Solving The Differential Equation Y' - 3y = 6 General Solution Explained

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Hey guys! Today, we're diving deep into the fascinating world of differential equations, specifically focusing on a first-order ordinary differential equation (ODE). We'll be tackling the equation y' - 3y = 6, exploring its solution, and understanding the underlying concepts. So, buckle up and let's get started!

Understanding the Problem

Our main task is to find the general solution to the given first-order ODE:

y' - 3y = 6

Where y' represents the first derivative of the function y with respect to the independent variable (which we'll assume is x here). In simpler terms, we're looking for a function y(x) that satisfies this equation. Now, one common technique to solve such equations is separation of variables. However, the prompt suggests we explore other methods, making this a more interesting challenge. Before we jump into the solution, it's crucial to understand what makes this equation tick.

This particular ODE is a linear first-order ODE. Linearity means that the dependent variable (y) and its derivatives appear only to the first power and are not multiplied together. The first-order designation comes from the highest order derivative present in the equation (which is y'). Recognizing these characteristics is key because it guides us towards appropriate solution strategies. Linear first-order ODEs have a standard form, which we'll leverage to crack this problem.

Rewriting the Equation

The first step in tackling this beast is to rewrite the equation into its standard form. The standard form for a first-order linear ODE is:

y' + P(x)y = Q(x)

Where P(x) and Q(x) are functions of x. Our given equation, y' - 3y = 6, is already pretty close. We can directly see that P(x) = -3 and Q(x) = 6. This simple rearrangement is a crucial step because it sets the stage for using the integrating factor method, a powerful technique for solving these types of equations. The ability to identify and transform equations into standard forms is a fundamental skill in differential equations, and it's something you'll use time and time again.

The Integrating Factor Method

Now comes the fun part – applying the integrating factor method. This method is a clever way to turn a non-exact differential equation into an exact one, which we can then solve. The key idea is to multiply the entire equation by a special function called the integrating factor, denoted by μ(x) (mu of x). This integrating factor is chosen specifically to make the left-hand side of the equation a derivative of a product.

The formula for the integrating factor is:

μ(x) = e^(∫P(x) dx)

Where e is the base of the natural logarithm, and the integral is taken with respect to x. Remember, P(x) is the coefficient of y in the standard form of the ODE. In our case, P(x) = -3, so we can calculate the integrating factor:

μ(x) = e^(∫-3 dx) = e^(-3x)

So, our integrating factor is e^(-3x). This seemingly simple function is the magic ingredient that will unlock the solution to our ODE. We're not done yet, though. We need to multiply both sides of our standard form equation by this integrating factor. Doing so transforms the equation in a way that makes it solvable by direct integration.

Multiplying the entire equation y' - 3y = 6 by e^(-3x), we get:

e^(-3x)y' - 3e^(-3x)y = 6e^(-3x)

This might look more complicated, but trust me, it's a step in the right direction. The left-hand side of this equation is now the derivative of a product, which is the whole point of using the integrating factor method. Can you see what product it is the derivative of? Think about the product rule for differentiation!

Recognizing the Product Rule

The left-hand side of the equation, e^(-3x)y' - 3e^(-3x)y, is actually the derivative of the product y(x)e^(-3x). This is a crucial observation because it allows us to rewrite the equation in a much simpler form. Let's verify this using the product rule. The product rule states that the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x). In our case, let u(x) = y(x) and v(x) = e^(-3x).

Then, u'(x) = y'(x) and v'(x) = -3e^(-3x). Applying the product rule, we get:

d/dx [y(x)e^(-3x)] = y'(x)e^(-3x) + y(x)(-3e^(-3x)) = e^(-3x)y' - 3e^(-3x)y

This is exactly the left-hand side of our equation! So, we can rewrite the equation as:

d/dx [y(x)e^(-3x)] = 6e^(-3x)

This is a massive simplification. We've transformed a differential equation into a statement about the derivative of a product. Now, we can simply integrate both sides with respect to x to get rid of the derivative.

Solving the Equation

Integrating Both Sides

Now that we have the equation in the form d/dx [y(x)e^(-3x)] = 6e^(-3x), we can integrate both sides with respect to x. The integral of a derivative is just the original function, so the left-hand side becomes y(x)e^(-3x). On the right-hand side, we need to integrate 6e^(-3x). This is a straightforward integration using a simple u-substitution (or recognizing the pattern directly).

The integral of 6e^(-3x) is –2e^(-3x) + C, where C is the constant of integration. Remember, when we integrate, we always need to add a constant of integration because the derivative of a constant is zero. So, our equation now looks like this:

y(x)e^(-3x) = -2e^(-3x) + C

We're almost there! We just need to isolate y(x) to get the general solution.

Isolating y(x)

To isolate y(x), we simply multiply both sides of the equation by e^(3x). This will cancel out the e^(-3x) term on the left-hand side. Doing so, we get:

y(x) = -2 + Ce^(3x)

And there you have it! This is the general solution to the first-order ODE y' - 3y = 6. This solution represents a family of functions, each differing by the value of the constant C. The constant C is determined by initial conditions, if any are given. If we were given an initial condition, such as y(0) = 1, we could plug in x = 0 and y = 1 into the general solution and solve for C. This would give us a particular solution, which is a single function that satisfies both the ODE and the initial condition.

The General Solution

Therefore, the general solution to the ordinary differential equation y' - 3y = 6 is:

y(x) = -2 + Ce^(3x)

Where C is an arbitrary constant. This solution encompasses all possible solutions to the differential equation. Each value of C corresponds to a unique solution curve. When solving differential equations, it’s crucial to express the solution in its most general form, including the arbitrary constant, unless specific initial conditions are provided to determine a particular solution. The general solution provides a complete picture of the behavior of the system described by the differential equation.

Conclusion

So guys, we successfully navigated the world of first-order ODEs and found the general solution to the equation y' - 3y = 6 using the integrating factor method. We started by understanding the problem, rewriting the equation in standard form, calculating the integrating factor, and then integrating both sides to arrive at the solution. This journey highlights the power and elegance of the integrating factor method in solving linear first-order ODEs. Remember, the key is to recognize the standard form, calculate the integrating factor correctly, and then carefully perform the integration steps. With practice, you'll become a pro at solving these types of equations. Keep exploring, keep learning, and keep solving!