Problem 26 - Advanced Engineering Mathematics

Question

In Problems \(15-28\), find the general solution of the given higher-order differential equation. $$ \frac{d^{4} y}{d x^{4}}-7 \frac{d^{2} y}{d x^{2}}-18 y $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 e^{3x} + C_2 e^{-3x} + C_3 \cos(\sqrt{2}x) + C_4 \sin(\sqrt{2}x) \).

Step 1: Recognize the Equation Type

The given equation \( \frac{d^4 y}{d x^4} - 7 \frac{d^2 y}{d x^2} - 18y = 0 \) is a fourth-order linear homogeneous differential equation with constant coefficients.

Step 2: Assume a Solution Form

For a differential equation with constant coefficients, assume a solution of the form \( y = e^{rx} \). This transforms the differential equation into an algebraic equation, known as the characteristic equation.

Step 3: Formulate the Characteristic Equation

Substitute \( y = e^{rx} \) into the differential equation: \( r^4 e^{rx} - 7r^2 e^{rx} - 18 e^{rx} = 0 \). Factoring out \( e^{rx} \) (since \( e^{rx} eq 0 \)), we get the characteristic equation \( r^4 - 7r^2 - 18 = 0 \).

Step 4: Solve the Characteristic Equation

Attempt to factor the characteristic equation \( r^4 - 7r^2 - 18 = 0 \). Let \( u = r^2 \), transforming it into a quadratic equation: \( u^2 - 7u - 18 = 0 \). Solve for \( u \) using the quadratic formula: \( u = \frac{7 \pm \sqrt{7^2 + 4 \, \cdot 18}}{2} = \frac{7 \pm \sqrt{121}}{2} \).

Step 5: Calculate the Roots

Solve \( u = \frac{7 \pm 11}{2} \), giving roots \( u = 9 \) and \( u = -2 \). Remembering \( u = r^2 \), this gives us \( r^2 = 9 \) and \( r^2 = -2 \).

Step 6: Determine Individual Roots

For \( r^2 = 9 \), \( r = \pm 3 \). For \( r^2 = -2 \), \( r = \pm i \sqrt{2} \). This provides the roots \( r = 3, -3, i \sqrt{2}, -i \sqrt{2} \).

Step 7: Formulate the General Solution

The general solution is a combination of exponential and trigonometric terms: \( y(x) = C_1 e^{3x} + C_2 e^{-3x} + C_3 \cos(\sqrt{2}x) + C_4 \sin(\sqrt{2}x) \), where \( C_1, C_2, C_3, C_4 \) are arbitrary constants.

Key Concepts

Characteristic Equation

The characteristic equation is a crucial concept when solving higher-order linear differential equations, especially those with constant coefficients. Its primary purpose is to turn a differential equation into an easier-to-manage algebraic form.
In our specific problem, we started with the fourth-order differential equation: \[ \frac{d^4 y}{d x^4} - 7 \frac{d^2 y}{d x^2} - 18y = 0 \] To transform it, we assumed a solution of the form \( y = e^{rx} \). This approach helps translate derivatives into powers of \( r \). After substitution, the differential equation becomes:\[ r^4 e^{rx} - 7r^2 e^{rx} - 18 e^{rx} = 0 \] By factoring out \( e^{rx} \), which is never zero, we are left with the characteristic equation: \[ r^4 - 7r^2 - 18 = 0 \]. This algebraic expression allows us to find the roots, which are fundamental in constructing the general solution.

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a type of differential equation where a combination of derivatives of a function and the function itself equals zero. "Linear" indicates that each term is either a constant, a derivative of the function, or the function itself raised only to the first power. "Homogeneous" signifies that there isn't any standalone constant or function present on the other side of the equation; it equals zero.

• For example, the equation \( \frac{d^4 y}{d x^4} - 7 \frac{d^2 y}{d x^2} - 18y = 0 \) is linear because all terms involve either derivatives or the function \( y \), raised to the power of one.
• It is homogeneous because it equates to 0 without any external function or constant.

These properties are vital as they influence the structure of the solution, determining how we approach solving them and what assumptions we can make for finding solutions.

Constant Coefficients

Whenever a differential equation has constant coefficients, it means that these coefficients remain fixed rather than varying with respect to the independent variable, typically \( x \). This constancy simplifies the process of finding the general solution.
In our exercise's differential equation: \[ \frac{d^4 y}{d x^4} - 7 \frac{d^2 y}{d x^2} - 18y = 0 \] The coefficients "1," "-7," and "-18" remain constant. This stability allows us to assume solutions of the form \( y = e^{rx} \) since each derivative of \( e^{rx} \) brings down a power of \( r \) without affecting the coefficient values.
This approach wouldn't work as straightforwardly if the coefficients were functions of \( x \), as such cases often require different solution techniques.

General Solution

The general solution of a differential equation encompasses all possible solutions, constructed by considering both the exponential and trigonometric forms as guided by the roots of the characteristic equation.
For our equation, we derived the roots: 3, -3, \( i \sqrt{2} \), and \( -i \sqrt{2} \). To form the general solution, these roots guide us to:

• Exponential terms come from real roots \( r = 3 \) and \( r = -3 \), forming \( C_1 e^{3x} \) and \( C_2 e^{-3x} \).
• Trigonometric terms arise from complex roots \( i \sqrt{2} \) and \( -i \sqrt{2} \), resulting in \( C_3 \cos(\sqrt{2}x) \) and \( C_4 \sin(\sqrt{2}x) \).

Thus, the full general solution is:\[ y(x) = C_1 e^{3x} + C_2 e^{-3x} + C_3 \cos(\sqrt{2}x) + C_4 \sin(\sqrt{2}x) \]where \( C_1, C_2, C_3, C_4 \) are arbitrary constants showing the solution's degree of freedom reflecting initial/boundary conditions.