Solving (D^2 - 2D + 4)y = E^x Cos X + Sin X Cos 3x A Comprehensive Guide
This article delves into the solution of the second-order linear non-homogeneous differential equation: (D^2 - 2D + 4)y = e^x cos x + sin x cos 3x. We will systematically break down the problem, starting with finding the auxiliary equation, then determining the complementary function, and finally, obtaining the particular integral. This involves employing various techniques such as trigonometric identities and the method of undetermined coefficients. Understanding the process of solving such differential equations is crucial in many fields of science and engineering, as they often model physical phenomena. From electrical circuits to mechanical oscillations, differential equations provide a powerful tool for analyzing and predicting system behavior. Therefore, mastering these solution techniques is an invaluable asset for anyone working in these disciplines.
1. Auxiliary Equation and Complementary Function
The first step in solving this differential equation is to find the auxiliary equation (AE). The auxiliary equation is obtained by replacing the differential operator D with the variable m in the homogeneous part of the given equation. In this case, the homogeneous equation is (D^2 - 2D + 4)y = 0. Therefore, the auxiliary equation becomes:
m^2 - 2m + 4 = 0
To solve this quadratic equation, we can use the quadratic formula:
m = [-b ± √(b^2 - 4ac)] / 2a
Where a = 1, b = -2, and c = 4. Substituting these values, we get:
m = [2 ± √((-2)^2 - 4 * 1 * 4)] / 2 * 1 m = [2 ± √(4 - 16)] / 2 m = [2 ± √(-12)] / 2 m = [2 ± 2i√3] / 2 m = 1 ± i√3
Since the roots are complex conjugates of the form α ± iβ, where α = 1 and β = √3, the complementary function (CF), which is the solution to the homogeneous equation, is given by:
y_c = e^(αx) (C_1 cos(βx) + C_2 sin(βx))
Substituting the values of α and β, we obtain:
y_c = e^x (C_1 cos(√3x) + C_2 sin(√3x))
Here, C_1 and C_2 are arbitrary constants that will be determined by initial or boundary conditions if provided. The complementary function represents the natural response of the system, the part of the solution that arises from the system's inherent properties rather than any external forcing function. Understanding the complementary function is crucial, as it dictates the stability and oscillatory behavior of the system. A decaying exponential term indicates stability, while oscillatory terms imply the system will oscillate, potentially leading to instability if not properly damped. This section provides a solid foundation for further analysis and solving the complete differential equation.
2. Particular Integral: Addressing the Non-Homogeneous Part
Now, we need to find the particular integral (PI), which is a particular solution to the non-homogeneous equation:
(D^2 - 2D + 4)y = e^x cos x + sin x cos 3x
The right-hand side (RHS) of the equation consists of two terms: e^x cos x and sin x cos 3x. We will find the particular integral for each term separately and then add them together, using the principle of superposition. This method significantly simplifies the problem, as we can focus on each term independently and combine the results. The principle of superposition is a powerful tool in solving linear differential equations, allowing us to break down complex forcing functions into simpler components and find the corresponding responses. It's a direct consequence of the linearity of the differential operator and is a fundamental technique in solving such equations.
2.1 Particular Integral for e^x cos x
Let y_p1 be the particular integral for e^x cos x. We assume a solution of the form:
y_p1 = e^x (A cos x + B sin x)
Where A and B are constants to be determined. This form is chosen based on the structure of the forcing function, e^x cos x, and its derivatives. The derivatives of this form will also involve e^x cos x and e^x sin x, ensuring that the assumed solution can effectively cancel out the forcing function when substituted into the differential equation. This method of assuming a particular solution based on the form of the forcing function is known as the method of undetermined coefficients, a widely used technique for solving non-homogeneous linear differential equations.
