Problem 62 - Advanced Engineering Mathematics

Question

Verify that \(y \quad \sinh x-2 \cos (x+\pi / 6)\) is a particular solution of \(y^{(4)}-y \quad 0\). Reconcile this particular solution with the general solution of the \(\mathrm{DE}\).

Step-by-Step Solution

Answer

The function is a particular solution since it satisfies the equation; it's reconcilable as a specific case of the general solution.

Step 1: Differentiate the given function

The given function is \( y = \sinh x - 2\cos(x + \pi/6) \). We need its fourth derivative: First derivative: \( y' = \cosh x + 2\sin(x + \pi/6) \)Second derivative: \( y'' = \sinh x + 2\cos(x + \pi/6) \)Third derivative: \( y''' = \cosh x - 2\sin(x + \pi/6) \)Fourth derivative: \( y^{(4)} = \sinh x - 2\cos(x + \pi/6) \)

Step 2: Plug the function into the differential equation

Substitute the function \( y \) and its fourth derivative \( y^{(4)} \) into the differential equation \( y^{(4)} - y = 0 \).\[ y^{(4)} - y = (\sinh x - 2\cos(x + \pi/6)) - (\sinh x - 2\cos(x + \pi/6)) = 0 \]Since this equation holds true, \( y = \sinh x - 2\cos(x + \pi/6) \) is indeed a particular solution.

Step 3: Find the general solution of the differential equation

The general solution of a linear homogeneous differential equation with constant coefficients, such as \( y^{(4)} - y = 0 \), can be found using characteristic equations.The characteristic equation associated with this differential equation is \( r^4 - 1 = 0 \). Solving this, we get the roots: \( r = 1, -1, i, -i \).So, the general solution is:\[ y_g = C_1 e^x + C_2 e^{-x} + C_3 \cos(x) + C_4 \sin(x) \]

Step 4: Reconcile the particular solution with the general solution

Notice that the function \( y = \sinh x - 2\cos(x + \pi/6) \) can be expressed in terms of the solutions found in the general solution. Specifically, \( \sinh x = \frac{e^x - e^{-x}}{2} \) matches the terms in the general solution \( e^x \) and \( e^{-x} \), and \(-2\cos(x + \pi/6)\) can be represented using trigonometric identities based on \( \cos(x) \) and \( \sin(x) \).Thus, the particular solution can be expressed as a specific case of the general solution when the constants \( C_1, C_2, C_3, \) and \( C_4 \) are chosen appropriately.

Key Concepts

Particular Solution

A particular solution to a differential equation is a solution that satisfies the equation but does not cover all possible solutions. In the context of a differential equation like \(y^{(4)} - y = 0\), a particular solution is one where specific conditions or forms are considered.

This equation suggests that when you substitute a particular solution, it should hold the equation as true across its domain.

• The key purpose of finding a particular solution is to understand specific behaviors of the system modeled by the differential equation.
• It helps in representing real-world scenarios where boundary or initial conditions are known.

When you find that \(y = \sinh x - 2\cos(x + \pi/6)\) is a particular solution, it means that this specific form satisfies the differential equation for a known set of variables.

Particular solutions often form the basis for building up more complex solutions along with the general solution.

General Solution

The general solution of a differential equation encapsulates all possible solutions that can be derived from it. It includes an arbitrary constant or constants that cater to various initial conditions. For a differential equation like \(y^{(4)} - y = 0\), the general solution is represented in terms of its characteristic equation solutions. By finding the roots of the characteristic equation, we can express the complete general solution:\[y_g = C_1 e^x + C_2 e^{-x} + C_3 \cos(x) + C_4 \sin(x)\]Here, \(C_1, C_2, C_3,\) and \(C_4\) are constants that need to be determined based on additional information or conditions.

• The general solution represents the family of all solutions of the given differential equation.
• Every particular solution is a specific case within the general solution where these constants take particular values.

In practical scenarios, once we have the general solution, specific cases or behaviors in the form of particular solutions can be calculated by assigning appropriate values to these constants.

Characteristic Equation

The characteristic equation is a crucial step in solving linear differential equations with constant coefficients. It translates the differential equation into an algebraic equation. This helps find solutions as exponential functions.In the equation \(y^{(4)} - y = 0\), the characteristic equation is formed by assuming solutions of the form \(y = e^{rx}\), leading to:\[r^4 - 1 = 0\]This simplifies into finding the roots of the polynomial \(r^4 - 1 = 0\). The solutions are \(r = 1, -1, i, -i\), providing different types of functions (exponential and trigonometric) which can be recombined to form the general solution.

• The characteristic equation method allows for systematic solving of differential equations.
• Once roots are found, they dictate the terms that will compose the general solution.

The beauty of this approach is its ability to handle complex roots, which result in the trigonometric functions part of the solution, giving a full view of the behavior of the differential equation across different conditions.