Solving Y' - 3y = 6 A Step-by-Step Guide
Hey guys! Let's dive into the exciting world of differential equations! Today, we're going to tackle a classic problem: solving a first-order ordinary differential equation (ODE). Specifically, we'll be working through the equation y' - 3y = 6 step by step. Don't worry if you're new to this – we'll break it down so it's super easy to follow.
Understanding First-Order ODEs
First, let's make sure we're all on the same page. A first-order ordinary differential equation is an equation that involves an unknown function and its first derivative. In our case, the unknown function is y, and its first derivative is y'. The term "ordinary" means that we're dealing with functions of a single independent variable, which is usually x in these problems. So, y is a function of x, and y' represents the rate of change of y with respect to x (dy/dx). Understanding the basics of first-order ODEs is crucial before jumping into solving them. They pop up all over the place in science and engineering, from modeling population growth to describing the motion of objects.
The general form of a linear first-order ODE is: y' + P(x)y = Q(x). Comparing this to our equation, y' - 3y = 6, we can see that P(x) = -3 and Q(x) = 6. This is a linear equation because y and y' appear to the first power and are not multiplied together. Recognizing the form of the equation helps us choose the right solution method. There are different types of first-order ODEs, such as separable equations, exact equations, and linear equations. Each type requires a slightly different approach, so it's essential to identify the type you're dealing with. For example, separable equations can be solved by directly integrating both sides after separating the variables, while exact equations involve checking for a specific condition related to partial derivatives.
Our equation falls into the category of linear first-order ODEs, which are often solved using an integrating factor. This method is powerful because it transforms the equation into a form that can be easily integrated. Before we jump into the integrating factor method, it's helpful to understand why it works. The idea is to multiply the entire equation by a function (the integrating factor) that makes the left-hand side the derivative of a product. This allows us to integrate both sides with respect to x and solve for y. Linear first-order ODEs are particularly important in many applications because they often arise in models where the rate of change of a quantity depends linearly on the quantity itself and some external input. For example, in electrical circuits, the current in a circuit with a resistor and an inductor can be modeled by a linear first-order ODE.
Step 1: Identifying the Integrating Factor
The magic ingredient for solving this type of equation is the integrating factor. The integrating factor, often denoted by μ(x), is calculated using the formula: μ(x) = e^(∫P(x) dx). Remember, P(x) is the coefficient of the y term in our equation. In our case, P(x) = -3. So, let's plug that into the formula and calculate our integrating factor. We need to find the integral of -3 with respect to x, which is simply -3x. Then, we take e to the power of that result. This might seem a bit abstract at first, but the integrating factor is really just a clever trick that makes the equation solvable. It's like a special key that unlocks the solution.
The integral of -3 with respect to x is -3x. Therefore, our integrating factor μ(x) = e^(-3x). This integrating factor is the function we'll multiply our entire equation by. Now, why do we do this? Well, the magic happens because multiplying by the integrating factor transforms the left-hand side of the equation into the derivative of a product. This is the key to simplifying the equation and ultimately solving for y. It's like finding the perfect ingredient that makes a recipe work. Without the integrating factor, solving the equation would be much more difficult. The concept of an integrating factor is not unique to differential equations; it's also used in other areas of mathematics and physics to simplify problems. For example, it's used in solving linear partial differential equations and in some areas of control theory.
Calculating the integrating factor correctly is crucial. A small mistake here can throw off the entire solution. So, double-check your work and make sure you've properly identified P(x) and performed the integration. Once you have the integrating factor, the rest of the process becomes much smoother. Think of the integrating factor as a tool that streamlines the solution process. It's not just a random mathematical trick; it's a carefully designed method that leverages the properties of derivatives and integrals. Understanding the purpose of the integrating factor can help you remember the formula and apply it correctly. It's not just about memorizing steps; it's about understanding why those steps work.
Step 2: Multiplying the Equation by the Integrating Factor
Now that we've found our integrating factor, μ(x) = e^(-3x), we're going to multiply both sides of the original equation (y' - 3y = 6) by it. This is a crucial step, so let's do it carefully. Multiplying both sides by e^(-3x) gives us: e^(-3x)y' - 3e^(-3x)y = 6e^(-3x). This might look more complicated, but trust me, it's actually simplifying things. The left-hand side is now in a special form that we can recognize. This step is all about transforming the equation into a form that's easier to integrate. We're essentially preparing the equation for the final integration step.
Notice the left-hand side: e^(-3x)y' - 3e^(-3x)y. This is the derivative of the product of y and our integrating factor, e^(-3x). In other words, it's the derivative of (ye^(-3x)). This is the magic of the integrating factor at work! It transforms the left-hand side into a simple derivative. We can rewrite the equation as: d/dx (ye^(-3x)) = 6e^(-3x). This is a key realization because now we have the derivative of something equal to something else. To find that
