Solving (D^3 - 7D^2 + 14D - 8)y = E^x Cos 2x A Comprehensive Guide
This article delves into the process of solving the given third-order linear homogeneous differential equation with constant coefficients and a non-homogeneous term. We will explore the concepts of auxiliary equations, complementary functions, particular integrals, and the method of undetermined coefficients to arrive at the general solution. Understanding these concepts is crucial for solving various problems in physics, engineering, and other scientific disciplines where differential equations play a vital role in modeling real-world phenomena.
Understanding the Differential Equation
The given differential equation is:
(D³ - 7D² + 14D - 8)y = eˣ cos 2x
Where D represents the differential operator d/dx. This is a third-order linear non-homogeneous differential equation with constant coefficients. To solve this type of equation, we generally follow these steps:
- Find the auxiliary equation by replacing D with 'm' and setting the left-hand side to zero.
- Solve the auxiliary equation to find the roots, which determine the form of the complementary function (y_c).
- Determine the particular integral (y_p) based on the form of the non-homogeneous term (eˣ cos 2x in this case).
- The general solution is then the sum of the complementary function and the particular integral (y = y_c + y_p).
Let's break down each step in detail.
Step 1: Finding the Auxiliary Equation
To begin, we need to form the auxiliary equation. This is achieved by replacing the differential operator D with the variable 'm' in the homogeneous part of the differential equation and setting the expression equal to zero. This gives us:
m³ - 7m² + 14m - 8 = 0
This cubic equation is the auxiliary equation, and its roots will dictate the form of the complementary function, which is a part of the general solution. Solving this cubic equation is crucial, and we can use various methods such as factoring, synthetic division, or numerical methods to find the roots. Factoring is often the most straightforward approach if we can identify integer roots. In this case, we can try integer values that are factors of the constant term (-8). By trying m = 1, we find that it is a root of the equation:
(1)³ - 7(1)² + 14(1) - 8 = 1 - 7 + 14 - 8 = 0
Therefore, (m - 1) is a factor of the cubic polynomial. Now, we can perform polynomial division or synthetic division to find the remaining quadratic factor. After dividing (m³ - 7m² + 14m - 8) by (m - 1), we obtain the quadratic factor:
m² - 6m + 8
Thus, the auxiliary equation can be rewritten as:
(m - 1)(m² - 6m + 8) = 0
Next, we need to solve the quadratic equation m² - 6m + 8 = 0. This can be done by factoring, completing the square, or using the quadratic formula. Factoring is usually the quickest method if the quadratic expression factors easily. In this case, we can factor it as:
(m - 2)(m - 4) = 0
So, the roots of the quadratic equation are m = 2 and m = 4. Combining these with the root m = 1 that we found earlier, we have all three roots of the auxiliary equation:
m₁ = 1, m₂ = 2, m₃ = 4
These distinct real roots are essential for constructing the complementary function, which represents the general solution of the homogeneous part of the differential equation. The next step is to use these roots to write the complementary function.
Step 2: Determining the Complementary Function (y_c)
Now that we have the roots of the auxiliary equation (m₁ = 1, m₂ = 2, m₃ = 4), we can construct the complementary function (y_c). Since we have three distinct real roots, the complementary function will take the form:
y_c = C₁e^(m₁x) + C₂e^(m₂x) + C₃e^(m₃x)
Where C₁, C₂, and C₃ are arbitrary constants. Substituting the values of the roots, we get:
y_c = C₁e^(1x) + C₂e^(2x) + C₃e^(4x)
Which simplifies to:
y_c = C₁eˣ + C₂e^(2x) + C₃e^(4x)
This is the complementary function, representing the general solution of the homogeneous part of the differential equation. It contains three arbitrary constants because the original differential equation is of third order. These constants will be determined by initial conditions if they are given. However, to find the general solution of the non-homogeneous differential equation, we also need to find the particular integral (y_p). The particular integral is a solution that satisfies the entire non-homogeneous differential equation, including the term eˣ cos 2x. The form of the particular integral depends on the form of the non-homogeneous term, and we will use the method of undetermined coefficients to find it. In this case, the non-homogeneous term is of the form e^(ax)cos(bx), which suggests a particular integral of the form e^(ax)(Acos(bx) + Bsin(bx)), where A and B are constants to be determined. Therefore, the next step is to find the particular integral.
