State relevant domain restrictions
A curve that runs off the page
The equation is \(\frac{dy}{dx} = 1 + y^2\). Separating and integrating gives \(\arctan y = x + C\), and with \(y(0) = 0\) the constant is 0, so the solution is \(y = \tan x\).
Drag the orange point to the right. At \(x = 1\), \(y\) is about 1.56. At \(x = 1.5\), \(y\) is about 14.1. The curve rises faster and faster.
The red dashed line at \(x = \frac{\pi}{2}\), about 1.571, is a vertical asymptote. As \(x\) approaches it, \(y\) grows without bound, so the solution has no value there and does not continue past the line.
It exists on one unbroken stretch, \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), that contains the start \(x = 0\).
The interval that holds the start
A solution is a function that is defined and differentiable on an interval, and the interval must contain the starting x-value.
The equation or its solution formula can fail at some values of x, for example where a denominator is 0. The red dashed lines mark these values, and they cut the x-axis into pieces.
Move the start. The solution is defined only on the piece that holds the start, so it cannot cross a red line. The faint curves on the other pieces are other solutions.
Notation
Find the values of \(x\) where the solution formula or the equation is undefined. Look for a zero denominator, a logarithm of a nonpositive number, or an even root of a negative number. Here \(y = \tan x = \frac{\sin x}{\cos x}\) is undefined at \(x = \pm\frac{\pi}{2}\), where \(\cos x = 0\).
Mark the start \((x_0, y_0)\), here \((0, 0)\). The red lines divide the x-axis into pieces, and one piece holds the start.
Keep that piece only. If it runs from \(a\) to \(b\), the solution is the formula on \(a < x < b\), here \(-\frac{\pi}{2} < x < \frac{\pi}{2}\).
\[x_0 \in (a, b)\]Use strict inequalities at each end. The solution has no value where the formula fails, and the equation gives no slope where it fails.
The equation can fail where the formula does not. For example, \(\dfrac{dy}{dx} = \dfrac{1}{2y}\) fails where \(y = 0\).
When it fails
The formula \(y = \tan x\) is also defined for \(\frac{\pi}{2} < x < \frac{3\pi}{2}\), and it satisfies the same equation there.
That stretch is a different solution. The two are not joined, because \(y\) is unbounded at the line between them. The solution through \((0, 0)\) is only the stretch on its side of the line.
Example 1: a denominator that is 0
Solve \(\dfrac{dy}{dx} = \dfrac{y}{x - 3}\) with \(y(4) = 2\). State the interval on which the solution is valid.
Separate and integrate, then write \(\pm e^{C}\) as \(A\). The curve passes through \((4, 2)\), so substitute to find \(A\).
\[\begin{aligned} \ln|y| &= \ln|x - 3| + C \\ y &= A(x - 3), \quad A = 2 \end{aligned}\]The formula \(y = 2(x - 3)\) is defined for every \(x\), shown faint to the left of 3. But the equation has \(x - 3\) in a denominator, so it fails at \(x = 3\), the red line.
\[y = 2(x - 3)\]The solution cannot cross \(x = 3\), because the equation gives no slope there. The start \(x = 4\) is to the right of 3, so the valid interval is \(x > 3\).
\[x > 3\]
Example 2: a logarithm
Solve \(\dfrac{dy}{dx} = e^{y}\) with \(y(0) = 0\). State the interval on which the solution is valid.
Separate and integrate. The curve passes through \((0, 0)\), so substitute to find \(C\).
\[\begin{aligned} e^{-y}\,dy &= dx \\ -e^{-y} &= x + C, \quad C = -1 \end{aligned}\]Multiply both sides by \(-1\). Then take the logarithm of both sides to find \(y\).
\[e^{-y} = 1 - x, \qquad y = -\ln(1 - x)\]A tempting wrong move: State that the solution is valid for every \(x\).
\[-\infty < x < \infty\]The curve rises without bound toward the red line at \(x = 1\), and there is no curve to the right of it. The formula takes the logarithm of \(1 - x\), which is undefined when \(1 - x \le 0\).
The logarithm needs \(1 - x > 0\), so the solution is valid on \(x < 1\), the piece that holds the start \(x = 0\).
\[x < 1\]
Practice
Solve \(\dfrac{dy}{dx} = \dfrac{1}{2y}\) with \(y(1) = 2\). State the interval on which the solution is valid.
Move the point to the left end of the curve. There \(y\) is 0, and the equation divides by \(2y\).
Solution
Separate and integrate. The curve passes through \((1, 2)\), so substitute to find \(C\).
\[\begin{aligned} 2y\,dy &= dx \\ y^2 &= x + C, \quad C = 3 \end{aligned}\]The start has \(y > 0\), so take the positive root.
\[y = \sqrt{x + 3}\]The root needs \(x + 3 \ge 0\). At \(x = -3\) the root is 0, so the equation divides by 0 and fails there, marked by the red ring. The solution is valid on \(x > -3\).
\[x > -3\]
Questions
The graph shows the slope field of \(\dfrac{dy}{dx} = y^2\). The chalk curve is the solution \(y = \dfrac{1}{1 - x}\) through the orange start \((0, 1)\). The faint curve is the same formula for \(x > 1\), and it passes through \((2, -1)\).
At which \(x\)-value does the solution through the orange point \((0, 1)\) stop?
- \(x = 1\)
- \(x = 0\)
- \(x = 2\)
Answer
\(x = 1\)
The same formula gives the faint curve through \((2, -1)\) to the right of the line. Is it part of the solution through \((0, 1)\)?
- No. The solution cannot cross the red line, so the faint curve is a different solution.
- Yes, because both pieces use the same formula.
- Yes, because the two pieces join at the line.
Answer
No. The solution cannot cross the red line, so the faint curve is a different solution.
Integrating \(\dfrac{dy}{dx} = \dfrac{1}{x - 2}\) with \(y(3) = 1\) gives \(y = \ln|x - 2| + 1\). On which interval is this the solution?
- \(x > 2\)
- \(x \ge 2\)
- \(x \ne 2\)
- All real x
Answer
\(x > 2\)
Summary
You find where the formula and the equation fail, and you keep the one stretch of x that contains the start.
This works because a solution must be defined and differentiable on an unbroken interval, so it cannot cross an asymptote or an undefined point.
You use it when you state a solution of an initial value problem.
This is one of five skills in topic 7.6 of Which & Why: AP Calculus.