Check the equation throughout an interval
Matching the slope everywhere
A candidate is a function that someone proposes as a solution. It solves a differential equation when its slope matches the slope that the equation asks for. The faint segments show that slope for \(dy/dx = 2y/x\).
The curve is the candidate \(y = x^2\), for \(x > 0\). Drag the slider to move the dot. The orange tangent shows the slope of the candidate, and the dashed line shows the slope that the equation asks for. The two lines lie on top of each other.
The lower graph compares the two slopes at every \(x\). The orange curve is the candidate's slope \(2x\), and the dashed curve is \(2y/x\), which on this curve is \(2x^2/x = 2x\). No red region opens between them, because the slopes match at every \(x\).
This is what makes \(y = x^2\) a solution: the equation must be true on a whole interval, not at one point only.
The two sides of the equation
Call the candidate \(g(x)\). The left side of the equation is its slope \(g'(x)\), the orange curve below. The right side is the dashed curve, the slope that the equation asks for, found by writing \(g(x)\) in place of \(y\).
A red region fills any gap between the curves. The candidate is a solution on an interval when no red region appears there.
Press "Change the candidate". The segments on the new curve turn red, and the red region opens at most points. A point where the curves touch proves nothing.
Notation
A function is a solution on an interval if the equation is true for every x in the interval.
Write the derivative on the left side, and \(g(x)\) in place of \(y\) on the right side.
\[\cy{g'(x)} = 2x, \qquad \dfrac{2\,g(x)}{x} = \dfrac{2x^2}{x}\]Simplify each side by itself, until it is an expression in x.
\[\dfrac{2x^2}{x} = 2x\]If both sides are the same expression, the equation is true wherever both have a value. If they differ, the candidate fails.
Keep a letter such as \(C\) as a letter, so that one comparison covers every value of \(C\).
When it fails
The equation divides by \(x\), so it has no value at \(x = 0\). Press "Move the point to x = 0". The red dashed line marks the break, and the dot disappears because there is no slope to compare.
A solution must stay on one side of the break, so here the interval is x > 0, the orange bar on the axis. State the interval whenever the equation has a break.
Example 1: a candidate with a constant
Show that \(y = Ce^{3x}\) is a solution of \(\dfrac{dy}{dx} = 3y\) for every real \(x\) and every constant \(C\).
Differentiate the candidate: the chain rule gives the factor 3, and \(C\) stays as a factor.
\[\cy{\dfrac{dy}{dx}} = 3Ce^{3x}\]Write the candidate in the right side of the equation.
\[3y = 3Ce^{3x}\]Both sides are the same expression for every \(x\) and every \(C\). As \(C\) changes, the curve moves and no red region opens, so each \(C\) gives a solution.
Example 2: a candidate that fails
Is \(y = x^2\) a solution of \(\dfrac{dy}{dx} = x + y\) on the interval from 0 to 3?
The left side is the derivative of \(x^2\). The right side has \(x^2\) in place of \(y\).
\[\cy{2x}, \qquad x + x^2\]A tempting wrong move: At \(x = 1\), both sides equal 2, so \(y = x^2\) is a solution.
\[2 = 2\]The equation must hold at every \(x\) of the interval, and at \(x = 2\) the left side is 4 and the right side is 6. The red region fills the gap except where the curves touch.
The sides are equal only when \(2x = x + x^2\). This means \(x^2 - x = 0\), so \(x = 0\) or \(x = 1\).
\[x(x - 1) = 0\]On the interval from 0 to 3, the sides differ at every other point, so the candidate is not a solution there.
Practice
Is \(y = \sqrt{25 - x^2}\) a solution of \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)? Find the interval on which it holds.
Compare the two sides as expressions. Both curves below fall steeply toward the ends, so ask where the denominator becomes 0.
Solution
Differentiate the candidate with the chain rule.
\[\cy{g'(x)} = \dfrac{-2x}{2\sqrt{25 - x^2}} = \dfrac{-x}{\sqrt{25 - x^2}}\]Write \(g(x)\) in place of \(y\) in the right side.
\[\dfrac{-x}{y} = \dfrac{-x}{\sqrt{25 - x^2}}\]The two sides are the same expression. At \(x = \pm 5\) the denominator is 0, and beyond them \(25 - x^2\) is negative, so neither side has a value there. The candidate is a solution for \(-5 < x < 5\).
Questions
The graph compares the two sides of \(\dfrac{dy}{dx} = \dfrac{y}{x}\) for the candidate \(y = x^2 + 1\), for \(x\) from 0 to 3. The orange curve is the left side, and the dashed curve is the right side.
At which value of x do the two sides have the same value?
- x = 1
- x = 2
- x = 0
Answer
x = 1
Is the candidate a solution on the interval from 0 to 3?
- No, because the sides differ at every x except x = 1.
- Yes, because the sides are equal at x = 1.
- Yes, because the candidate is a smooth curve.
Answer
No, because the sides differ at every x except x = 1.
For \(\dfrac{dy}{dx} = -3y\), which statement about \(y = Ce^{-3x}\) is correct?
- Both sides equal \(-3Ce^{-3x}\) for every \(x\) and every \(C\).
- Both sides equal \(Ce^{-3x}\), so it is a solution.
- It is a solution only when \(C = 1\).
Answer
Both sides equal \(-3Ce^{-3x}\) for every \(x\) and every \(C\).
Summary
You write the candidate and its derivative in the equation, simplify each side, and compare the two expressions on the whole interval.
This works because a solution must match the slope that the equation asks for at every point, so one matching point proves nothing.
Use it to decide whether a proposed function is a solution, and state the interval on which it holds.
This is one of four skills in topic 7.2 of Which & Why: AP Calculus.