Verifying Solutions for Differential Equations - AP Calculus BC Study Guide

The Core Idea: Verifying Solutions for Differential Equations

A differential equation is an equation that contains an unknown function and one or more of its derivatives. Unlike an algebraic equation where the solution is typically a number, the solution to a differential equation is a function. The core idea of this topic is the process of verification: determining if a given function, $y=f(x)$, is indeed a valid solution to a given differential equation.

This verification process is analogous to checking a solution in algebra. To check if $x=3$ is a solution to $2x-1=5$, you substitute $3$ for $x$ and see if the equation holds true. Similarly, to verify a functional solution, you substitute the function and its required derivatives into the differential equation. If the substitution results in a true statement (i.e., the left side of the equation equals the right side), then the function is confirmed to be a solution. This process is a fundamental skill that confirms the relationship between a function and the differential equation it satisfies.

Key Process for Verification

The Essential Knowledge for this topic outlines a clear, procedural method for verifying a solution to a differential equation. There are no new formulas to memorize; rather, you must apply your existing knowledge of differentiation and algebra. The process is as follows:

1. Identify the Proposed Solution and the Differential Equation: You will be given a function, such as $y=f(x)$, and a differential equation involving $y$, $x$, and derivatives of $y$ (e.g., $\frac{dy}{dx}$, $\frac{d^{2}y}{d{x}^2}$).

2. Calculate Necessary Derivatives: Find all the derivatives of the proposed function that appear in the differential equation. If the equation has a $\frac{dy}{dx}$, you must find the first derivative. If it has a $\frac{d^{2}y}{d{x}^2}$, you must find the first and second derivatives.

3. Substitute into the Differential Equation: Replace every instance of $y$ with the given function and every instance of its derivatives (e.g., $\frac{dy}{dx}$) with the expressions you calculated in the previous step.

4. Simplify and Verify: Use algebraic manipulation to simplify both sides of the equation. If the left side of the equation simplifies to be identical to the right side, you have successfully verified that the function is a solution. The goal is to show the equation is satisfied and results in an identity, such as $0=0$ or $3x^2=3x^2$.

Understanding the Nature of a Solution

The critical nuance of this topic is understanding that a "solution" is a function, not a value. A function $y=f(x)$ is a solution to a differential equation if it satisfies the equation for all values of $x$ within its domain. The verification process is a formal proof of this relationship.

This means that the verification must hold true algebraically, not just for a single point. You are not plugging in a number for $x$; you are substituting entire functional expressions. The final step of the verification, where you show that the two sides of the equation are identical, is the confirmation that the relationship holds for any valid $x$. This process fundamentally connects the concept of a derivative as a rate of change to the function itself, showing that the function's behavior (its derivatives) is precisely described by the differential equation.

Core Concepts & Rules

• A Solution is a Function: The solution to a differential equation is a function, $y=f(x)$, that makes the differential equation a true statement when the function and its derivatives are substituted into it.

• Verification is a Process of Substitution: To verify a solution, you must substitute the function $y$ and its derivative(s) (e.g., $\frac{dy}{dx}$, $\frac{d^{2}y}{d{x}^2}$) into the differential equation.

• The Goal is an Identity: The verification is successful if, after substitution and algebraic simplification, the left side of the equation is identical to the right side.

• Mastery of Derivatives is Essential: This process relies heavily on accurate computation of derivatives. You must be proficient with all differentiation rules, including the power, product, quotient, and chain rules.

• Algebraic Precision is Crucial: After substitution, the problem becomes one of algebraic manipulation. Careful and accurate simplification is required to show that the two sides of the equation are equal.

Step-by-Step Example 1: Verifying a First-Order Solution

Problem: Show that $y=4e^{-3x}$ is a solution to the differential equation $\frac{dy}{dx}+3y=0$.

Step 1: Find the necessary derivative.

The differential equation contains $\frac{dy}{dx}$, so we need to find the first derivative of the proposed solution $y=4e^{-3x}$. Using the chain rule:

$$\frac{dy}{dx}=\frac{d}{dx}(4e^{-3x})=4e^{-3x}\cdot (-3)=-12e^{-3x}$$

Step 2: Substitute into the differential equation.

Replace $\frac{dy}{dx}$ with $-12e^{-3x}$ and $y$ with $4e^{-3x}$ in the equation $\frac{dy}{dx}+3y=0$.

$$(-12e^{-3x})+3(4e^{-3x})=0$$

Step 3: Simplify and verify.

Perform the algebraic simplification on the left side of the equation.

$$-12e^{-3x}+12e^{-3x}=0$$

$$0=0$$

Conclusion: Since the substitution leads to the true statement $0=0$, the function $y=4e^{-3x}$ is a solution to the differential equation.

Step-by-Step Example 2: Verifying a Second-Order Solution

Problem: Determine if $y=x\ln (x)$ is a solution to the second-order differential equation $x^2{y}^{\prime \prime }-x{y}^{\prime }+y=0$ for $x>0$.

Step 1: Find the first derivative, $y^{\prime }$.

The proposed solution is $y=x\ln (x)$. We use the product rule: $u{v}^{\prime }+v{u}^{\prime }$.

