynomial Derivative
Find \(y(x)\) if \(y'(x)=3x^2-4x+5\) and \(y(2)=7\).
First integrate each term. The power rule gives
Use the initial condition by substituting \(x=2\) and \(y(2)=7\):
Therefore \(C=-3\), and the particular function is
Check the derivative:
which is the given derivative. Check the initial value:
Both conditions hold.
Worked Example: An Exponential and Trigonometric Derivative
Find \(y(t)\) if \(y'(t)=\cos t+2e^t\) and \(y(0)=4\).
An antiderivative of \(\cos t\) is \(\sin t\), and an antiderivative of \(2e^t\) is \(2e^t\). Therefore
Substitute \(t=0\). Since \(\sin 0=0\) and \(e^0=1\),
Thus \(C=2\), so
Differentiate to check the derivative:
Evaluate the function at the specified input:
The derivative and initial condition are both satisfied.
Worked Example: A Logarithmic Antiderivative on a Restricted Domain
Find \(y(x)\) if \(y'(x)=2x+\frac{1}{x}\), for \(x>0\), and \(y(1)=3\).
On the stated domain \(x>0\), an antiderivative of \(\frac{1}{x}\) is \(\ln x\). Integrating the two terms gives
Apply the condition at \(x=1\). Since \(\ln 1=0\),
It follows that \(C=2\), so the answer is
Differentiate:
Then check the given value:
The domain condition matters: the logarithm and the given derivative are being used on \(x>0\), which contains the initial input \(x=1\).
Connection with Accumulated Change
There is another way to express the same result. Suppose \(y'(x)=q(x)\), \(y(a)=b\), and \(q\) is continuous on the interval from \(a\) to \(x\). The Fundamental Theorem of Calculus gives
Substituting \(y(a)=b\) yields \(y(x)=b+\int_a^x q(t)\,dt\). In words, the function’s value at \(x\) equals its starting value plus the accumulated change from \(a\) to \(x\). This agrees with the antiderivative method: if \(F'=q\), then the definite integral is \(F(x)-F(a)\), so
That is exactly the particular antiderivative found by solving for the constant. This connection is useful when a problem describes a rate of change and an initial amount, such as a position with a known velocity or a quantity with a known rate of accumulation.
A Common Mistake: Setting the Constant Equal to the Initial Value
The initial condition specifies the value of the entire function at \(x=a\), not the value of the constant by itself. A common incorrect calculation is to integrate \(y'(x)=3x^2-4x+5\), obtain \(y=x^3-2x^2+5x+C\), and then use \(y(2)=7\) to write \(C=7\). That skips the other terms in the function’s value at \(x=2\).
The wrong choice \(C=7\) gives
not \(7\). The correction is to substitute the input into every term before solving:
A reliable check is to evaluate the final function at the given input. If that evaluation does not reproduce the stated value, the constant has not been chosen correctly.
What the Initial Condition Does—and Does Not—Tell You
For a derivative given on an interval, one initial condition ordinarily selects one function from the family of antiderivatives on that interval. It does not change the derivative rule; adding a constant changes function values but has derivative zero. Nor should the stated value be substituted into the derivative equation as though it were a function value there: \(y'(a)\) describes the slope at \(a\), whereas \(y(a)\) describes the function’s value at \(a\).
When solving, keep those two pieces of information distinct. Integrate the derivative to form the family, and use the function value only to determine \(C\). Then check the result in both places: its derivative must match the given rate, and its value at the initial input must match the initial condition.
Check Your Understanding
For each problem, find the particular function and verify it by checking both its derivative and its stated value.
- Find \(y(x)\) if \(y'(x)=4x-3\) and \(y(1)=2\).
- Find \(y(t)\) if \(y'(t)=e^t\) and \(y(0)=5\).
- For \(x>0\), find \(y(x)\) if \(y'(x)=\frac{1}{x}\) and \(y(1)=2\).
- In one sentence, explain why setting \(C=b\) is generally incorrect when the condition is \(y(a)=b\).
- Why can two functions with the same derivative on an interval differ only by a constant?
