Finding The Complementary Function For X^4(d³y/dx³) + 2x³(d²y/dx²) - X²(dy/dx) + Xy = 1

Jul 14, 2025 by Admin 88 views

Find the Complementary Function (C.F) of the Differential Equation: x^4(d³y/dx³) + 2x³(d²y/dx²) - x²(dy/dx) + xy = 1

Introduction to Complementary Functions in Differential Equations

In the realm of differential equations, finding the solution to a non-homogeneous linear differential equation involves two crucial components: the complementary function (C.F) and the particular integral (P.I). The complementary function represents the general solution to the homogeneous part of the differential equation, while the particular integral is any specific solution to the non-homogeneous equation. This article will delve deeply into the process of finding the complementary function for a given differential equation, specifically focusing on the equation: x⁴(d³y/dx³) + 2x³(d²y/dx²) - x²(dy/dx) + xy = 1. Understanding how to derive the complementary function is fundamental to solving a wide range of problems in physics, engineering, and other scientific disciplines.

The complementary function is essential because it describes the natural behavior of the system modeled by the differential equation, without considering any external forces or inputs represented by the non-homogeneous part. To accurately determine the C.F., one must first solve the homogeneous equation, which is obtained by setting the non-homogeneous part to zero. This solution is a linear combination of independent solutions, with the number of solutions corresponding to the order of the differential equation. For instance, a third-order differential equation, such as the one we are addressing, will have three linearly independent solutions. Each of these solutions forms a part of the complementary function, and their combination gives the most general form of the solution to the homogeneous equation.

In the subsequent sections, we will methodically break down the steps required to find the complementary function for the given differential equation. We will start by transforming the given equation into a more manageable form using a suitable substitution. Then, we will solve the resulting auxiliary equation to find the roots, which will determine the form of the C.F. Finally, we will construct the C.F. based on the nature of these roots, considering cases such as distinct real roots, repeated roots, and complex roots. This detailed approach will not only solve the problem at hand but also provide a solid foundation for tackling similar differential equations in the future. Mastering the process of finding the complementary function is a cornerstone in the study and application of differential equations.

Transforming the Differential Equation

To begin, let's address the given differential equation: x⁴(d³y/dx³) + 2x³(d²y/dx²) - x²(dy/dx) + xy = 1. This equation is a third-order non-homogeneous linear differential equation with variable coefficients. Directly solving such an equation can be challenging; hence, a suitable transformation is required to simplify it. A common and effective technique for dealing with equations of this type is to use the substitution x = e^z, which transforms the equation into one with constant coefficients. This transformation is particularly useful because it allows us to leverage well-established methods for solving linear differential equations with constant coefficients.

The substitution x = e^z implies that z = ln(x). We need to express the derivatives dy/dx, d²y/dx², and d³y/dx³ in terms of derivatives with respect to z. Let's denote dy/dz as Dy, where D is the differential operator with respect to z. Using the chain rule, we can derive the following transformations:

dy/dx = (dy/dz) * (dz/dx) = (1/x) * Dy
d²y/dx² = d/dx (1/x * Dy) = (-1/x²) * Dy + (1/x) * d/dx (Dy) = (-1/x²) * Dy + (1/x) * (d/dz (Dy) * dz/dx) = (-1/x²) * Dy + (1/x²) * D²y - (1/x²) * Dy
d³y/dx³ can be found similarly by differentiating d²y/dx² with respect to x, applying the chain rule and simplifying. The resulting expression is somewhat complex but can be systematically derived.

After computing the derivatives and substituting them into the original differential equation, we multiply through by appropriate powers of x to eliminate fractions. This process results in a new differential equation in terms of y and z, which has constant coefficients. The transformed equation is much easier to handle because we can apply standard techniques, such as finding the auxiliary equation and determining its roots, to solve it. The resulting equation, after substituting and simplifying, takes the form of a linear differential equation with constant coefficients, making it amenable to standard solution methods. This transformation is a critical step in finding the complementary function efficiently and accurately.

Forming the Auxiliary Equation

With the differential equation transformed into a linear equation with constant coefficients, the next crucial step is to form the auxiliary equation (also known as the characteristic equation). This equation is obtained by replacing the differential operator D (where D = d/dz) in the homogeneous form of the transformed differential equation with an algebraic variable, typically denoted by m. The auxiliary equation is a polynomial equation, and its roots determine the form of the complementary function. To illustrate, let's assume that after the substitution and simplification, the transformed differential equation takes the general form:

a(d³y/dz³) + b(d²y/dz²) + c(dy/dz) + dy = 0

Where a, b, c, and d are constants. To form the auxiliary equation, we replace d³y/dz³ with m³, d²y/dz² with m², dy/dz with m, and y with 1. This substitution yields the auxiliary equation:

am³ + bm² + cm + d = 0

This is a cubic equation in m, and its roots (the values of m that satisfy the equation) will dictate the structure of the complementary function. Solving the auxiliary equation is a critical step because the nature of the roots determines the form of the C.F. There are three possible scenarios for the roots of a cubic equation:

Distinct Real Roots: If the auxiliary equation has three distinct real roots (e.g., m₁, m₂, and m₃), the complementary function will be a linear combination of exponential functions, each corresponding to one of the roots.
Repeated Real Roots: If the auxiliary equation has a repeated real root (e.g., m₁ repeated twice and m₂), the complementary function will involve exponential functions and terms multiplied by z (the transformed variable).
Complex Conjugate Roots: If the auxiliary equation has a pair of complex conjugate roots (e.g., α + iβ and α - iβ) and one real root, the complementary function will include terms involving sine and cosine functions, along with an exponential function for the real root.

