Let's dive into solving a classic differential equation and proving a fundamental result. We're going to tackle the equation y' = y, where y' represents the derivative of y with respect to t. Our goal is to show that if y(t) is a solution to this equation, then it must be of the form y(t) = c * e^t, where 'c' is an arbitrary constant. Buckle up, folks, it's gonna be a smooth ride!

Understanding the Problem

First, let's make sure we're all on the same page. The equation y' = y is a first-order, linear, homogeneous differential equation. That's a mouthful, I know, but it just means that the rate of change of y (y') is directly proportional to y itself. Think of it like this: the bigger y is, the faster it grows (or shrinks, if y is negative). This kind of relationship pops up all over the place in science and engineering, from population growth to radioactive decay.

What we want to prove is that every solution to this equation looks like c * e^t. In other words, the function e^t (multiplied by any constant) is the only kind of function that satisfies y' = y. That's a pretty powerful statement, and we're going to show how to prove it rigorously.

Before we jump into the proof, let's think about why this result makes intuitive sense. The exponential function e^t has the special property that its derivative is equal to itself. That's why it's a natural fit for the equation y' = y. The constant 'c' just scales the exponential function up or down, depending on the initial value of y. So, if y(0) = c, then the solution is simply y(t) = c * e^t.

Now, let's get down to the nitty-gritty and prove this result. We'll use a clever trick that involves multiplying both sides of the equation by a carefully chosen function. This will allow us to rewrite the equation in a form that's easy to integrate. Trust me, it's easier than it sounds!

The Proof: A Step-by-Step Guide

Here's how we can formally prove that if y(t) is a solution of y' = y, then y(t) = c * e^t:

1. Start with the differential equation: We are given that y' = y.
2. Multiply by an integrating factor: To solve this, we'll multiply both sides of the equation by e^(-t). This might seem like it's coming out of nowhere, but trust me, it's a standard technique for solving linear differential equations. Multiplying by e^(-t) gives us: e^(-t) * y' = e^(-t) * y. Now, rearrange this to get: e^(-t) * y' - e^(-t) * y = 0.
3. Recognize the product rule: Notice that the left-hand side of the equation looks suspiciously like the result of applying the product rule in reverse. Recall that the product rule states that (u * v)' = u' * v + u * v'. If we let u = e^(-t) and v = y, then u' = -e^(-t) and v' = y'. So, (e^(-t) * y)' = e^(-t) * y' - e^(-t) * y. Therefore, we can rewrite our equation as: (e^(-t) * y)' = 0.
4. Integrate both sides: Now we have a very simple equation: the derivative of e^(-t) * y is zero. That means that e^(-t) * y must be a constant. Let's call that constant 'c'. So, we have: e^(-t) * y = c.
5. Solve for y: Finally, to solve for y, we just multiply both sides of the equation by e^(t): y = c * e^(t).

And that's it! We've shown that if y(t) is a solution to the differential equation y' = y, then it must be of the form y(t) = c * e^(t), where 'c' is an arbitrary constant. QED (quod erat demonstrandum – which was to be demonstrated!).

Why This Matters

Okay, so we've proven a mathematical result. But why should you care? Well, this result is incredibly useful in many areas of science and engineering. Here are just a few examples:

• Population growth: In simple models of population growth, the rate of change of the population is assumed to be proportional to the population size. This leads to the differential equation y' = ky, where k is a constant. The solution to this equation is y(t) = c * e^(kt), which shows that the population grows exponentially over time (if k is positive).
• Radioactive decay: Radioactive decay is a process in which the amount of a radioactive substance decreases over time. The rate of decay is proportional to the amount of substance present, which leads to the differential equation y' = -ky, where k is a positive constant. The solution to this equation is y(t) = c * e^(-kt), which shows that the amount of substance decreases exponentially over time.
• Compound interest: If you invest money in an account that earns compound interest, the amount of money in the account grows exponentially over time. The differential equation that describes this is similar to the population growth equation, and the solution is also an exponential function.

These are just a few examples, but the basic idea is the same: whenever you have a situation where the rate of change of something is proportional to the amount of that thing, you're likely to encounter the differential equation y' = ky and its solution y(t) = c * e^(kt).

General Solutions and Initial Conditions

It's crucial to understand the role of the constant 'c' in the general solution y(t) = c * e^t. This constant is determined by the initial condition of the problem. An initial condition is simply the value of y at a specific time, usually t = 0. For example, if we know that y(0) = 5, then we can plug this into the general solution to find the value of 'c':

5 = c * e^(0) 5 = c * 1 c = 5

So, the specific solution to the differential equation y' = y with the initial condition y(0) = 5 is y(t) = 5 * e^t. Notice that there are infinitely many solutions to the differential equation y' = y (one for each value of 'c'), but only one solution that satisfies a given initial condition.

The general solution y(t) = c * e^t represents a family of curves, each of which satisfies the differential equation. The initial condition picks out one particular curve from this family. Think of it like this: the differential equation tells you the shape of the curve, while the initial condition tells you where to start the curve.

Alternative Approaches

While we've used the integrating factor method to solve the differential equation y' = y, there are other approaches you can take. One common method is called separation of variables. Here's how it works:

1. Rewrite the equation: Start with y' = y and rewrite it as dy/dt = y.
2. Separate the variables: Divide both sides by y and multiply both sides by dt to get: dy/y = dt.
3. Integrate both sides: Integrate both sides of the equation: ∫(dy/y) = ∫dt. This gives you: ln|y| = t + k, where k is an arbitrary constant.
4. Solve for y: Exponentiate both sides of the equation to get: |y| = e^(t+k) = e^k * e^t. Since e^k is just another constant, we can write: y = c * e^t, where c = ±e^k.

Notice that we get the same solution as before: y(t) = c * e^t. The separation of variables method is often simpler than the integrating factor method, but it's not always applicable. It only works if you can separate the variables (i.e., get all the y's on one side of the equation and all the t's on the other side).

Common Mistakes to Avoid

When solving differential equations, it's easy to make mistakes. Here are a few common pitfalls to watch out for:

• Forgetting the constant of integration: When you integrate both sides of an equation, you must add a constant of integration. If you forget this constant, you'll get the wrong solution.
• Dividing by zero: Be careful when dividing by a variable. If the variable could be zero, you need to consider that case separately. In the separation of variables method, we divided by y. We need to check what happens if y = 0. In this case, y' = 0, so y(t) = 0 is a solution. This solution is included in the general solution y(t) = c * e^t when c = 0.
• Incorrectly applying the product rule: Make sure you understand the product rule and how to apply it in reverse. This is crucial for the integrating factor method.
• Not checking your solution: Always check your solution by plugging it back into the original differential equation. This will help you catch any mistakes you may have made.

Conclusion

So there you have it, guys! We've successfully proven that if y(t) is a solution of y' = y, then y(t) = c * e^t. We've also discussed why this result is important, how to solve the equation using different methods, and common mistakes to avoid. I hope this has been helpful. Keep practicing, and you'll be a differential equation whiz in no time!