7.2 Verifying Solutions for Differential Equations

To verify a solution to a differential equation, you plug the proposed function and its derivatives into the equation and check that both sides match for all $x$. If the equation reduces to a true statement, the function is a solution. For AP Calculus, show the substitution clearly so the verification is more than a yes-or-no claim.

Why This Matters for the AP Calculus Exam

Verifying solutions is a quick, reliable skill that shows up across the differential equations unit on the AP Calculus exam. You will often be handed a function and asked to confirm whether it satisfies a given equation, which means computing a derivative and substituting carefully. This connects directly to later skills like separation of variables, particular solutions from initial conditions, and reading slope fields. Getting comfortable with the plug-and-check method also builds your confidence with derivative rules under time pressure, which helps on both multiple-choice and free-response problems where you set up and solve differential equations.

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Key Takeaways

• A function is a solution to a differential equation when substituting it (and its derivatives) makes the equation true for every x in the interval.
• Use the right derivative rules to find the needed derivative, then plug it back in and simplify both sides.
• A differential equation can have infinitely many general solutions that differ only by a constant of integration.
• A particular solution is the single curve that also passes through a given point or satisfies an initial condition.
• Verifying does not require solving the equation, so it is usually faster and lower-risk than finding the solution from scratch.

How a Solution Verification Works

Solving a differential equation from scratch can feel like a lot of steps, but verifying a given solution is much simpler. You are not searching for the answer; you are checking whether a proposed answer fits.

Differential equations often have not just one solution but infinitely many. These are called general solutions, and they usually differ only by a constant. Picture a family of curves on a graph, each shifted slightly, but all following the same pattern described by the differential equation. When you add an initial condition or a point on the curve, you narrow that family down to one particular solution.

The verification process is built on derivatives. When you are given a differential equation and a proposed solution, your job is to differentiate the proposed solution and check whether it fits the original equation. Think of it like testing a key in a lock.

The Verification Process

Suppose you are given a differential equation and a proposed solution. To check it:

1. Find the derivative (or derivatives) of the proposed solution.
2. Substitute the function and its derivatives back into the differential equation.
3. Simplify both sides. If the equation holds true for all x, the solution is verified.

For example, consider the differential equation $dy/dx = 3x^2$. If you are given the proposed solution $y=x^3$, first find the derivative of $y$, which is $dy/dx = 3x^2$. Substitute it back into the original equation. Since $3x^2 = 3x^2$, the solution checks out.

Worked Example: Product Rule Check

Verify whether $y = x^2sin(x)$ is a solution to the differential equation $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$.

You need to confirm that $y = x^2sin(x)$ satisfies the equation. Differentiate $y$ and check whether it matches the right side.

Differentiate $y = x^2sin(x)$ using the product rule. With $u = x^2$ and $v=sin(x)$, the rule gives $y'=u'v +uv'$.

1. $u' = 2x$ and $v' = cos(x)$
2. Apply the product rule: $y' = (x^2)(cos(x)) + (2x)(sin(x))$

Now check whether $y'=\frac {dy}{dx}$. Since $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$ and $y' = (x^2)(cos(x)) + (2x)(sin(x))$, both sides match, so the solution is verified.

How to Use This on the AP Calculus Exam

MCQ

Many multiple-choice problems hand you a function and ask which differential equation it satisfies, or vice versa. Differentiate the candidate function, substitute, and look for the choice where both sides reduce to the same expression. Watch for answer choices that match only at one x value but not for all x.

Free Response

When a free-response problem asks you to verify a solution, show the derivative step clearly and then show the substitution that makes both sides equal. Writing out the derivative and the final comparison makes your reasoning easy to follow, which is important for clear exam work.

Problem Solving

• Differentiate carefully using the correct rule (power, product, quotient, or chain).
• For a second-order equation, you may need the first and second derivatives.
• Substitute every required derivative, not just the first one.
• Simplify until you can clearly say the two sides are equal for all x in the interval.

Practice Problems

Try these to build fluency with the plug-and-check method:

1. Verify whether the function $y = e^{2x}$ is a solution to the differential equation $\frac{dy}{dx} = 2e^{2x}$.
2. Verify whether the function $y = 3x^3$ is a solution to the differential equation $\frac{dy}{dx} = 9x^2$.

Step-By-Step Solution: Example 1

First, understand the task. You need to verify whether $y = e^{2x}$ satisfies $\frac{dy}{dx} = 2e^{2x}$. To do this, find the derivative of $y$ and check if it matches the right side.

Next, differentiate the given function. Differentiate $y = e^{2x}$ with respect to $x$. Using the chain rule, the derivative is the function times the derivative of the exponent:

Calculate the derivative of $2x$:

Therefore,

Finally, verify against the differential equation. Since both sides are equal ($\frac{dy}{dx} = 2e^{2x}$), the proposed solution $y = e^{2x}$ is verified.

Step-By-Step Solution: Example 2

Follow the same steps.

1. Understand the task: verify whether $y = 3x^3$ satisfies $\frac{dy}{dx} = 9x^2$. Find the derivative of $y$ and check if it matches the right side.
2. Differentiate the given function: differentiate $y = 3x^3$ with respect to $x$. Using the power rule, $\frac {dy}{dx} = \frac{d}{dx}(3x^3) = 3*3x^{3-1} = 9x^2$.
3. Verify with the differential equation: $\frac{dy}{dx} = 9x^2$ matches the right side, which is also $9x^2$, so $y = 3x^3$ is a verified solution.

Common Misconceptions

• Verifying a solution is not the same as solving the equation. You only need to differentiate and substitute, not isolate variables or integrate.
• A match at one x value does not verify a solution. The equation must hold for every x in the interval.
• Not every differential equation has a single solution. General solutions form a family that differs by a constant, while a particular solution is pinned down by an initial condition or a point on the curve.
• Do not assume differential equations with fractions always lead to logarithmic solutions. Some do, but many do not, so verify with the actual derivative.
• For higher-order equations, remember to compute and substitute every derivative the equation asks for, not just the first.

Related AP Calculus Guides

• Unit 7 Overview: Differential Equations
• 7.3 Sketching Slope Fields
• 7.6 Finding General Solutions Using Separation of Variables
• 7.5 Approximating Solutions Using Euler’s Method
• 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables
• 7.4 Reasoning Using Slope Fields