Keep solutions lost through division
The curve the formula misses
The slope field shows \(\frac{dy}{dx} = x\,(y + 1)\). Along the dashed row \(y = -1\), every segment is flat, because the factor \(y + 1\) is 0.
To separate the variables, you divide by \(y + 1\). You cannot divide by 0, so this step assumes that \(y\) is not \(-1\). It gives the curves \(y = -1 + Ae^{x^2/2}\) with \(A\) not 0.
Drag \(A\) to a positive or negative value. The curves lie above or below the dashed line, and none of them is the dashed line itself.
The line \(y = -1\) is still a solution, because its slope is 0 and \(x(-1 + 1)\) is 0 for every \(x\). Drag \(A\) to 0. The formula gives \(y = -1\), so the lost solution returns if you let \(A\) be 0.
Check the zeros of the divisor
To separate \(\frac{dy}{dx} = g(x)\,h(y)\), you divide by \(h(y)\). The division is allowed only where \(h(y)\) is not 0.
On the dashed rows \(h(y) = 0\), so the slope \(g(x)\,h(y)\) is 0 and every segment there is flat. Drag the orange point onto a dashed row. The solution through it is the row itself, a constant solution.
The faint curves come from the division, which assumed h(y) is not 0, so none of them touches a row. Solve h(y) = 0 before you divide, and keep each root as a constant solution.
Notation
Before you divide by \(h(y)\), solve \(h(y) = 0\). Each root \(c\) may give a constant solution \(y = c\), a flat row of the slope field.
\[h(c) = 0\]Check each root. The derivative of the constant function \(y = c\) is 0, and the right side is also 0, so \(y = c\) is a solution.
\[\dfrac{d}{dx}(c) = 0 = g(x)\,h(c)\]Where \(h(y)\) is not 0, separate and integrate as before. The faint curves are this family of solutions.
\[\frac{1}{h(y)}\,dy = g(x)\,dx\]Compare the two results. If the family gives \(y = c\) for some value of its constant, nothing is lost. If it does not, write \(y = c\) as a separate solution.
Here \(A = 0\) gives the row \(y = -1\), so one formula holds every solution.
When it fails
The family does not always include the constant solution. For \(\dfrac{dy}{dx} = y^2\), dividing by \(y^2\) gives the faint curves \(y = -\dfrac{1}{x + C}\).
No value of \(C\) makes this formula 0, but the dashed row \(y = 0\) is a solution, because both sides are 0 there. Check the constant solutions even when the family looks complete.
Example 1: the family includes the line
Find all solutions of \(\dfrac{dy}{dx} = (y - 4)\,e^x\).
Set \(y - 4 = 0\) before you divide. The constant function \(y = 4\) gives 0 on both sides, so it is a solution.
\[\dfrac{d}{dx}(4) = 0 = (4 - 4)\,e^x\]For \(y\) not 4, separate and integrate. The faint curves are these solutions.
\[\begin{aligned} \dfrac{1}{y - 4}\,dy &= e^x\,dx \\ \ln|y - 4| &= e^x + C \end{aligned}\]Raise \(e\) to the power of both sides, so \(|y - 4| = e^C e^{e^x}\). Write \(\pm e^C\) as \(A\), which is any nonzero number because \(e^C\) is never 0.
\[y = 4 + A\,e^{e^x}\]If \(A = 0\), the formula gives \(y = 4\), the constant solution, so one formula describes every solution when \(A\) is any real number.
\[\begin{aligned} y &= 4 + A\,e^{e^x} \\ A &\text{ is any real number} \end{aligned}\]
Example 2: the family misses the line
Find the solution of \(\dfrac{dy}{dx} = 2xy^2\) with \(y(0) = 0\).
For \(y\) not 0, separate and integrate. Then solve for \(y\). The faint curves are these solutions.
\[\begin{aligned} \dfrac{1}{y^2}\,dy &= 2x\,dx \\ -\dfrac{1}{y} &= x^2 + C \\ y &= -\dfrac{1}{x^2 + C} \end{aligned}\]A tempting wrong move: Use \(y(0) = 0\) in the formula. The formula \(y = -\dfrac{1}{x^2 + C}\) is never 0, so there is no solution.
\[0 = -\dfrac{1}{C} \;\Rightarrow\; \text{impossible}\]The slope on the row \(y = 0\) is \(2x \cdot 0^2 = 0\), so the orange segments there are flat. The horizontal line through \((0, 0)\) follows every one of them, so a solution exists.
Check the constant function \(y = 0\), which the division left out. Both sides are 0, so it is a solution.
\[\dfrac{d}{dx}(0) = 0 = 2x\cdot 0^2\]The constant function \(y = 0\) also satisfies \(y(0) = 0\), so it is the solution.
\[y = 0\]
Practice
Find all constant solutions of \(\dfrac{dy}{dx} = x(y - 1)(y + 2)\). Then give a solution with \(y(0) = 1\).
A constant solution makes the right side 0 for every \(x\). Set each factor that contains only \(y\) equal to 0. Slide the orange line until it follows the segments.
Solution
The factor \(x\) is 0 only at \(x = 0\), so the right side is 0 for every \(x\) only if \((y - 1)(y + 2)\) is 0. Solve this equation.
\[\begin{aligned} (y - 1)(y + 2) &= 0 \\ y = 1 &\text{ or } y = -2 \end{aligned}\]Check \(y = 1\). Both sides are 0, so it is a solution. Check \(y = -2\) in the same way.
\[\dfrac{d}{dx}(1) = 0 = x(1 - 1)(1 + 2)\]The constant solution \(y = 1\) satisfies \(y(0) = 1\), so it is a solution of the problem.
\[y = 1\]
Questions
The graph shows the slope field of \(\dfrac{dy}{dx} = y^2 - 4\). The segments on some horizontal rows are flat.
Which horizontal lines are solutions?
- \(y = 2\) and \(y = -2\)
- \(y = 4\) and \(y = -4\)
- \(y = 0\)
Answer
\(y = 2\) and \(y = -2\)
A student divides by \(y^2 - 4\) and finds curves that never touch the row \(y = 2\). What does the graph show about \(y = 2\)?
- It is a solution that the division left out.
- It is not a solution, because the curves do not touch it.
- It is a solution only for positive \(x\).
Answer
It is a solution that the division left out.
Which constant solutions does \(\dfrac{dy}{dx} = e^x(y + 1)(y - 6)\) have?
- \(y = -1\) and \(y = 6\)
- \(y = 1\) and \(y = -6\)
- \(y = 6\) only
- There are none, because \(e^x\) is never 0.
Answer
\(y = -1\) and \(y = 6\)
Summary
You set the expression you divide by equal to 0 first, check each root as a constant solution, and see whether the family includes it.
This works because dividing by an expression assumes that it is not 0, and a constant solution is a flat row of the slope field.
You use it each time you separate by dividing by an expression in y.
This is one of five skills in topic 7.6 of Which & Why: AP Calculus.