Solving Differential Equations: A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of differential equations. Specifically, we're going to tackle the equation (x^2D2 + 4xD + 1)y = 1/x^3, where D represents dy/dx. This might look a bit intimidating at first, but don't worry, we'll break it down step by step. This article aims to provide a comprehensive guide on how to solve this particular type of differential equation, ensuring you grasp every concept along the way. So, let's jump right in and get our hands dirty with some math!

Understanding the Problem

Before we start crunching numbers, let's make sure we understand what we're dealing with. This is a second-order Cauchy-Euler equation, also known as an equidimensional equation. These types of equations have a specific form, and recognizing this form is the first step in solving them. In this section, we'll dissect the equation, identify its key components, and discuss why the Cauchy-Euler approach is the perfect method for this problem. Understanding the underlying structure will make the solution process much smoother and more intuitive. So, let’s put on our detective hats and analyze this equation together!

The given differential equation is (x^2D2 + 4xD + 1)y = 1/x^3. Here, D represents the differential operator dy/dx. This type of equation falls under the category of second-order linear differential equations with variable coefficients. Specifically, it's a Cauchy-Euler equation due to the form of the coefficients of the derivatives. Notice how the power of x matches the order of the derivative (x^2 with D^2, x with D, and a constant term). This is the hallmark of a Cauchy-Euler equation.

Why is this important? Well, Cauchy-Euler equations have a standard method of solution, which involves making a clever substitution to transform the equation into one with constant coefficients. Equations with constant coefficients are much easier to solve, using techniques we'll explore later. Ignoring this structure and trying other methods would be like trying to fit a square peg in a round hole – it's possible, but definitely not efficient! By recognizing the equation type, we've already significantly simplified our approach.

The right-hand side of the equation, 1/x^3, is also crucial. It signifies that we're dealing with a non-homogeneous equation. This means the solution will have two parts: the homogeneous solution (the solution to the equation when the right-hand side is zero) and the particular solution (a specific solution that satisfies the non-homogeneous equation). We'll tackle both of these components in detail as we move forward. For now, just remember that the presence of 1/x^3 adds an extra layer to our solution process.

In summary, we've identified our equation as a non-homogeneous Cauchy-Euler equation. This knowledge provides a roadmap for the rest of the solution. We know we'll need a substitution to transform it, and we know we'll need to find both a homogeneous and a particular solution. With this solid foundation, we're ready to move on to the next step: making the crucial substitution that will unlock the solution.

The Substitution: x = e^z

The magic trick for solving Cauchy-Euler equations is a simple yet powerful substitution: x = e^z. This substitution transforms the variable coefficient differential equation into a constant coefficient one, which is much easier to solve. This step is the cornerstone of our solution strategy, and understanding its purpose and execution is vital. In this section, we'll not only explain the substitution but also demonstrate how it affects the derivatives and transforms the original equation. Get ready to witness some mathematical wizardry!

So, why does this substitution work? The core idea is to replace the variable x with an exponential function of a new variable z. This cleverly manipulates the derivatives, eliminating the variable coefficients in front of them. Let's delve into the math to see how this happens. If x = e^z, then z = ln(x). Now, we need to express our derivatives in terms of z instead of x. We'll use the chain rule for this.

Let's start with the first derivative, dy/dx. Using the chain rule, we have:

dy/dx = (dy/dz) * (dz/dx)

Since z = ln(x), dz/dx = 1/x. Therefore,

dy/dx = (1/x) * (dy/dz)

Now, let's introduce a new notation to simplify things. Let D = d/dx and θ = d/dz. Then, the above equation can be written as:

Dy = (1/x)θy

This gives us a relationship between the derivative with respect to x (D) and the derivative with respect to z (θ). But we need to deal with the second derivative as well. Let's find D^2y:

D^2y = D(Dy) = D((1/x)θy)

Applying the product rule, we get:

D^2y = D(1/x) * θy + (1/x) * D(θy)

We know D(1/x) = -1/x^2. Also, D(θy) = d/dx (dy/dz) = (d/dz (dy/dz)) * (dz/dx) = (d^2y/dz2) * (1/x) = (1/x)θ^2y. Substituting these back into the equation, we have:

D^2y = (-1/x^2)θy + (1/x) * (1/x)θ^2y

D^2y = (1/x^2)(θ2y - θy)

Now we have expressions for Dy and D^2y in terms of θ and x. This is the key to transforming our original equation.

