From a Family of Curves to One Initial-Value Solution
A separable differential equation can often be integrated to obtain a family of curves containing an arbitrary constant. An initial condition supplies a particular point that the required solution must pass through. Applying that condition determines the constant, so the family becomes a specific candidate solution.
This step is more than substituting numbers into a final formula. The integrated equation may involve a logarithm, an absolute value, or a choice of sign. The initial value determines which constant—and sometimes which sign—is compatible with the specified point.
In this tutorial, we focus on using the initial condition after separating and integrating the equation. We will check each resulting formula by evaluating it at \(x_0\). This check establishes that the curve passes through the specified point; it is distinct from a full verification that the formula satisfies the differential equation.
How the Initial Condition Determines the Constant
Suppose separation and integration have produced an equation of the form \(F(y)=G(x)+C\), where \(C\) is an arbitrary constant. If the initial condition is \(y(x_0)=y_0\), substitute \(x=x_0\) and \(y=y_0\) into that equation. The result is \(F(y_0)=G(x_0)+C\), so
This is the basic calculation. It depends on the integrated equation being valid at the initial point. In particular, the initial value must lie in the range allowed by the steps used to separate and integrate, and the resulting expression must use a branch compatible with \(y_0\).
Why does this calculation work? The integrated equation describes the members of a family, with \(C\) distinguishing one member from another. Every member that passes through \((x_0,y_0)\) must give the same value of \(C\) when that point is substituted. Conversely, setting \(C\) to this value makes the integrated equation hold at the specified point. Thus substitution is both necessary and sufficient for selecting a member of this integrated family that passes through the point, subject to the validity conditions just stated.
Keep the arbitrary constant in the integrated relation until the initial condition is applied.
Replace \(x\) by \(x_0\) and \(y\) by \(y_0\) in the integrated relation.
Use the resulting equation to find \(C\), keeping logarithms and signs as written.
Evaluate the resulting formula at \(x_0\) and confirm that its value is \(y_0\).
Worked Examples
Worked Example: A Positive Initial Value in a Logarithmic Equation
Solve \(y'=xy\) subject to \(y(0)=2\), using separation and the initial condition.
For a solution with \(y\ne 0\), separate variables:
Integrate both sides:
Apply \(y(0)=2\). Substituting \(x=0\) and \(y=2\) gives
The integrated relation is therefore \(\ln|y|=x^2/2+\ln 2\). Since the initial value is positive, the corresponding curve has positive \(y\) near the initial point, so we use the positive sign when solving for \(y\):
Check the initial condition by substitution:
The formula passes through \((0,2)\), as required.
Worked Example: Determining a Constant That Is a Logarithm
Solve \(y'=\dfrac{2x}{1+x^2}y\) subject to \(y(1)=6\).
Separate variables where \(y\ne0\):
Integrating gives
At the initial point \(x=1,\ y=6\), substitute into the integrated relation:
Subtract \(\ln 2\) from both sides. Since \(\ln 6-\ln 2=\ln(6/2)=\ln 3\),
The integrated relation becomes \(\ln|y|=\ln(1+x^2)+\ln 3=\ln(3(1+x^2))\). The initial value is positive, so choose the positive value of \(y\):
Check at \(x=1\):
The curve satisfies the specified initial condition. Notice that the constant in the logarithmic equation is \(\ln 3\), not \(3\); exponentiating the equation produces the factor \(3\) in the explicit formula.
Worked Example: A Negative Initial Value and the Sign of the Solution
Solve \(y'=xy^2\) subject to \(y(0)=-1\).
Separate variables where \(y\ne0\):
Integrate each side, using \(\int y^{-2}\,dy=-y^{-1}\):
Substitute the initial values \(x=0\) and \(y=-1\):
Thus \(-1/y=x^2/2+1\). Solving for \(y\) gives
Check the point by substitution:
The result passes through \((0,-1)\). Its negative sign is essential: the initial value is negative, and the formula retains that sign.
A Common Pitfall: Confusing the Initial Value with the Constant
The initial value \(y_0\) is not generally the integration constant \(C\). It must be substituted into the integrated equation, and then that equation must be solved for \(C\). For example, in the first worked example, setting \(C=2\) because \(y(0)=2\) would be incorrect. The integrated equation is \(\ln|y|=x^2/2+C\), so substituting the point gives \(C=\ln 2\), not \(2\).
There is a related sign error to avoid. If the integrated equation contains \(\ln|y|\), replacing it with \(\ln y\) without checking the sign may make the calculation invalid for a negative initial value. For instance, with \(y(0)=-1\), \(\ln y(0)=\ln(-1)\) is not a real number. The correct expression is \(\ln|y(0)|=\ln 1=0\). After using the initial condition, choose the sign of the explicit solution to agree with the specified value.
When the Substitution Must Be Reconsidered
The constant calculation assumes that the integrated relation can be used at the initial point. If a step in separation divided by an expression that is zero at \(y_0\), that step may not describe a solution through the specified point. Similarly, an initial value outside the range of a proposed explicit formula cannot be made valid simply by choosing a different constant. In either situation, check the validity of the separation and the selected formula before proceeding.
It is also useful to keep the scope of the calculation clear. Finding \(C\) and confirming \(y(x_0)=y_0\) shows that the resulting curve passes through the required point. By itself, this substitution does not establish that the formula satisfies the original differential equation at every point of an interval, nor does it establish that no other solution is possible. Those are separate questions.
Summary
After separating and integrating, an arbitrary constant labels the resulting family. Apply \(y(x_0)=y_0\) by substituting both initial values into the integrated relation and solving for that constant. Preserve absolute values in logarithms, select signs consistent with the initial value, and check the final formula directly at \(x_0\).
Check Your Understanding
For each question, focus on how the initial condition determines the member of the integrated family.
- If integration gives \(F(y)=G(x)+C\), what equation results from applying \(y(x_0)=y_0\), and how can it be solved for \(C\)?
- For \(\ln|y|=x^2/2+C\) with \(y(0)=4\), what is \(C\)?
- Why is it incorrect to set \(C=y_0\) without substituting into the integrated equation?
- What does the absolute value in \(\ln|y|\) allow you to handle, and what must you still determine from the initial value?
- What does checking \(y(x_0)=y_0\) confirm, and what does that check alone not establish?