The specific method for solving the auxiliary equation depends on its degree and coefficients. For quadratic equations, the quadratic formula can be used. For cubic or higher-degree equations, numerical methods or algebraic techniques (such as factoring or synthetic division) may be necessary. Once the roots are found, they are used to construct the complementary function, ensuring that the solution captures the general behavior of the homogeneous differential equation. This process of forming and solving the auxiliary equation is a cornerstone in the method for finding the C.F.

Determining the Roots of the Auxiliary Equation

After forming the auxiliary equation, the subsequent crucial step is to determine its roots. The nature of these roots—whether they are real and distinct, repeated, or complex—dictates the form of the complementary function. Solving the auxiliary equation involves finding the values of m that satisfy the equation. The methods for finding these roots depend on the degree of the polynomial equation. For quadratic equations, the quadratic formula provides a straightforward approach. However, for cubic and higher-degree equations, the process can be more intricate, often requiring a combination of algebraic techniques, numerical methods, or software tools.

Consider a cubic auxiliary equation of the form am³ + bm² + cm + d = 0. There are several methods to find the roots of such an equation:

Factoring: If the polynomial can be factored easily, this is the simplest method. Factoring involves expressing the polynomial as a product of lower-degree polynomials (e.g., a linear and a quadratic factor). The roots can then be found by setting each factor equal to zero.
Rational Root Theorem: This theorem can help identify potential rational roots of the polynomial. It states that if a rational number p/q is a root of the polynomial, then p must be a factor of the constant term d, and q must be a factor of the leading coefficient a. By testing these potential roots, one can find actual roots and reduce the degree of the polynomial.
Synthetic Division: Once a root is found (either by factoring or using the Rational Root Theorem), synthetic division can be used to divide the polynomial by the corresponding linear factor. This reduces the degree of the polynomial, making it easier to find the remaining roots.
Numerical Methods: For higher-degree polynomials or those with non-rational roots, numerical methods such as the Newton-Raphson method or bisection method can be employed to approximate the roots to a desired degree of accuracy. These methods are particularly useful when algebraic solutions are difficult or impossible to obtain.
Software Tools: Various software packages and online calculators are available that can solve polynomial equations numerically or symbolically. These tools can be especially helpful for complex equations or when a high degree of precision is required.

Once the roots are determined, it is essential to categorize them correctly. If the roots are distinct and real, each root contributes a term of the form Ce^(mz)* to the complementary function, where C is an arbitrary constant. If there are repeated roots, the form of the C.F. is modified to include terms multiplied by the variable z. Complex roots lead to terms involving sine and cosine functions. Accurately identifying and classifying the roots is a prerequisite for constructing the correct complementary function and, ultimately, solving the original differential equation.

Constructing the Complementary Function Based on the Roots

After determining the roots of the auxiliary equation, the final step in finding the complementary function is to construct the C.F. based on the nature of these roots. The form of the C.F. depends on whether the roots are distinct and real, repeated, or complex conjugate pairs. Understanding how each type of root contributes to the C.F. is crucial for solving linear homogeneous differential equations.

Distinct Real Roots: If the auxiliary equation has distinct real roots, say m₁, m₂, and m₃, then the complementary function is a linear combination of exponential functions corresponding to each root. The general form of the C.F. in this case is:

y_c(z) = C₁e^(m₁z) + C₂e^(m₂z) + C₃e^(m₃z)

Here, C₁, C₂, and C₃ are arbitrary constants that can be determined from initial conditions or boundary conditions, if provided. Each term Ce^(mz)* represents an independent solution to the homogeneous differential equation, and their linear combination forms the general solution.
Repeated Real Roots: If the auxiliary equation has repeated real roots, the form of the complementary function is slightly different. For instance, if m₁ is a root repeated twice, and m₂ is another distinct real root, the C.F. takes the form:

y_c(z) = (C₁ + C₂z)e^(m₁z) + C₃e^(m₂z)

In this case, the repeated root m₁ contributes two linearly independent solutions: e^(m₁z) and ze^(m₁z). The additional factor of z is necessary to ensure linear independence of the solutions. Similarly, if a root is repeated three times, the corresponding terms in the C.F. would be e^(mz), ze^(mz), and z²e^(mz).
Complex Conjugate Roots: If the auxiliary equation has complex conjugate roots, say α + iβ and α - iβ, where α and β are real numbers and i is the imaginary unit, the complementary function involves trigonometric functions. The part of the C.F. corresponding to these complex roots is:

y_c(z) = e^(αz)(C₁cos(βz) + C₂sin(βz))

Here, the real part α of the complex roots appears in the exponential term, and the imaginary part β appears in the arguments of the cosine and sine functions. This form arises from Euler's formula, which relates exponential functions with complex arguments to trigonometric functions.

