Choose the valid branch and interval
A circle holds two solutions
The equation \(\frac{dy}{dx} = -\frac{x}{y}\) is not defined where \(y = 0\), so the x-axis is a red dashed line.
The orange dot is the start \((3, -4)\). Separating and integrating gives \(x^2 + y^2 = C\), and \(3^2 + (-4)^2 = 25\), so the solution lies on the faint circle \(x^2 + y^2 = 25\).
A circle is not the graph of a function. The chalk curve is the lower half, because the dot is below the axis. It stops at the red rings on the x-axis, where the slope is not defined.
Press the button for the start \((3, 4)\). The solution becomes the upper half, so the same circle holds two solutions, and the starting point picks one.
One piece between two breaks
A solution is one function \(y(x)\) on one interval that contains the starting input \(x_0\), and its curve passes through the starting point.
The red dashed lines are breaks, places where the equation is not defined. A solution needs a slope at every point of its interval, so it cannot cross a break.
Drag the dot. The solution runs left and right from it toward the nearest break on each side, and these breaks are the ends of its interval. Where a side has no break, the interval has no end there.
The ends do not belong to the interval, because the derivative does not exist there.
Notation
Solve \(x^2 + y^2 = 25\) for \(y\). The square root gives two branches, one for each sign, and each branch is one function of \(x\).
\[y = \pm\sqrt{25 - x^2}\]Choose the sign from the starting point. A solution is continuous, so it cannot change sign without passing through \(y = 0\), where the equation breaks, and it keeps the sign of \(y_0\).
\[\begin{aligned} y(3) &= -4 < 0 \\ \Rightarrow\ y &= -\sqrt{25 - x^2} \end{aligned}\]Find the inputs where the formula or the equation breaks. There a denominator is 0, a root has a negative number inside, or a logarithm has a number that is not positive. At \(x = \pm 5\), \(y = 0\) and the equation divides by 0, so the ends are left out.
\[25 - x^2 > 0 \ \Rightarrow\ -5 < x < 5\]The interval of the solution is the longest open interval that contains \(x_0\) and none of the breaks.
\[x_0 \in (a, b)\]The piece of the formula beyond a break is a different solution, not part of this one.
When it fails
A solution cannot join the two halves of the circle. At the red rings the tangent is vertical, so the slope is not defined and the curve cannot continue across the x-axis.
Press the button to draw the upper half in red. It solves the same equation, but it passes through \((3, 4)\), not the starting point, so it is another solution.
Example 1: A hyperbola
Find the solution of \(\dfrac{dy}{dx} = -\dfrac{y}{x}\) with \(y(-2) = 3\), and give its interval.
Separate and integrate. This assumes that \(y \ne 0\).
\[\ln|y| = -\ln|x| + C\]Exponentiate both sides. One constant \(A = \pm e^C\) covers both signs of \(y\), and the faint curves show some values of \(A\).
\[y = \frac{A}{x}\]Substitute the input \(x = -2\) and the output \(y = 3\).
\[3 = \frac{A}{-2} \ \Rightarrow\ A = -6\]Write the particular solution.
\[y = -\frac{6}{x}\]The formula and the equation both break at \(x = 0\), where they divide by 0. The start \(x_0 = -2\) lies to the left of 0, so the interval is \(x < 0\). The faint piece on the right is a different solution.
\[x \in (-\infty, 0)\]
Example 2: A root with a sign
Find the solution of \(\dfrac{dy}{dx} = \dfrac{x}{y}\) with \(y(4) = -3\), and give its interval.
Separate, integrate, and multiply both sides by 2. Call \(2C\) the constant \(K\). The faint curves show some values of \(K\).
\[y\,dy = x\,dx \ \Rightarrow\ y^2 = x^2 + K\]Substitute the input \(x = 4\) and the output \(y = -3\).
\[9 = 16 + K \ \Rightarrow\ K = -7\]A tempting wrong move: Take the square root of both sides. The root sign gives the positive branch.
\[y = \sqrt{x^2 - 7}\]At \(x = 4\) this gives \(y = \sqrt{9} = 3\). The point is \((4, -3)\), below the axis, so the red curve passes above it.
Choose the minus sign, because the starting output \(y_0 = -3\) is negative.
\[y = -\sqrt{x^2 - 7}\]The root needs \(x^2 - 7 > 0\), because at \(x = \pm\sqrt7\) the output is \(y = 0\) and the equation divides by 0. The interval must contain \(x_0 = 4\), so it is \(x > \sqrt7\).
\[x \in (\sqrt7, \infty)\]
Practice
Find the solution of \(\dfrac{dy}{dx} = 2xy^2\) with \(y(0) = 1\), and give its interval.
Find the inputs where the denominator of your formula is 0. Slide the red lines onto those inputs. The interval is the part that contains \(x = 0\).
Solution
Separate and integrate. This assumes that \(y \ne 0\).
\[\frac{dy}{y^2} = 2x\,dx \ \Rightarrow\ -\frac{1}{y} = x^2 + C\]Substitute the input \(x = 0\) and the output \(y = 1\).
\[-1 = C\]Solve for \(y\).
\[y = \frac{1}{1 - x^2}\]The denominator is 0 at \(x = -1\) and at \(x = 1\). The start \(x_0 = 0\) lies between them, so the interval is \(-1 < x < 1\).
\[x \in (-1, 1)\]
Questions
Questions 1 and 2 use the curve \(y = \ln|x - 2| + 1\), which solves \(\frac{dy}{dx} = \frac{1}{x - 2}\). The orange point is the start \((3, 1)\).
On which interval is the solution through \((3, 1)\) defined?
- \(x > 2\)
- All \(x\) except 2
- \(x > 3\)
- \(2 < x < 3\)
Answer
\(x > 2\)
What does the solution through \((3, 1)\) do as \(x\) approaches 2 from the right?
- It falls without bound.
- It approaches 1, the height of the point.
- It rises without bound.
Answer
It falls without bound.
The solution of \(\frac{dy}{dx} = -y^2\) with \(y(0) = \frac12\) is \(y = \frac{1}{x + 2}\). On which interval is it the solution?
- \(x > -2\)
- All real x
- \(x < -2\)
- \(-2 < x < 2\)
Answer
\(x > -2\)
Summary
You choose the sign that matches the starting point, and you stop the solution at the nearest break on each side.
This works because a solution is continuous and has a derivative on its whole interval, so it cannot change branch or cross a break.
Use it when the solution formula has a square root, a fraction or a logarithm.
This is one of four skills in topic 7.7 of Which & Why: AP Calculus.