Let $u=x$ and $v=\ln (x)$. Then $u^{\prime }=1$ and $v^{\prime }=\frac{1}{x}$.

$$y^{\prime }=\frac{dy}{dx}=(x)\left(\frac{1}{x}\right)+(\ln x)(1)=1+\ln (x)$$

Step 2: Find the second derivative, $y^{\prime \prime }$.

Now, we differentiate $y^{\prime }=1+\ln (x)$.

$$y^{\prime \prime }=\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(1+\ln (x))=0+\frac{1}{x}=\frac{1}{x}$$

Step 3: Substitute $y$, $y^{\prime }$, and $y^{\prime \prime }$ into the differential equation.

The equation is $x^2{y}^{\prime \prime }-x{y}^{\prime }+y=0$.

Substitute $y^{\prime \prime }=\frac{1}{x}$, $y^{\prime }=1+\ln (x)$, and $y=x\ln (x)$.

$$x^2\left(\frac{1}{x}\right)-x(1+\ln (x))+(x\ln (x))=0$$

Step 4: Simplify the left side.

Carefully perform the algebraic operations.

$$\frac{x^2}{x}-x-x\ln (x)+x\ln (x)=0$$

$$x-x-x\ln (x)+x\ln (x)=0$$

Combine like terms.

$$(x-x)+(-x\ln (x)+x\ln (x))=0$$

$$0+0=0$$

$$0=0$$

Conclusion: The substitution results in the identity $0=0$. Therefore, $y=x\ln (x)$ is a solution to the differential equation for $x>0$.

Using Your Calculator

Verifying solutions for differential equations is an analytical process that must be demonstrated with by-hand calculations of derivatives and algebraic simplification. A calculator cannot perform the symbolic manipulation required for a formal verification. Therefore, this topic is typically tested on the no-calculator portion of the AP exam.

However, a graphing calculator can be used to check your work or build confidence in your answer.

Method: Graphical Check

You can check if the two sides of the differential equation are equal by graphing them separately.

Example: To check the verification from Example 1 ($\frac{dy}{dx}=-3y$ for $y=4e^{-3x}$), you can graph the left side and the right side.

1. Graph the Left Side (the derivative): In your calculator's graphing menu, enter Y1 = nDeriv(4e^(-3X), X, X). The nDeriv function (often found in the MATH menu) numerically calculates the derivative of the expression for each value of X.

2. Graph the Right Side (the expression): In Y2, enter the right side of the rewritten differential equation, $-3y$. Substitute the expression for y`: `Y2 = -3 * (4e^(-3X))`, which simplifies to `Y2 = -12e^(-3X)`. 3. **Compare the Graphs:** If your by-hand derivative calculation was correct, the graph of `Y1` (the numerical derivative) and `Y2` (the analytical expression for the derivative) should be identical. If they are, it provides strong evidence that your differentiation was correct, which is the most critical step in the verification process. This method does not replace the need for the analytical steps but serves as a powerful tool for confirming your calculations. ## AP Exam Quick Hit ### Common Question Types - **Direct Verification (Free Response):** You will be given a function and a differential equation and asked to "Show that" or "Verify that" the function is a solution. This requires you to write out all the steps of differentiation and substitution clearly. - *Example:* "Show that $y = 2\sin(x) + \cos(x) is a solution to $\frac{d^{2}y}{d{x}^2}+y=0$."

• Multiple-Choice Selection: You will be given a differential equation and several functions as options. You must test each option until you find the one that satisfies the equation. This can be time-consuming, so efficient differentiation is key.

  • Example: "Which of the following is a solution to $x{y}^{\prime }-2y=0$? (A) $y=x^3$ (B) $y=e^{2x}$ (C) $y=x^2$ (D) $y=\ln (x)$"

• Finding a Parameter (Multiple Choice or Free Response): You will be given a general form of a solution with an unknown constant and asked to find the value of that constant that makes it a valid solution.

  • Example: "For what value of $k$ is $y=e^{kt}$ a solution to $2y^{\prime \prime }+y^{\prime }-y=0$?"

Common Mistakes

• Chain Rule Errors: Forgetting to apply the chain rule is the single most common error. For a function like $y=e^{5x}$, students might incorrectly state $y^{\prime }=e^{5x}$ instead of the correct $y^{\prime }=5e^{5x}$.

• Product/Quotient Rule Errors: Incorrectly applying or forgetting to use the product or quotient rule when necessary. For $y=x^2{e}^x$, one must use the product rule.

• Algebraic Simplification Errors: After correctly substituting the function and its derivatives, simple mistakes in distribution, combining terms, or sign errors can lead to an incorrect conclusion that a valid solution does not work.

• Substitution Errors: Accidentally swapping $y$ and $y^{\prime }$ during the substitution step. It is critical to be organized and substitute each component into its correct place in the original differential equation.

• Assuming a Solution Works: On a "Show that" question, you cannot assume the statement is true. You must show every step of the process that leads to the final identity (e.g., $0=0$). Simply writing the function and its derivative and then jumping to the conclusion is not sufficient.