Solving The Linear Equation (D⁵ - D)y = 2⁻ˣ: A Math Discussion
Hey math enthusiasts! Let's dive into solving this interesting linear differential equation: (D⁵ - D)y = 2⁻ˣ. This equation falls under the category of linear, non-homogeneous differential equations, and we're going to break down the steps to find its solution. We'll explore the characteristic equation, find the complementary function, determine a particular integral, and finally, combine these to get the general solution. So, buckle up and let's get started!
Understanding the Problem
Before we jump into the solution, let's make sure we understand what we're dealing with. The equation (D⁵ - D)y = 2⁻ˣ is a fifth-order linear differential equation. The 'D' here represents the differential operator d/dx, meaning D⁵y is the fifth derivative of y with respect to x, and Dy is the first derivative. The equation is linear because the dependent variable y and its derivatives appear only to the first power and are not multiplied together. It's non-homogeneous because of the 2⁻ˣ term on the right-hand side, which is a function of x. Solving such equations involves finding a general solution that satisfies the equation for all values of x. This general solution will consist of two parts: the complementary function (yc) and the particular integral (yp). The complementary function is the solution to the homogeneous equation (D⁵ - D)y = 0, while the particular integral is a specific solution to the non-homogeneous equation. We'll find each of these and then add them together to get the complete solution. So, let's dive into the first step: finding the complementary function.
Finding the Complementary Function (yc)
The first step in solving this differential equation is to find the complementary function, often denoted as yc. To do this, we need to solve the homogeneous equation, which is obtained by setting the right-hand side of our original equation to zero. So, we consider:
(D⁵ - D)y = 0
To solve this, we convert the differential equation into its characteristic equation. We replace the differential operator D with a variable, usually 'm' or 'r'. In this case, our characteristic equation becomes:
m⁵ - m = 0
Now, we need to find the roots of this equation. We can factor out an 'm' from the equation:
m(m⁴ - 1) = 0
This gives us one root immediately: m = 0. Next, we can further factor the term (m⁴ - 1) using the difference of squares:
m(m² - 1)(m² + 1) = 0
And again:
m(m - 1)(m + 1)(m² + 1) = 0
Now we have the roots m = 0, m = 1, and m = -1. The term (m² + 1) = 0 gives us complex roots. Solving for m, we get:
m² = -1
m = ±i
So, our roots are m = 0, 1, -1, i, and -i. These roots determine the form of the complementary function. For each real root 'r', we have a term of the form Ce^(rx), where C is a constant. For each pair of complex conjugate roots a ± bi, we have terms of the form e^(ax)(C₁cos(bx) + C₂sin(bx)).
Therefore, the complementary function yc is:
yc = C₁e^(0x) + C₂e^(x) + C₃e^(-x) + C₄cos(x) + C₅sin(x)
Simplifying, since e^(0x) = 1, we have:
yc = C₁ + C₂e^(x) + C₃e^(-x) + C₄cos(x) + C₅sin(x)
Where C₁, C₂, C₃, C₄, and C₅ are arbitrary constants. This is the complementary function, which represents the general solution to the homogeneous equation. Now, we move on to finding the particular integral, which will give us a specific solution to the non-homogeneous equation. This part involves a bit more cleverness and often involves making an educated guess about the form of the solution.
Finding the Particular Integral (yp)
Now that we've found the complementary function yc, the next crucial step is to determine the particular integral, denoted as yp. The particular integral is a specific solution to the non-homogeneous equation: (D⁵ - D)y = 2⁻ˣ. To find yp, we'll use the method of undetermined coefficients.
This method involves making an educated guess about the form of the particular solution based on the form of the non-homogeneous term, which in our case is 2⁻ˣ. Since 2⁻ˣ can be written as (1/2)ˣ, it's an exponential function. Therefore, we'll assume that the particular integral has the form:
yp = Ae⁻ˣ
Where A is a constant we need to determine. Now, we need to find the derivatives of yp up to the fifth order since our differential equation is fifth-order.
