Solving Initial Value Problem: Dy/dx = 1/x^2 - X, Y(6) = 1

by Alex Johnson 59 views

Let's dive into solving an interesting initial value problem! This problem involves finding a function y(x) that satisfies a given differential equation and an initial condition. Specifically, we're tasked with solving the differential equation dy/dx = 1/x^2 - x where x > 0, and the initial condition y(6) = 1. This means we need to find the particular solution to the differential equation that passes through the point (6, 1).

Understanding the Problem

Before we jump into the solution, let's break down what we're dealing with. The equation dy/dx = 1/x^2 - x is a first-order differential equation. It tells us how the rate of change of the function y with respect to x is related to x itself. The initial condition, y(6) = 1, gives us a specific point on the solution curve, allowing us to find a unique solution rather than a family of solutions.

In essence, we're looking for a function y(x) whose derivative is 1/x^2 - x, and which has a value of 1 when x is 6. This type of problem is fundamental in many areas of mathematics, physics, and engineering, as it allows us to model and understand dynamic systems where rates of change are important.

Why is this important? Initial value problems like this appear in various real-world scenarios. For example, they can model the motion of an object under a certain force, the decay of a radioactive substance, or the growth of a population. By solving these problems, we can make predictions and gain insights into the behavior of these systems.

Step-by-Step Solution

Now, let's get our hands dirty and solve this problem step-by-step.

1. Separate the Variables and Integrate: The first step is to recognize that this differential equation is separable. This means we can rearrange the equation so that all the y terms are on one side and all the x terms are on the other. In this case, it's already set up nicely for us:

dy = (1/x^2 - x) dx

Next, we integrate both sides of the equation. Remember, integration is the reverse process of differentiation, so it will help us find y(x) from its derivative.

∫ dy = ∫ (1/x^2 - x) dx

The integral of dy is simply y. On the right side, we need to find the antiderivative of 1/x^2 - x. We can rewrite 1/x^2 as x^(-2) to make it easier to integrate.

y = ∫ (x^(-2) - x) dx

Now we apply the power rule for integration, which states that ∫ x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.

y = -x^(-1) - (x^2)/2 + C

We can rewrite -x^(-1) as -1/x, giving us:

y = -1/x - (x^2)/2 + C

2. Apply the Initial Condition: We've found the general solution to the differential equation, but we still need to find the particular solution that satisfies the initial condition y(6) = 1. This means when x = 6, y = 1. We can plug these values into our general solution to solve for C.

1 = -1/6 - (6^2)/2 + C

1 = -1/6 - 18 + C

Now, we solve for C:

C = 1 + 1/6 + 18

C = 19 + 1/6

C = 115/6

3. Write the Particular Solution: Now that we've found the value of C, we can plug it back into our general solution to get the particular solution that satisfies the initial condition:

y = -1/x - (x^2)/2 + 115/6

This is the solution to the initial value problem. It's a function y(x) that satisfies the differential equation dy/dx = 1/x^2 - x and the initial condition y(6) = 1.

Verification and Interpretation

It's always a good idea to verify our solution. We can do this by differentiating our solution and checking if it matches the original differential equation. Also, we can plug in x = 6 to see if we get y = 1.

Differentiating our solution:

dy/dx = d/dx (-1/x - (x^2)/2 + 115/6)

dy/dx = 1/x^2 - x

This matches the original differential equation, so our solution is likely correct.

Now let's check the initial condition:

y(6) = -1/6 - (6^2)/2 + 115/6

y(6) = -1/6 - 18 + 115/6

y(6) = (-1 - 108 + 115)/6

y(6) = 6/6

y(6) = 1

The initial condition is also satisfied, so we can be confident in our solution.

What does this solution tell us? The function y(x) = -1/x - (x^2)/2 + 115/6 describes the relationship between x and y under the given conditions. It tells us how y changes as x changes, starting from the point (6, 1). This information could be useful in various applications, depending on what x and y represent.

Common Mistakes and How to Avoid Them

Solving initial value problems involves several steps, and it's easy to make mistakes along the way. Here are some common pitfalls and how to avoid them:

  1. Forgetting the Constant of Integration: This is a classic mistake. When you integrate, always remember to add the constant of integration, C. This constant is crucial for finding the particular solution.

    • How to avoid it: Make it a habit to always write + C after every indefinite integral.

  2. Incorrectly Applying the Power Rule: The power rule for integration is straightforward, but it's easy to make a mistake with the exponents or coefficients.

    • How to avoid it: Double-check your work and remember the formula: ∫ x^n dx = (x^(n+1))/(n+1) + C. Pay special attention to negative exponents.

  3. Algebra Errors: Solving for the constant of integration, C, often involves algebraic manipulation. Mistakes in this step can lead to an incorrect solution.

    • How to avoid it: Write out each step clearly and double-check your arithmetic. If possible, use a calculator or computer algebra system to verify your calculations.

  4. Not Verifying the Solution: It's always a good practice to verify your solution by differentiating it and checking if it matches the original differential equation and initial condition.

    • How to avoid it: Take the time to plug your solution back into the original problem. This can help you catch errors and build confidence in your answer.

By being aware of these common mistakes and taking steps to avoid them, you can improve your accuracy and efficiency in solving initial value problems.

Practice Problems

To solidify your understanding of solving initial value problems, here are a few practice problems you can try:

  1. Solve the initial value problem: dy/dx = 2x - 1, y(0) = 3
  2. Solve the initial value problem: dy/dx = cos(x), y(Ï€) = 0
  3. Solve the initial value problem: dy/dx = e^x, y(1) = e

Working through these problems will give you valuable practice and help you develop your problem-solving skills.

Conclusion

Solving initial value problems is a fundamental skill in calculus and differential equations. By understanding the steps involved and practicing regularly, you can master this important technique. Remember to separate the variables, integrate, apply the initial condition, and verify your solution. With practice, you'll become more confident and proficient in solving these types of problems. This journey will not only enhance your mathematical abilities but also equip you to tackle real-world scenarios where understanding dynamic systems is crucial.

For further exploration and a deeper understanding of differential equations, consider visiting Khan Academy's Differential Equations section. This is an excellent resource for learning more about this fascinating topic.