Determine a parameter and valid domain
A constant and a break
The curves \(y = 1/(C - x)\) solve \(dy/dx = y^2\) for every constant \(C\). Each one is a candidate, a function proposed as a solution. The initial condition \(y(0) = \tfrac12\) is the orange dot at \((0, \tfrac12)\).
Drag the slider for \(C\). The red segment is the gap between the dot and the curve at \(x = 0\). It closes only at \(C = 2\), so the initial condition fixes the constant.
The red dashed line marks a break. At \(x = C\) the denominator is 0, so the candidate has no value there, and the curve climbs without limit beside the line.
A solution cannot cross a break, so it stays on the side of the dot. The faint piece beyond the break is not part of it. The orange bar on the x-axis shows where the solution holds: all \(x\) less than \(C\), which is \(x < 2\) when \(C = 2\).
What limits a solution
Move the slider for \(C\). The break and the end of the orange bar move together, because the break bounds the interval of the solution.
A break is an input where the candidate has no value or no derivative. A solution needs a derivative at every input of its interval, so it cannot cross a break.
Drag the dot to the other side of the break. The orange bar moves with it, because the valid interval is the longest interval that contains the starting input and no break.
Notation
A parameter is a letter in the candidate that stands for an unknown number, such as C.
\[y = \dfrac{1}{C - x}\]To find a parameter from the initial condition, write the starting input and output in the candidate. Then solve for the parameter.
\[\tfrac12 = \dfrac{1}{C - 0} \ \Rightarrow\ C = 2\]To find a parameter that makes the equation true, write the candidate and its derivative in the equation. The two sides must match for every x.
Find the breaks. These are the inputs where a denominator is 0 or where a logarithm or a square root has no value.
\[\cr{2 - x = 0} \ \Rightarrow\ x = 2\]The valid interval is the longest one that contains the starting input 0 and no break.
\[\cy{x < 2}\]
Example 1: a parameter in the exponent
Find the value of \(k\) for which \(y = e^{kx} - 2\) is a solution of \(\dfrac{dy}{dx} = 3y + 6\).
Differentiate the candidate. The letter \(k\) is a constant, so the chain rule gives \(ke^{kx}\).
\[\cy{\dfrac{dy}{dx}} = ke^{kx}\]Write the candidate in the right side and simplify. The dashed curve below is this right side.
\[3(e^{kx} - 2) + 6 = 3e^{kx}\]The sides must match for every x. The red region below marks where they differ.
\[ke^{kx} = 3e^{kx}\]The factor \(e^{kx}\) is never 0, so divide both sides by it. The red region closes at this value.
\[k = 3\]
Example 2: a logarithm
The equation is \(\dfrac{dy}{dx} = e^{-y}\), the candidate is \(y = \ln(x + C)\), and the initial condition is \(y(0) = 0\). Find \(C\) and the interval on which the solution holds.
Write \(x = 0\) and \(y = 0\) in the candidate. A logarithm is 0 only when its input is 1.
\[0 = \ln(0 + C) \ \Rightarrow\ C = 1\]A tempting wrong move: The formula \(\ln(x + 1)\) has a value for every \(x\) except \(x = -1\), so the solution holds for all \(x\) except \(-1\).
\[x \neq -1\]A logarithm has no value when its input is 0 or negative. To the left of the red dashed line there is no curve, because \(\ln(x + 1)\) has no value for \(x < -1\).
The logarithm needs a positive input, so the solution holds where \(x + 1 > 0\), which contains the starting input 0.
\[x + 1 > 0 \ \Rightarrow\ \cy{x > -1}\]The solution is \(y = \ln(x + 1)\) on the interval \(x > -1\).
Practice
The equation is \(\dfrac{dy}{dx} = 2xy^2\). The candidate is \(y = \dfrac{1}{C - x^2}\), and the initial condition is \(y(0) = \tfrac14\). Find \(C\) and the interval on which the solution holds.
Use the initial condition to find \(C\). Then find where the denominator is 0, and keep the piece between the breaks that contains the dot.
Solution
Write \(x = 0\) and \(y = \tfrac14\) in the candidate.
\[\tfrac14 = \dfrac{1}{C - 0} \ \Rightarrow\ C = 4\]The denominator \(4 - x^2\) is 0 at two inputs, which are the breaks.
\[\cr{4 - x^2 = 0} \ \Rightarrow\ x = -2 \text{ or } x = 2\]The starting input 0 lies between the two breaks, so the solution holds on this interval only.
\[\cy{-2 < x < 2}\]
Questions
The graph shows the candidate \(y = \dfrac{1}{C - x}\) for the equation \(\dfrac{dy}{dx} = y^2\). The initial condition is \(y(1) = -1\).
What is the largest interval on which the solution through the dot holds?
- x > 0
- All x except 0
- x > 1
Answer
x > 0
The break is at x = C. What is C?
- 0
- 1
- −1
Answer
0
For which \(k\) is \(y = 3e^{kx} + 1\) a solution of \(\dfrac{dy}{dx} = 2y - 2\)?
- k = 2
- k = 6
- k = 3
Answer
k = 2
Summary
You find a parameter from the initial condition or from the equation, and keep the longest interval around the starting input with no break.
This works because the solution passes through the starting point, matches the equation at every input, and has a derivative on its whole interval.
Use it when a candidate has an unknown letter, or when its formula has a break.
This is one of four skills in topic 7.2 of Which & Why: AP Calculus.