Substituting x = e^z, Dy = (1/x)θy, and D^2y = (1/x^2)(θ2y - θy) into the original equation (x^2D2 + 4xD + 1)y = 1/x^3, we get:

x^2 * (1/x^2)(θ2y - θy) + 4x * (1/x)θy + y = 1/x^3

Simplifying, we obtain:

(θ^2y - θy) + 4θy + y = 1/x^3

θ^2y + 3θy + y = 1/x^3

But remember, x = e^z, so 1/x^3 = e^(-3z). Our transformed equation now looks like this:

θ^2y + 3θy + y = e^(-3z)

Notice the magic! The coefficients are now constants. We've successfully transformed our Cauchy-Euler equation into a linear, second-order differential equation with constant coefficients. This is a huge win, as we have standard methods to solve such equations. The substitution x = e^z was the key, and now we're well-positioned to find the solution. Next, we'll tackle the homogeneous part of this equation.

Solving the Homogeneous Equation

Now that we've transformed our original equation, it's time to solve the homogeneous part. This means we'll be looking at the equation θ^2y + 3θy + y = 0. Solving the homogeneous equation is a crucial step because it gives us the general structure of the solution. The particular solution we find later will simply add a specific piece to this general structure. In this section, we'll walk through the process of finding the homogeneous solution, which involves forming the auxiliary equation, finding its roots, and constructing the solution based on those roots. Let's get started!

To solve this homogeneous differential equation with constant coefficients, we assume a solution of the form y = e^(mz), where m is a constant. This is a standard technique for these types of equations. We'll substitute this assumed solution into the equation and see what conditions m must satisfy.

First, we need to find the derivatives θy and θ^2y. If y = e^(mz), then:

θy = dy/dz = me^(mz)

θ^2y = d^2y/dz2 = m^2e(mz)

Now, substitute these into our homogeneous equation θ^2y + 3θy + y = 0:

m^2e(mz) + 3me^(mz) + e^(mz) = 0

We can factor out e^(mz) from each term:

e^(mz)(m2 + 3m + 1) = 0

Since e^(mz) is never zero, the equation is satisfied only if the quadratic term is zero:

m^2 + 3m + 1 = 0

This quadratic equation is called the auxiliary equation (or characteristic equation). The roots of this equation will determine the form of our homogeneous solution. We can solve this using the quadratic formula:

m = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 1, b = 3, and c = 1. Plugging these values into the quadratic formula, we get:

m = (-3 ± √(3^2 - 4 * 1 * 1)) / 2 * 1

m = (-3 ± √5) / 2

So, we have two distinct real roots:

m_1 = (-3 + √5) / 2

m_2 = (-3 - √5) / 2

Since we have two distinct real roots, the general solution to the homogeneous equation is a linear combination of the corresponding exponential functions:

y_h = c_1e^(m_1z) + c_2e^(m_2z)

Where c_1 and c_2 are arbitrary constants. Now, we substitute back the values of m_1 and m_2:

y_h = c_1e^(((-3 + √5) / 2)z) + c_2e^(((-3 - √5) / 2)z)

Remember that z = ln(x). So, we can rewrite this in terms of x using the property e^(a ln(x)) = x^a:

y_h = c_1x^((-3 + √5) / 2) + c_2x^((-3 - √5) / 2)

This is the homogeneous solution to our differential equation. It represents a family of solutions that satisfy the equation when the right-hand side is zero. Now, we need to find a particular solution that satisfies the original non-homogeneous equation. This particular solution will add a specific function to our homogeneous solution, giving us the complete general solution. We're one step closer to the final answer!

