7.2 Verifying Solutions for Differential Equations

7.2 Verifying Solutions for Differential Equations

Topic 7.1 turned a verbal description into a differential equation. The next question is whether a proposed function really satisfies that equation. Answering it needs no solving technique: differentiate, substitute, and check that the two sides agree. The same check shows that one differential equation can be satisfied by a whole family of functions.

Investigative question: If you know a quantity’s rate of change, how can you describe the quantity itself—and how can you prove that a proposed function really works?

Verifying a proposed solution

The learning objective FUN-7.B: Verify solutions to differential equations requires checking, not guessing. The essential knowledge is FUN-7.B.1: Derivatives can be used to verify that a function is a solution to a given differential equation.

To verify a proposed function, differentiate it as many times as the differential equation requires, substitute the derivatives and the function into the equation, and determine whether the resulting statement is true on the relevant domain.

Worked verification. Test whether

$$ y=4e^{3t} $$

is a solution of

$$ \frac{dy}{dt}=3y. $$

Differentiate:

$$ \frac{dy}{dt}=12e^{3t}. $$

Now substitute $y=4e^{3t}$ into the right-hand side:

$$ 3y=3(4e^{3t})=12e^{3t}. $$

Because $\dfrac{dy}{dt}=3y$, the proposed function is a solution.

For a second-order equation, compute the second derivative too. For example, if the equation contains $\dfrac{d^2y}{dt^2}$, checking only the first derivative is incomplete.

Why one equation can have infinitely many solutions

The essential knowledge FUN-7.B.2: There may be infinitely many solutions to a differential equation is visible in the equation $\dfrac{dy}{dt}=ky$. Antidifferentiation gives

$$ \frac{1}{y},dy=k,dt, $$

so

$$ \ln|y|=kt+C. $$

Rewriting the constant produces the general solution

$$ y=Ce^{kt}. $$

Every value of $C$ gives a different solution. Thus $y=2e^{kt}$, $y=10e^{kt}$, and $y=-e^{kt}$ all satisfy the same differential equation when their domains are appropriate. An initial condition, such as $y(0)=6$, selects one particular solution:

$$ 6=Ce^0 \quad\Longrightarrow\quad C=6, $$

so $y=6e^{kt}$.

Slope fields as a first visual estimate

A slope field is a graphical representation of a differential equation on a finite set of points in the plane. At each point $(t,y)$, a short line segment shows the slope required by the differential equation.

For $\dfrac{dy}{dt}=ky$, points with the same $y$-value have the same slope because the right side depends only on $y$. A solution curve can be estimated by tracing a smooth path whose tangent follows the small segments. The exact sketching and reasoning procedures belong to Topics 7.3 and 7.4; here, the key connection is that the differential equation supplies local direction while the solution supplies the global curve.

AP skills in action

Topic 7.2 develops Skill 3.G: Confirm that solutions are accurate and appropriate. In written work, show the derivative of the proposed function, substitute it and the function into the original equation, and state that the two sides agree on the relevant domain. Claiming that a function works, without that substitution, is not a verification.

Misconception check — “A differential equation has one answer.” A differential equation may describe a family of functions. Constants such as $C$ remain until an initial condition or other restriction identifies a particular member.

Retrieval check: Does $y=7e^{-2t}$ satisfy $\dfrac{dy}{dt}=-2y$? Differentiate: $\dfrac{dy}{dt}=-14e^{-2t}$. Since $-2y=-14e^{-2t}$, it does.