AP Calculus AB: Differential Equations Verification

Verifying a solution to a differential equation is a foundational skill that transforms the abstract concept of a "solution" into a concrete, checkable procedure. On the AP exam, you will frequently be asked to confirm that a given function satisfies a differential equation, a task that tests your differentiation skills and logical substitution. Mastering this process not only guarantees accuracy on these specific questions but also deepens your understanding of what differential equations represent and builds a critical habit of checking your work.

What Does It Mean to Verify a Solution?

A differential equation is an equation that relates a function with one or more of its derivatives. A solution to a differential equation is a function that, when substituted into the equation along with its required derivatives, makes the equation true for all values in its domain. Verification is the process of proving this truth.

Think of the differential equation as a rule or a condition. A proposed solution is a candidate that claims to follow that rule. Your job as the verifier is to audit this candidate by performing two core steps:

1. Differentiate: Find the necessary derivative(s) of the proposed solution.
2. Substitute: Plug both the original function and its derivative(s) into the original differential equation and simplify.

If the left-hand side (LHS) simplifies identically to the right-hand side (RHS) for all $x$ (or $t$), the verification is successful. This process does not find the solution; it confirms a given function's validity.

The Verification Process for Explicit Functions

The most straightforward cases involve explicit functions, where $y$ is written solely in terms of $x$, such as $y=x^3+\cos (2x)$. The process here is a direct application of calculus and algebra.

Example 1: First-Order Verification Verify that $y=2e^{3x}$ is a solution to the differential equation $\frac{dy}{dx}=3y$.

Step 1: Differentiate. Given $y=2e^{3x}$, find its derivative using the chain rule: $$\frac{dy}{dx}=2\cdot e^{3x}\cdot 3=6e^{3x}.$$

Step 2: Substitute. The differential equation states: $\frac{dy}{dx}\stackrel{?}{=}3y$. Substitute your expressions for $y$ and $\frac{dy}{dx}$ into this equation: LHS: $\frac{dy}{dx}=6e^{3x}$. RHS: $3y=3(2e^{3x})=6e^{3x}$. Since $6e^{3x}=6e^{3x}$ for all $x$, the equation is satisfied. Verification complete.

Example 2: Second-Order Verification Verify that $y=4\sin (2x)-5\cos (2x)$ is a solution to $\frac{d^{2}y}{d{x}^2}+4y=0$.

Step 1: Differentiate (Twice). First derivative: $\frac{dy}{dx}=8\cos (2x)+10\sin (2x)$. Second derivative: $\frac{d^{2}y}{d{x}^2}=-16\sin (2x)+20\cos (2x)$.

Step 2: Substitute. Substitute $y$ and $\frac{d^{2}y}{d{x}^2}$ into the left-hand side of the equation: $$\frac{d^{2}y}{d{x}^2}+4y=[-16\sin (2x)+20\cos (2x)]+4[4\sin (2x)-5\cos (2x)]$$ Simplify: $$=-16\sin (2x)+20\cos (2x)+16\sin (2x)-20\cos (2x)=0.$$ This matches the right-hand side of $0$. The function is a verified solution.

Verification with Implicitly Defined Solutions

Sometimes, the proposed solution is given implicitly as an equation relating $x$ and $y$, like $x^2+y^2=25$. Here, you must use implicit differentiation to find $\frac{dy}{dx}$ before substituting.

Example 3: Implicit Verification Verify that $x^2+y^2=C$, where $C$ is a constant, is an implicit solution to the differential equation $\frac{dy}{dx}=-\frac{x}{y}$.

Step 1: Differentiate Implicitly. Differentiate both sides of $x^2+y^2=C$ with respect to $x$: $$\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(C)$$ $$2x+2y\frac{dy}{dx}=0.$$

Step 2: Solve for the derivative. Solve the resulting equation for $\frac{dy}{dx}$: $$2y\frac{dy}{dx}=-2x$$ $$\frac{dy}{dx}=-\frac{x}{y}.$$

The expression we derived for $\frac{dy}{dx}$ matches the given differential equation exactly. Therefore, any function $y$ defined implicitly by $x^2+y^2=C$ (for $y\ne 0$) satisfies the equation. This verifies the entire family of solutions.

Verifying General and Particular Solutions

You must distinguish between a general solution, which includes an arbitrary constant $C$ (representing a family of curves), and a particular solution, where $C$ has a specific value determined by an initial condition (e.g., $y(0)=2$).

The verification process is identical: you differentiate and substitute, treating $C$ as a constant. For a general solution, your substitution should result in an identity (like $0=0$) regardless of the value of $C$. For a particular solution, you simply use the given $C$ value.

Example 4: General Solution Verify that $y=C{e}^{x^2}$ is a general solution to $\frac{dy}{dx}=2xy$.

1. Differentiate: $\frac{dy}{dx}=C\cdot e^{x^2}\cdot 2x=2xC{e}^{x^2}$.
2. Substitute into RHS: $2xy=2x(C{e}^{x^2})=2xC{e}^{x^2}$.

Since LHS ($\frac{dy}{dx}$) equals RHS, the function satisfies the differential equation for any real value of the constant $C$.

Common Pitfalls

1. Incorrect Differentiation: This is the most frequent error. Pay close attention to the chain rule, product rule, and implicit differentiation. For example, the derivative of $e^{2x}$ is $2e^{2x}$, not $e^{2x}$. Always double-check your calculus before substituting.

• Correction: Practice derivative rules in isolation. Before substituting, ask yourself, "Is this derivative absolutely correct?"

1. Algebraic Errors During Substitution: After substituting, simplifying the expression is crucial. Mistakes in distributing negatives, combining like terms, or canceling factors can lead you to incorrectly conclude a function is not a solution.

• Correction: Simplify methodically, one step at a time. For complex verifications, write out all terms before combining.

1. Ignoring the Domain of the Solution: A proposed solution must satisfy the equation for its entire domain. A function like $y=\sqrt{x}$ cannot be a solution to an equation requiring its derivative at $x=0$, as it is not differentiable there. Similarly, in implicit solutions like Example 3, the verification $\frac{dy}{dx}=-x/y$ is only valid where $y\ne 0$.

• Correction: After verifying the algebra, consider if there are any values (like division by zero or even roots of negatives) that would break the equality.

1. Confusing Verification with Solving: Verification is a check of a proposed answer. It is not a method for finding the solution from scratch. Trying to "reverse-engineer" the verification step into a solution technique will lead to confusion on exams.

• Correction: Mentally separate the two processes. Verification is a straightforward "plug-and-chug" procedure following the given steps.

Summary

• To verify a solution to a differential equation, you must differentiate the proposed function and substitute it and its derivatives into the original equation. If the equation holds true (simplifies to an identity like $0=0$ or $f(x)=f(x)$), the function is a solution.
• The process is identical for explicit functions ($y=f(x)$), implicit functions ($F(x,y)=C$), general solutions (with $C$), and particular solutions (with a specific $C$). For implicit functions, use implicit differentiation.
• This skill is vital for the AP exam, both for direct verification questions and for checking your answers to free-response differential equation problems, ensuring you earn maximum points.
• The most common errors involve calculus (wrong derivatives) and algebra (improper simplification). Careful, step-by-step work mitigates these risks.
• Ultimately, mastering verification reinforces the core concept: a solution to a differential equation is a function whose behavior, defined by its derivatives, perfectly matches the relationship prescribed by the equation.