Step 3: Finding the Particular Integral (y_p)
To find the particular integral (y_p) for the differential equation (D³ - 7D² + 14D - 8)y = eˣ cos 2x, we will use the method of undetermined coefficients. The non-homogeneous term is eˣ cos 2x, which is of the form e^(ax)cos(bx), where a = 1 and b = 2. This suggests a particular integral of the form:
y_p = eˣ(A cos 2x + B sin 2x)
Where A and B are constants that we need to determine. To find these constants, we need to differentiate y_p three times and substitute y_p and its derivatives into the original differential equation. First, let's find the first derivative (y_p'):
y_p' = d/dx [eˣ(A cos 2x + B sin 2x)]
Using the product rule, we get:
y_p' = eˣ(A cos 2x + B sin 2x) + eˣ(-2A sin 2x + 2B cos 2x)
y_p' = eˣ[(A + 2B) cos 2x + (B - 2A) sin 2x]
Now, let's find the second derivative (y_p''):
y_p'' = d/dx [eˣ((A + 2B) cos 2x + (B - 2A) sin 2x)]
Again, using the product rule, we get:
y_p'' = eˣ[(A + 2B) cos 2x + (B - 2A) sin 2x] + eˣ[-2(A + 2B) sin 2x + 2(B - 2A) cos 2x]
y_p'' = eˣ[(A + 2B + 2B - 4A) cos 2x + (B - 2A - 2A - 4B) sin 2x]
y_p'' = eˣ[(-3A + 4B) cos 2x + (-4A - 3B) sin 2x]
Next, we find the third derivative (y_p'''):
y_p''' = d/dx [eˣ((-3A + 4B) cos 2x + (-4A - 3B) sin 2x)]
Using the product rule again, we get:
y_p''' = eˣ[(-3A + 4B) cos 2x + (-4A - 3B) sin 2x] + eˣ[-2(-3A + 4B) sin 2x - 2(-4A - 3B) cos 2x]
y_p''' = eˣ[(-3A + 4B - 8A - 6B) cos 2x + (-4A - 3B + 6A - 8B) sin 2x]
y_p''' = eˣ[(-11A - 2B) cos 2x + (2A - 11B) sin 2x]
Now, we substitute y_p, y_p', y_p'', and y_p''' into the original differential equation:
(D³ - 7D² + 14D - 8)y_p = eˣ cos 2x
Substituting the derivatives and y_p, we have:
eˣ[(-11A - 2B) cos 2x + (2A - 11B) sin 2x] - 7eˣ[(-3A + 4B) cos 2x + (-4A - 3B) sin 2x] + 14eˣ[(A + 2B) cos 2x + (B - 2A) sin 2x] - 8eˣ(A cos 2x + B sin 2x) = eˣ cos 2x
Now, we can divide both sides by eˣ and group the terms with cos 2x and sin 2x:
[(-11A - 2B) - 7(-3A + 4B) + 14(A + 2B) - 8A] cos 2x + [(2A - 11B) - 7(-4A - 3B) + 14(B - 2A) - 8B] sin 2x = cos 2x
Simplifying the coefficients, we get:
[(-11A - 2B + 21A - 28B + 14A + 28B - 8A) cos 2x + (2A - 11B + 28A + 21B + 14B - 28A - 8B) sin 2x] = cos 2x
[16A - 2B] cos 2x + [2A + 16B] sin 2x = cos 2x
Now, we can equate the coefficients of cos 2x and sin 2x on both sides of the equation:
16A - 2B = 1
2A + 16B = 0
We now have a system of two linear equations with two unknowns, A and B. We can solve this system to find the values of A and B. From the second equation, we can express A in terms of B:
2A = -16B
A = -8B
Substitute this expression for A into the first equation:
16(-8B) - 2B = 1
-128B - 2B = 1
-130B = 1
B = -1/130
Now, substitute the value of B back into the expression for A:
A = -8(-1/130)
A = 8/130
A = 4/65
So, we have found A = 4/65 and B = -1/130. Now we can write the particular integral y_p:
y_p = eˣ(A cos 2x + B sin 2x)
y_p = eˣ((4/65) cos 2x + (-1/130) sin 2x)
This is the particular integral of the given differential equation. Finally, we can find the general solution by adding the complementary function and the particular integral.
Step 4: Forming the General Solution
Now that we have both the complementary function (y_c) and the particular integral (y_p), we can form the general solution (y) by adding them together:
y = y_c + y_p
We found that:
y_c = C₁eˣ + C₂e^(2x) + C₃e^(4x)
And:
y_p = eˣ((4/65) cos 2x + (-1/130) sin 2x)
Therefore, the general solution is:
y = C₁eˣ + C₂e^(2x) + C₃e^(4x) + eˣ((4/65) cos 2x - (1/130) sin 2x)
This is the complete general solution of the given differential equation (D³ - 7D² + 14D - 8)y = eˣ cos 2x. It includes three arbitrary constants (C₁, C₂, and C₃) that can be determined if initial conditions are provided. The solution consists of two parts: the complementary function, which represents the general solution of the homogeneous part of the equation, and the particular integral, which is a specific solution that accounts for the non-homogeneous term (eˣ cos 2x). This solution is essential for understanding the behavior of systems modeled by this differential equation, as it provides a complete description of all possible solutions.
Conclusion
In conclusion, we have successfully solved the third-order linear non-homogeneous differential equation (D³ - 7D² + 14D - 8)y = eˣ cos 2x. We accomplished this by first finding the auxiliary equation and its roots, which allowed us to construct the complementary function. Then, we used the method of undetermined coefficients to find the particular integral. Finally, we combined the complementary function and the particular integral to obtain the general solution:
y = C₁eˣ + C₂e^(2x) + C₃e^(4x) + eˣ((4/65) cos 2x - (1/130) sin 2x)
This general solution represents all possible solutions to the differential equation and is a fundamental result in the study of differential equations and their applications. Understanding the steps involved in solving such equations is crucial for various fields, including engineering, physics, and applied mathematics, where differential equations are used to model a wide range of phenomena. The techniques discussed in this article provide a solid foundation for tackling more complex differential equations and their applications in real-world problems.