Now we need to calculate the first and second derivatives of y_p1:
y_p1' = e^x (A cos x + B sin x) + e^x (-A sin x + B cos x) y_p1' = e^x [(A + B) cos x + (B - A) sin x]
y_p1'' = e^x [(A + B) cos x + (B - A) sin x] + e^x [-(A + B) sin x + (B - A) cos x] y_p1'' = e^x [2B cos x - 2A sin x]
Substitute y_p1, y_p1', and y_p1'' into the left-hand side (LHS) of the differential equation:
(D^2 - 2D + 4)y_p1 = y_p1'' - 2y_p1' + 4y_p1
Substitute the expressions for y_p1, y_p1', and y_p1'' that we derived:
e^x [2B cos x - 2A sin x] - 2e^x [(A + B) cos x + (B - A) sin x] + 4e^x [A cos x + B sin x] = e^x cos x
Divide throughout by e^x:
2B cos x - 2A sin x - 2(A + B) cos x - 2(B - A) sin x + 4(A cos x + B sin x) = cos x
Group the cos x and sin x terms together:
(2B - 2A - 2B + 4A) cos x + (-2A - 2B + 2A + 4B) sin x = cos x
Simplify the equation:
2A cos x + 2B sin x = cos x
Now, equate the coefficients of cos x and sin x on both sides:
2A = 1 => A = 1/2 2B = 0 => B = 0
Substitute the values of A and B back into the expression for y_p1:
y_p1 = e^x (1/2 cos x + 0 sin x) y_p1 = (1/2)e^x cos x
2.2 Particular Integral for sin x cos 3x
Now, we need to find the particular integral for the second term, sin x cos 3x. Before directly applying the method of undetermined coefficients, we can simplify this term using the trigonometric identity:
sin A cos B = (1/2) [sin(A + B) + sin(A - B)]
Applying this identity to sin x cos 3x, we get:
sin x cos 3x = (1/2) [sin(x + 3x) + sin(x - 3x)] sin x cos 3x = (1/2) [sin 4x + sin(-2x)] sin x cos 3x = (1/2) [sin 4x - sin 2x]
This simplification is crucial as it allows us to deal with simpler trigonometric functions, making the application of the method of undetermined coefficients more straightforward. Without this simplification, the complexity of the assumed solution would significantly increase, leading to a more cumbersome calculation. This highlights the importance of leveraging trigonometric identities and other mathematical tools to simplify complex expressions before proceeding with the solution process.
Let y_p2 be the particular integral for (1/2) [sin 4x - sin 2x]. We can further break this down into two parts and find the particular integral for each separately. Let y_p2a be the particular integral for (1/2)sin 4x and y_p2b be the particular integral for -(1/2)sin 2x. Then y_p2 = y_p2a + y_p2b.
2.2.1 Particular Integral for (1/2)sin 4x
Assume a solution of the form:
y_p2a = C cos 4x + D sin 4x
Where C and D are constants to be determined. This form is chosen because the derivatives of cos 4x and sin 4x will also involve cos 4x and sin 4x, allowing for a linear combination to match the forcing function (1/2)sin 4x.
Calculate the first and second derivatives of y_p2a:
y_p2a' = -4C sin 4x + 4D cos 4x y_p2a'' = -16C cos 4x - 16D sin 4x
Substitute y_p2a, y_p2a', and y_p2a'' into the LHS of the differential equation:
(D^2 - 2D + 4)y_p2a = y_p2a'' - 2y_p2a' + 4y_p2a
Substitute the expressions for y_p2a, y_p2a', and y_p2a'' that we derived:
(-16C cos 4x - 16D sin 4x) - 2(-4C sin 4x + 4D cos 4x) + 4(C cos 4x + D sin 4x) = (1/2)sin 4x
Group the cos 4x and sin 4x terms:
(-16C - 8D + 4C) cos 4x + (-16D + 8C + 4D) sin 4x = (1/2)sin 4x
Simplify the equation:
(-12C - 8D) cos 4x + (8C - 12D) sin 4x = (1/2)sin 4x
Equate the coefficients of cos 4x and sin 4x:
-12C - 8D = 0 8C - 12D = 1/2
Solve this system of equations. From the first equation, we can write:
C = (-8/12)D = (-2/3)D
Substitute this into the second equation:
8(-2/3)D - 12D = 1/2 (-16/3)D - 12D = 1/2 (-16 - 36)D / 3 = 1/2 (-52/3)D = 1/2 D = -3/104
Now, find C:
C = (-2/3)D = (-2/3)(-3/104) = 1/52
Substitute the values of C and D back into the expression for y_p2a:
y_p2a = (1/52) cos 4x - (3/104) sin 4x
2.2.2 Particular Integral for -(1/2)sin 2x
Assume a solution of the form:
y_p2b = E cos 2x + F sin 2x
Where E and F are constants to be determined.