Finding the Particular Solution

With the homogeneous solution in hand, the next step is to find a particular solution to the non-homogeneous equation θ^2y + 3θy + y = e^(-3z). The particular solution is a specific solution that satisfies the entire equation, not just the homogeneous part. We'll use the method of undetermined coefficients, a powerful technique for finding particular solutions when the non-homogeneous term has a specific form (like exponentials, polynomials, sines, and cosines). This section will guide you through the process of making an educated guess for the form of the particular solution, substituting it into the equation, and solving for the unknown coefficients. Let's dive in and uncover this crucial piece of the puzzle!

The method of undetermined coefficients involves making an educated guess about the form of the particular solution based on the form of the non-homogeneous term. In our case, the non-homogeneous term is e^(-3z), which is an exponential function. Therefore, we'll assume a particular solution of the form:

y_p = Ae^(-3z)

Where A is a constant we need to determine. Now, we need to find the derivatives θy_p and θ^2y_p:

θy_p = dy_p/dz = -3Ae^(-3z)

θ^2y_p = d^2y_p/dz2 = 9Ae^(-3z)

Substitute these into the non-homogeneous equation θ^2y + 3θy + y = e^(-3z):

9Ae^(-3z) + 3(-3Ae^(-3z)) + Ae^(-3z) = e^(-3z)

Simplifying, we get:

9Ae^(-3z) - 9Ae^(-3z) + Ae^(-3z) = e^(-3z)

Ae^(-3z) = e^(-3z)

Dividing both sides by e^(-3z), we have:

A = 1

So, our particular solution is:

y_p = e^(-3z)

Now, we need to express this in terms of x. Since z = ln(x), we have:

y_p = e^(-3ln(x)) = e^(ln(x(-3))) = x^(-3) = 1/x^3

This is our particular solution. It's a specific function that, when plugged into the original non-homogeneous equation, makes the equation true. We're now in the home stretch! We have both the homogeneous solution and the particular solution. The final step is to combine them to get the general solution.

The General Solution

We've reached the final step! We've found both the homogeneous solution (y_h) and the particular solution (y_p). Now, we simply combine them to obtain the general solution to the original differential equation. This general solution represents the complete family of solutions that satisfy the equation. In this section, we'll show you how to combine the two solutions and present the final answer. Let's wrap this up and celebrate our mathematical victory!

The general solution to a non-homogeneous linear differential equation is the sum of the homogeneous solution and the particular solution:

y = y_h + y_p

We found the homogeneous solution to be:

y_h = c_1x^((-3 + √5) / 2) + c_2x^((-3 - √5) / 2)

And the particular solution to be:

y_p = 1/x^3

Therefore, the general solution is:

y = c_1x^((-3 + √5) / 2) + c_2x^((-3 - √5) / 2) + 1/x^3

This is the final answer! It represents all possible solutions to the differential equation (x^2D2 + 4xD + 1)y = 1/x^3. The constants c_1 and c_2 can be determined if we have initial conditions (values of y and dy/dx at a specific point). Otherwise, this is the most general form of the solution.

Conclusion

Wow, guys! We did it! We successfully solved the differential equation (x^2D2 + 4xD + 1)y = 1/x^3. We started by recognizing it as a Cauchy-Euler equation, used the substitution x = e^z to transform it into an equation with constant coefficients, solved the homogeneous equation, found a particular solution using the method of undetermined coefficients, and finally combined these to obtain the general solution. This journey demonstrates the power of understanding different types of differential equations and applying the appropriate techniques.

Remember, solving differential equations is like solving a puzzle. Each step builds upon the previous one, and understanding the underlying concepts is key. Don't be afraid to break down complex problems into smaller, manageable steps. And most importantly, practice makes perfect! The more you work through these types of problems, the more comfortable and confident you'll become.

So, keep exploring the world of differential equations, and happy solving!