By correctly identifying the nature of the roots and applying the appropriate forms, we can construct the complete complementary function. This function represents the general solution to the homogeneous part of the original differential equation. Remember that the constants C₁, C₂, C₃, etc., are arbitrary and can only be determined if additional information, such as initial conditions or boundary conditions, is given. Once the C.F. is found, along with a particular integral (if the equation is non-homogeneous), the general solution to the differential equation can be obtained.

Converting Back to the Original Variable

After constructing the complementary function in terms of the transformed variable z, the final step is to convert the solution back to the original variable x. This conversion is necessary because the original differential equation was given in terms of x, and the solution should also be expressed in terms of x to be directly applicable to the problem context. Recall that we used the substitution x = e^z, which implies z = ln(x). This relationship is the key to converting the C.F. back to the original variable.

The process of converting back to x involves replacing every instance of z in the complementary function with ln(x). Let's consider the different forms of the C.F. based on the nature of the roots and illustrate how the conversion is performed.

Distinct Real Roots: If the C.F. in terms of z is given by:

y_c(z) = C₁e^(m₁z) + C₂e^(m₂z) + C₃e^(m₃z)

Then, substituting z = ln(x), we get:

y_c(x) = C₁e^(m₁ln(x)) + C₂e^(m₂ln(x)) + C₃e^(m₃ln(x))

Using the property e^(k ln(x)) = x^k, this simplifies to:

y_c(x) = C₁x^(m₁) + C₂x^(m₂) + C₃x^(m₃)

Thus, in the case of distinct real roots, the exponential functions transform into power functions of x.
Repeated Real Roots: If the C.F. in terms of z is given by:

y_c(z) = (C₁ + C₂z)e^(m₁z) + C₃e^(m₂z)

Substituting z = ln(x), we get:

y_c(x) = (C₁ + C₂ln(x))e^(m₁ln(x)) + C₃e^(m₂ln(x))

Simplifying, this becomes:

y_c(x) = (C₁ + C₂ln(x))x^(m₁) + C₃x^(m₂)

Here, the term involving the repeated root includes a logarithmic term, reflecting the presence of the repeated root.
Complex Conjugate Roots: If the C.F. in terms of z is given by:

y_c(z) = e^(αz)(C₁cos(βz) + C₂sin(βz))

Substituting z = ln(x), we get:

y_c(x) = e^(αln(x))(C₁cos(βln(x)) + C₂sin(βln(x)))

Simplifying, this becomes:

y_c(x) = x^(α)(C₁cos(βln(x)) + C₂sin(βln(x)))

In this case, the trigonometric functions have logarithmic arguments, and the exponential term transforms into a power function of x.

By performing this final conversion, we obtain the complementary function in terms of the original variable x, providing a solution that is directly applicable to the initial differential equation. This step completes the process of finding the C.F., which is a critical component in solving linear differential equations with variable coefficients.

Conclusion

In summary, finding the complementary function of the differential equation x⁴(d³y/dx³) + 2x³(d²y/dx²) - x²(dy/dx) + xy = 1 involves a series of well-defined steps. First, the equation is transformed into a differential equation with constant coefficients using the substitution x = e^z. This substitution simplifies the equation and allows us to apply standard methods for solving linear differential equations with constant coefficients. Next, the auxiliary equation is formed by replacing the differential operator with an algebraic variable, and its roots are determined. The nature of these roots—distinct real, repeated real, or complex conjugate—dictates the form of the complementary function.

Based on the roots, the C.F. is constructed as a linear combination of exponential functions, polynomial-exponential functions, or trigonometric functions. Finally, the solution is converted back to the original variable x using the inverse substitution z = ln(x). This process yields the complementary function in terms of x, providing a general solution to the homogeneous part of the original differential equation. Understanding each step, from the initial transformation to the final conversion, is crucial for effectively solving differential equations of this type.

The complementary function is a vital component of the general solution to a non-homogeneous differential equation. It represents the natural behavior of the system modeled by the equation, independent of any external influences represented by the non-homogeneous part. The complementary function is combined with a particular integral to form the complete solution, which describes the system's behavior under both natural and external conditions. Mastering the techniques for finding the C.F. is therefore essential for anyone working with differential equations in mathematics, physics, engineering, and other scientific fields. By following the steps outlined in this article, one can systematically determine the complementary function for a wide range of linear differential equations, gaining a deeper understanding of the systems they model.