Calculate the first and second derivatives of y_p2b:
y_p2b' = -2E sin 2x + 2F cos 2x y_p2b'' = -4E cos 2x - 4F sin 2x
Substitute y_p2b, y_p2b', and y_p2b'' into the LHS of the differential equation:
(D^2 - 2D + 4)y_p2b = y_p2b'' - 2y_p2b' + 4y_p2b
Substitute the expressions for y_p2b, y_p2b', and y_p2b'' that we derived:
(-4E cos 2x - 4F sin 2x) - 2(-2E sin 2x + 2F cos 2x) + 4(E cos 2x + F sin 2x) = -(1/2)sin 2x
Group the cos 2x and sin 2x terms:
(-4E - 4F + 4E) cos 2x + (-4F + 4E + 4F) sin 2x = -(1/2)sin 2x
Simplify the equation:
-4F cos 2x + 4E sin 2x = -(1/2)sin 2x
Equate the coefficients of cos 2x and sin 2x:
-4F = 0 => F = 0 4E = -1/2 => E = -1/8
Substitute the values of E and F back into the expression for y_p2b:
y_p2b = (-1/8) cos 2x + 0 sin 2x y_p2b = (-1/8) cos 2x
Now, combine y_p2a and y_p2b to get y_p2:
y_p2 = y_p2a + y_p2b y_p2 = (1/52) cos 4x - (3/104) sin 4x - (1/8) cos 2x
2.3 Complete Particular Integral
Finally, the complete particular integral (PI) is the sum of y_p1 and y_p2:
y_p = y_p1 + y_p2 y_p = (1/2)e^x cos x + (1/52) cos 4x - (3/104) sin 4x - (1/8) cos 2x
3. General Solution
The general solution to the differential equation is the sum of the complementary function (y_c) and the particular integral (y_p):
y = y_c + y_p
Substitute the expressions for y_c and y_p that we derived:
y = e^x (C_1 cos(√3x) + C_2 sin(√3x)) + (1/2)e^x cos x + (1/52) cos 4x - (3/104) sin 4x - (1/8) cos 2x
This is the general solution to the given differential equation. It represents a family of solutions, with C_1 and C_2 being arbitrary constants that can be determined if initial or boundary conditions are provided. This solution captures both the natural response of the system (represented by the complementary function) and the forced response due to the external driving function (represented by the particular integral). The combination of these two components provides a complete picture of the system's behavior.
Conclusion
In this comprehensive solution, we have meticulously solved the second-order linear non-homogeneous differential equation (D^2 - 2D + 4)y = e^x cos x + sin x cos 3x. We began by deriving the auxiliary equation and finding the complementary function, which represents the solution to the homogeneous part of the equation. Subsequently, we tackled the non-homogeneous part by employing the method of undetermined coefficients and the principle of superposition, carefully deriving the particular integral. Through trigonometric identities and systematic calculations, we simplified the problem into manageable steps. Finally, we combined the complementary function and the particular integral to obtain the general solution, a complete representation of all possible solutions to the given differential equation. This step-by-step approach provides a clear and thorough understanding of the solution process, highlighting the key techniques and concepts involved in solving such differential equations. This mastery is crucial for anyone working in fields where differential equations are used to model and analyze physical systems.
