In AP Calculus, one of the fascinating things we learn is how to solve differential equations. In this section, we will focus on verifying solutions to differential equations, a critical skill in both mathematics and real-world problem solving.

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✅ Verifying Solutions

While actually solving a differential equation may seem daunting, verifying a given solution is a piece of cake! 🍰

Differential equations often have not just one, but infinitely many solutions. These solutions are known as general solutions. Each of these solutions can be tweaked slightly by adding different constants, and yet, they still solve the original differential equation. Imagine a family of curves on a graph, each differing slightly from the others, but all fitting the same overall pattern described by the differential equation.

Image courtesy of Wolfram

This verification process is rooted in understanding derivatives and how they function. When you're given a differential equation and a potential solution, your job is to take the derivative of the proposed solution and see if it fits perfectly into the original equation. It's like having a key and checking to see if it fits the lock.

📝 The Verification Process

Let's say you're given a differential equation and a potential solution. How do you verify if this solution is correct? You start by taking the derivative of the proposed solution. Then, you substitute this derivative, along with the original solution, back into the differential equation. If all parts align and the equation holds true, then you've successfully verified the solution. 🥳

For example, consider the differential equation $dy/dx = 3x^2$. If you're given a potential solution $y=x^3$, you would first find the derivative of $y$ , which is $dy/dx = 3x^2$. Then, you substitute this back into the original equation. Since $3x^2 = 3x^2$, the solution is verified.

📍 Verifying a Solution Walkthrough Problem

Verify if the function $y = x^2sin(x)$ is a solution to the differential equation $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$. Let’s go through this! ⬇️

We need to verify if $y = x^2sin(x)$ satisfies the given differential equation. To do this, we'll first find the derivative of $y$ and then check if it matches the right-hand side of the differential equation.

Differentiate $y = x^2sin(x)$. To differentiate, we'll use the product rule since it's a product of two functions which states: if $u = x^2$ and $v=sin(x)$, then $y'=u'v +uv'$.

1. $u’ = 2x$ and $v’ = cos(x)$
2. Multiply based on the product rule: $y’ = (x^2)(cos(x)) + (2x)(sin(x))$
3. $y’ = (x^2)(cos(x)) + (2x)(sin(x))$

Finally, we just need to verify the solution by checking whether $y'=\frac {dy}{dx}$. Since $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$ and $y’ = (x^2)(cos(x)) + (2x)(sin(x))$, the solution is verified!

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✏️ Practice Problems

Let’s put your new skills to good use! Try out a couple practice problems:

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

✔️ Step-By-Step Solution: Example 1

First things first, we have to understand the given task.

We need to verify if the function $y = e^{2x}$ satisfies the given differential equation, which is $\frac{dy}{dx} = 2e^{2x}$. To do this, we'll find the derivative of y and check if it matches the right-hand side of the differential equation.

Now, we can differentiate the given function.

Differentiate $y = e^{2x}$ with respect to x. Since it's an exponential function, its derivative is simply itself times the derivative of the exponent:

Calculate the derivative of 2x:

Therefore,

Last but not least, verify with the differentiate equation!

Since both sides are equal ($\frac{dy}{dx} = 2e^{2x}$), the proposed solution $y = e^{2x}$ successfully verifies as a solution to the differential equation $\frac{dy}{dx} = 2e^{2x}$. ✅

✔️ Step-By-Step Solution: Example 2

Go through the same steps!

1. Understand the Task
   1. We need to verify if the function $y = 3x^3$ satisfies the given differential equation, which is $\frac{dy}{dx} = 9x^2$. To do this, we'll find the derivative of y and check if it matches the right-hand side of the differential equation.
2. Differentiate the Given Function
   1. Differentiate $y = 3x^3$ with respect to x. Since it's a polynomial, we can use the power rule directly: $\frac {dy}{dx} = \frac{d}{dx}(3x^3)$.
      1. Apply the power rule: $\frac {dy}{dx} = \frac{d}{dx}(3x^3)$ = $3*3x^{3-1}$ = $9x^2$.
3. Verify with Differential Equation
   1. We have found that $\frac{dy}{dx} = 9x^2$ for the function $y = 3x^3$, and this matches the right-hand side of the differential equation, which is also $9x^2$, making $y = 3x^3$ a verified solution to the differential equation.

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⭐️ Conclusion

Understanding how to verify solutions to differential equations opens up a world of infinite possibilities. It's not just about solving a math problem; it's about exploring a universe of potential solutions, each fitting the equation in its unique way. ✨