Solving Differential Equations: Y' - 4y = E^(5t)
Differential equations might seem daunting at first, but they're actually a powerful tool for modeling real-world phenomena. This article will walk you through the process of finding the general solution to a specific differential equation: y' - 4y = e^(5t). We'll break down each step in a clear, easy-to-understand way, so you can confidently tackle similar problems in the future.
Understanding the Problem
Before diving into the solution, let's first understand what we're dealing with. The equation y' - 4y = e^(5t) is a first-order linear ordinary differential equation. This means it involves the first derivative of the unknown function y (denoted as y') and the function y itself, and it's linear because y and y' appear only to the first power. The term e^(5t) on the right-hand side makes it a non-homogeneous equation.
To find the general solution, we need to find a function y(t) that satisfies this equation for all values of t. The general solution will typically involve an arbitrary constant, reflecting the fact that there are infinitely many solutions to a differential equation.
Why are Differential Equations Important?
You might be wondering why we bother solving these equations in the first place. Differential equations are crucial in various fields, including physics, engineering, economics, and biology. They help us model phenomena like population growth, radioactive decay, the motion of objects, and the flow of heat. By solving differential equations, we can predict how these systems will behave over time. In our example, the equation y' - 4y = e^(5t) could potentially represent a simple model of a system where the rate of change (y') is influenced by the current state (y) and an external driving force (e^(5t)).
Step 1: Finding the Integrating Factor
The first step in solving a linear first-order differential equation is to find the integrating factor. This is a function that we multiply the entire equation by, which helps us to rewrite the left-hand side as the derivative of a product. This is a crucial step to simplify the equation and allows us to integrate more easily. For a general first-order linear differential equation of the form:
y' + P(t)y = Q(t)
The integrating factor, often denoted by μ(t) (mu of t), is given by:
μ(t) = e^(∫P(t) dt)
In our case, the differential equation is y' - 4y = e^(5t). Comparing this with the general form, we can see that P(t) = -4 and Q(t) = e^(5t). Therefore, the integrating factor is:
μ(t) = e^(∫-4 dt) = e^(-4t)
So, the integrating factor for our equation is e^(-4t). This simple exponential function will be the key to unlocking the solution.
Step 2: Multiplying by the Integrating Factor
Now that we have the integrating factor, e^(-4t), we multiply both sides of the original differential equation, y' - 4y = e^(5t), by it. This gives us:
e^(-4t) * (y' - 4y) = e^(-4t) * e^(5t)
Distributing the integrating factor on the left-hand side, we get:
e^(-4t)y' - 4e^(-4t)y = e^(t)
The crucial observation here is that the left-hand side is now the derivative of the product of y(t) and the integrating factor e^(-4t). This is precisely why we chose this particular integrating factor. We can rewrite the left-hand side as:
d/dt [y(t)e^(-4t)] = e^(t)
This step is the heart of the integrating factor method. It transforms the original equation into a form that can be easily integrated. By recognizing the left-hand side as the derivative of a product, we've significantly simplified the problem. The next step will involve integrating both sides of the equation to find the general solution.
Step 3: Integrating Both Sides
In the previous step, we transformed our differential equation into:
d/dt [y(t)e^(-4t)] = e^(t)
Now, we integrate both sides of this equation with respect to t. This will effectively undo the differentiation on the left-hand side.
∫ [d/dt [y(t)e^(-4t)]] dt = ∫ e^(t) dt
The integral of the derivative on the left-hand side simply gives us the function inside the derivative:
y(t)e^(-4t) = ∫ e^(t) dt
The integral of e^(t) with respect to t is simply e^(t), plus a constant of integration, which we'll call C:
y(t)e^(-4t) = e^(t) + C
This constant of integration, C, is essential because it represents the family of solutions to the differential equation. Different values of C will give different particular solutions, but they all share the same underlying form.
Step 4: Solving for y(t)
We've made great progress! We now have the equation:
y(t)e^(-4t) = e^(t) + C
Our goal is to isolate y(t) to find the general solution. To do this, we simply multiply both sides of the equation by e^(4t):
y(t)e^(-4t) * e^(4t) = (e^(t) + C) * e^(4t)
On the left-hand side, e^(-4t) and e^(4t) cancel out, leaving us with just y(t). On the right-hand side, we distribute e^(4t):
y(t) = e^(t) * e^(4t) + C * e^(4t)
Using the rule of exponents (e^a * e^b = e^(a+b)), we can simplify the first term:
y(t) = e^(5t) + Ce^(4t)
And there we have it! This is the general solution to the differential equation y' - 4y = e^(5t).
The General Solution: y(t) = e^(5t) + Ce^(4t)
The general solution to the differential equation y' - 4y = e^(5t) is:
y(t) = e^(5t) + Ce^(4t)
Where C is an arbitrary constant. This solution represents a family of functions that satisfy the original differential equation. Each value of C corresponds to a specific solution curve.
Understanding the Components of the Solution
The general solution has two main components:
- e^(5t): This is a particular solution to the non-homogeneous equation. It's a specific function that satisfies the equation when we plug it in.
- Ce^(4t): This is the general solution to the corresponding homogeneous equation y' - 4y = 0. This part represents the natural behavior of the system without the external driving force (e^(5t)).
The combination of these two parts gives us the complete picture of all possible solutions to the original differential equation.
Conclusion
Solving differential equations might seem challenging at first, but by breaking down the problem into manageable steps, like finding the integrating factor and integrating both sides, we can arrive at the general solution. In this article, we successfully found the general solution to y' - 4y = e^(5t), which is y(t) = e^(5t) + Ce^(4t). Remember, practice makes perfect, so keep solving those differential equations!
For further exploration and more in-depth explanations of differential equations, you can visit reputable resources like Khan Academy's Differential Equations course. This can provide a more structured learning path and cover a wider range of topics in differential equations.
