Why the Interval Matters
A formula can make sense at many inputs without solving the differential equation at all of them. The equation itself may be undefined at some inputs or solution values, and an implicit relation may describe several branches. For an initial-value problem, the relevant question is therefore not just “What formula satisfies the equation?” but also “On what largest interval containing the initial input does this formula remain a valid solution?”
This interval tells us how far the solution can be followed in either direction from the initial input without leaving the conditions required by the differential equation. To find it, check both the formula and the equation, and select the interval that contains the initial input.
Solutions and Their Intervals
Consider a first-order differential equation written as \(y'=f(x,y)\), together with an initial condition \(y(x_0)=y_0\). Here \(f\) is the function that specifies the required slope at each allowed pair \((x,y)\).
The domain of a displayed formula may be larger than the interval on which it solves the equation. For instance, a formula might be differentiable at a point where \(f(x,y)\) is undefined. In that case, the formula cannot be a solution of the differential equation at that point.
In practice, the domain of a formula can be disconnected. The initial input determines which connected interval matters: use the interval containing \(x_0\), then check whether the solution and the right-hand side of the differential equation are both valid throughout it.
A Reliable Procedure
For an explicit formula \(y=\phi(x)\), first find where \(\phi\) is defined and differentiable. Then find where \(f(x,\phi(x))\) is defined. Check that the differential equation holds on the candidate interval, and select the interval containing the initial input. If an endpoint is excluded, determine whether the solution could nevertheless be extended there as a solution.
The interval must contain \(x_0\); other intervals where the formula works do not solve the same initial-value problem.
Check such features as square roots, logarithms, denominators, and points where the formula is not differentiable.
Check where \(f(x,y)\) is defined after substituting the solution, and ensure the equality \(y'=f(x,y)\) holds.
Take the interval containing \(x_0\). At an excluded finite endpoint, check whether a valid solution extension is possible.
For an implicit relation, the same checks apply, but there is an additional issue: the relation may describe more than one branch. Use the initial condition to identify the relevant branch, and verify that it defines a differentiable function on the interval being considered.
Worked Example: A Square-Root Formula
Worked Example: A Square-Root Formula
Find the maximal solution interval containing \(x_0=0\) for the initial-value problem \(y'=1/y,\ y(0)=2\), given the candidate solution \(y(x)=\sqrt{2x+4}\).
First, the square root is real when \(2x+4\geq 0\), so the formula is defined for \(x\geq -2\). However, at \(x=-2\) the formula gives \(y=0\), and the right-hand side \(1/y\) is undefined. The candidate interval for a solution is therefore \(x>-2\).
Differentiate, including the chain-rule factor from the inner function \(2x+4\):
The initial condition is satisfied because \(y(0)=\sqrt{4}=2\). The formula and equation are both valid for every \(x\in(-2,\infty)\), so this is a solution on that interval.
It cannot extend through \(x=-2\) as a solution of the same equation. Any continuous extension of the formula to \(x=-2\) must have \(y(-2)=0\), but \(1/y\) is undefined there. Thus the maximal interval containing \(0\) is \((-2,\infty)\). As a final check, substituting the derivative back into the equation gives \(y'=1/y\) throughout that interval.
Worked Example: A Singular Input in the Equation
Worked Example: A Singular Input in the Equation
Consider \(y'=1/(x-1)\) with \(y(0)=0\). Integrating the right-hand side gives the candidate formula \(y(x)=\ln|x-1|+C\), valid only for \(x\ne 1\). Apply the initial condition:
The formula is therefore \(y(x)=\ln|x-1|\). Its domain has two intervals, \((-\infty,1)\) and \((1,\infty)\). Since the initial input \(0\) lies in the first interval, that is the interval relevant to the initial-value problem.
For \(x<1\), \(|x-1|=1-x\), so \(y(x)=\ln(1-x)\). Differentiate and compare with the equation:
Also, \(y(0)=\ln(1)=0\), as required. The differential equation itself is undefined at \(x=1\), so no solution interval containing \(0\) can pass through that input. The maximal interval containing \(0\) is \((-\infty,1)\). The formula also works on \((1,\infty)\), but that separate interval does not contain the initial input.
Worked Example: Choosing a Branch of an Implicit Solution
Worked Example: Choosing a Branch of an Implicit Solution
Suppose an implicit solution is given by \(\ln y=x\), with initial condition \(y(0)=1\), for the differential equation \(y'=y\). The logarithm requires \(y>0\). Solving the relation for \(y\) gives \(y=e^x\), which is positive for every real \(x\).
Differentiate the explicit branch and check the equation:
The initial condition also holds: \(y(0)=e^0=1\). Equivalently, differentiating the implicit relation gives \(y'/y=1\); since \(y>0\), multiplying by \(y\) gives \(y'=y\). The function \(e^x\) is defined and differentiable for all real \(x\), and the right-hand side \(y\) is defined for all real \(y\). Thus the solution interval containing \(0\) is \((-\infty,\infty)\), which is maximal because it is already the entire real line.
Why an Excluded Endpoint Can Prevent Extension
An endpoint excluded by a formula is not automatically a barrier to a solution. A formula might have an algebraic restriction that disappears after rewriting it, or a different formula might describe a valid continuation. What matters is whether a differentiable extension can still satisfy the differential equation at and beyond the endpoint.
There is a straightforward obstruction when the equation itself is undefined at the endpoint’s required solution value. In the square-root example, continuity forced the endpoint value to be \(0\), where \(1/y\) is undefined. In the logarithm example, the independent-variable input \(x=1\) made \(1/(x-1)\) undefined regardless of the value assigned to \(y\). Neither solution can cross its excluded point as a solution of the stated equation.
This is why checking only the formula is insufficient. The formula and the differential equation impose separate conditions, and both must hold at every point of a solution interval.
Common Mistake: Ignoring the Chain Rule
For \(y(x)=\sqrt{2x+4}\), a common incorrect differentiation is
This omits the derivative of the inner function \(2x+4\), which is \(2\). The error is visible at \(x=0\): the incorrect expression gives \(\frac12\cdot 4^{-1/2}=\frac14\), whereas the differential equation requires \(1/y(0)=1/2\).
The correction is to include the chain-rule factor:
This correction verifies the equation at every \(x>-2\), not merely at the initial input. Checking at the initial input is useful for catching an error, but it does not replace checking the derivative identity throughout the proposed interval.
Keep the Three Checks Separate
When identifying a maximal interval, it helps to keep three questions distinct:
| Check | Question |
|---|---|
| Formula | Where is the proposed expression defined and differentiable? |
| Differential equation | Where is \(f(x,y(x))\) defined, and does \(y'=f(x,y(x))\) hold? |
| Initial condition and extension | Which interval contains \(x_0\), and can the solution extend past an excluded endpoint? |
The maximal solution interval is determined by all of these checks together. An endpoint may arise from a formula restriction, a singularity in the equation, or failure of a branch to remain a differentiable function. Always state which issue prevents extension rather than relying on the appearance of the formula alone.
Check Your Understanding
For each question, identify the restrictions that come from the formula separately from those that come from the differential equation.
- Why might a formula be defined at an input where it is not a solution of the differential equation?
- For \(y'=1/(x-2)\), why can a solution interval containing \(x_0=0\) not include \(x=2\)?
- If an implicit relation describes two branches, what determines which branch is relevant to an initial-value problem?
- For \(y'=1/y\) and \(y(x)=\sqrt{2x+4}\), what prevents extension through \(x=-2\)?
- Why is it not enough to report the full domain of a formula when that domain has more than one interval?
