Verifying Solutions to… — AP® Calculus AB Notes

A differential equation is a rule about rates. A solution to a differential equation is a function that obeys the rule. Before you learn to find solutions, you need to be able to check one. The good news: checking uses nothing new. You differentiate, substitute and compare.

What "solution" means here

In algebra, a solution of $x^2 = 9$ is a number. Here a solution is a whole function. The function $y = 3e^{2x}$ is a solution of $\frac{dy}{dx} = 2y$ because, for every $x$ , its slope equals twice its height:

$\frac{dy}{dx} = 6e^{2x} \qquad \text{and} \qquad 2y = 2\left(3e^{2x}\right) = 6e^{2x}.$

The two sides are the same function of $x$ . That is what you must show: equality for every $x$ in an interval, not at one lucky point.

The three-step check

Differentiate. Find every derivative the equation uses: $y'$ , and $y''$ if it appears.
Substitute. Put the function and its derivatives into the left side and into the right side. Work on each side separately.
Compare. Simplify both sides. If they are identical on an interval, the function is a solution there. If they differ anywhere in the interval, it is not.

Keep the two sides apart while you simplify. Writing a long chain like $\frac{dy}{dx}$ $= 1 + Ce^x$ $= x + 1 + Ce^x$ puts an equals sign between things that are not equal, a linkage error that costs points on the exam. Write "left side $= \dots$ " and "right side $= \dots$ ", then compare.

Worked example 1. Show that $y = x + 1 + 2e^x$ is a solution of $\frac{dy}{dx} = y - x$ .

Differentiate: $\frac{dy}{dx} = 1 + 2e^x$ . (The derivative of $x$ is 1, of the constant 1 is 0.)

Right side: $y - x$ $= (x + 1 + 2e^x) - x$ $= 1$ $+ 2e^x$ .

Compare: both sides are $1 + 2e^x$ for every $x$ . So the function is a solution on $(-\infty, \infty)$ .

Worked example 2. Is $y = x^2 + 1$ a solution of $\frac{dy}{dx} = \frac{2y}{x}$ for $x > 0$ ?

Left side: $\frac{dy}{dx} = 2x$ . Right side: $\frac{2(x^2 + 1)}{x} = 2x + \frac{2}{x}$ .

These differ by $\frac{2}{x}$ , which is never 0. So $y = x^2 + 1$ is not a solution. (Try $y = x^2$ instead: then the right side is $\frac{2x^2}{x} = 2x$ , and it works.)

Equations with a second derivative

The same check works when $y''$ appears. For $y = \sin(2x)$ : $y' = 2\cos(2x)$ and $y'' = -4\sin(2x) = -4y$ . So $y = \sin(2x)$ is a solution of $y'' = -4y$ .

Worked example 3. Find the positive constant $k$ for which $y = e^{kx}$ satisfies $y'' - y' - 6y = 0$ .

Differentiate: $y' = ke^{kx}$ and $y'' = k^2e^{kx}$ .

Substitute: $k^2e^{kx} - ke^{kx} - 6e^{kx}$ $= e^{kx}\left(k^2 - k - 6\right)$ .

Since $e^{kx}$ is never 0, the left side is 0 for every $x$ exactly when $k^2 - k - 6 = 0$ , that is $(k - 3)(k + 2) = 0$ . The positive value is $k = 3$ .

Many solutions, not one

An added constant has derivative 0. So if $y = x^2$ solves $\frac{dy}{dx} = 2x$ , so do $x^2 + 5$ and $x^2 - \pi$ : a differential equation usually has infinitely many solutions. The same happens with multiplied constants: every $y = Ce^{3x}$ satisfies $\frac{dy}{dx} = 3y$ , because $\frac{dy}{dx} = 3Ce^{3x} = 3y$ .

Five curves y = x + 1 + Ce^x for C = −1, −0.5, 0, 0.5 and 1 on −3 ≤ x ≤ 2. The C = 0 curve is the line y = x + 1, drawn in crimson. At x = 0.5 each curve carries a short black segment whose slope is y − 0.5, the value of y − x there, showing that every curve satisfies dy/dx = y − x. — **Figure 1.** Five members of the family $y = x + 1 + Ce^x$ . All of them satisfy $\frac{dy}{dx} = y - x$ ; at $x = 0.5$ each one's slope equals $y - x$ at that point. LevlPrep original diagram.

Picking one member of the family needs one more fact, such as a point the graph passes through. That comes in a later topic.

Solutions live on intervals

A solution must be differentiable wherever you claim it works. The function $y = \frac{1}{x}$ satisfies $\frac{dy}{dx} = -y^2$ , since $-\frac{1}{x^2} = -\left(\frac{1}{x}\right)^2$ . But it is undefined at $x = 0$ , so it is a solution on $x > 0$ and, separately, on $x < 0$ . When you state that a function is a solution, name the interval.

Common slips

Checking only one value of $x$ . Agreement at one point proves nothing.
Dropping a derivative: the derivative of $x$ is 1, and the chain rule factor in $e^{3x}$ is 3.
Comparing with the wrong side: in $x\frac{dy}{dx} = 2y$ the left side includes the factor $x$ .
Chaining unequal expressions with $=$ instead of simplifying each side on its own.