- yp' = -Ae⁻ˣ
- yp'' = Ae⁻ˣ
- yp''' = -Ae⁻ˣ
- yp'''' = Ae⁻ˣ
- yp''''' = -Ae⁻ˣ
Next, we substitute yp and its derivatives into the original non-homogeneous equation:
(D⁵ - D)yp = 2⁻ˣ
(-Ae⁻ˣ) - (-Ae⁻ˣ) = 2⁻ˣ
-Ae⁻ˣ + Ae⁻ˣ = (1/2)ˣ
Oops! We've run into a problem. The left side of the equation simplifies to 0, while the right side is (1/2)ˣ, which is not zero. This means our initial guess for yp was incorrect. This often happens when the guessed form of yp is a solution to the homogeneous equation (i.e., part of the complementary function). Notice that e⁻ˣ is already part of our yc. To fix this, we need to modify our guess by multiplying it by x:
yp = Axe⁻ˣ
Now, let's find the derivatives again (this will be a bit more involved):
- yp' = Ae⁻ˣ - Axe⁻ˣ
- yp'' = -Ae⁻ˣ - (Ae⁻ˣ - Axe⁻ˣ) = -2Ae⁻ˣ + Axe⁻ˣ
- yp''' = 2Ae⁻ˣ + (Ae⁻ˣ - Axe⁻ˣ) = 3Ae⁻ˣ - Axe⁻ˣ
- yp'''' = -3Ae⁻ˣ - (Ae⁻ˣ - Axe⁻ˣ) = -4Ae⁻ˣ + Axe⁻ˣ
- yp''''' = 4Ae⁻ˣ + (-Ae⁻ˣ + Axe⁻ˣ) = 5Ae⁻ˣ - Axe⁻ˣ
Now, substitute these into the original equation:
(D⁵ - D)yp = 2⁻ˣ
(5Ae⁻ˣ - Axe⁻ˣ) - (Ae⁻ˣ - Axe⁻ˣ) = (1/2)ˣ
4Ae⁻ˣ = (1/2)ˣ
To solve for A, we can equate the coefficients of e⁻ˣ:
4A = 1
A = 1/4
Therefore, the particular integral is:
yp = (1/4)xe⁻ˣ
We've now successfully found the particular integral. This, combined with the complementary function we found earlier, will give us the general solution to the original differential equation. Let's put it all together!
General Solution
We've reached the final step! We've found both the complementary function (yc) and the particular integral (yp). Now, to get the general solution (y) of the non-homogeneous differential equation (D⁵ - D)y = 2⁻ˣ, we simply add them together:
y = yc + yp
We found the complementary function to be:
yc = C₁ + C₂e^(x) + C₃e^(-x) + C₄cos(x) + C₅sin(x)
And the particular integral to be:
yp = (1/4)xe^(-x)
Therefore, the general solution is:
y = C₁ + C₂e^(x) + C₃e^(-x) + C₄cos(x) + C₅sin(x) + (1/4)xe^(-x)
Where C₁, C₂, C₃, C₄, and C₅ are arbitrary constants. This is the complete solution to the given linear differential equation. It represents a family of solutions, each corresponding to a different set of values for the constants. The complementary function accounts for the general behavior of the system, while the particular integral accounts for the specific forcing function (the 2⁻ˣ term).
Conclusion
Guys, we've successfully navigated through solving the linear differential equation (D⁵ - D)y = 2⁻ˣ! We tackled this problem step-by-step, starting with understanding the equation, finding the complementary function by solving the homogeneous equation, then cleverly determining the particular integral using the method of undetermined coefficients, and finally, combining them to obtain the general solution. Remember, the key to solving differential equations is practice and a good understanding of the underlying concepts. This example showcased a fifth-order linear non-homogeneous equation, but the principles we applied here can be extended to solve many other types of differential equations. Keep practicing, and you'll become a pro at solving these in no time! If you have any questions or want to discuss other math problems, feel free to drop them in the comments below. Let's keep the learning going!