Problem 18 - Advanced Engineering Mathematics

Question

In Problems 15-28, find the general solution of the given higher-order differential equation. $$ y^{\prime \prime \prime}+3 y^{\prime \prime}-4 y^{\prime}-12 y \quad 0 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

The given differential equation is a third-order linear homogeneous differential equation. Our task is to find the general solution of this equation.

Step 2: Write the Differential Equation

The given equation is: \[ y''' + 3y'' - 4y' - 12y = 0. \]

Step 3: Assume a Solution Form

Assume the solution has the form \( y = e^{rx} \). This means derivatives are \( y' = re^{rx} \), \( y'' = r^2 e^{rx} \), and \( y''' = r^3 e^{rx} \).

Step 4: Substitute into the Differential Equation

Substitute \( y = e^{rx} \), \( y' = re^{rx} \), \( y'' = r^2 e^{rx} \), and \( y''' = r^3 e^{rx} \) into the differential equation.\[ r^3 e^{rx} + 3r^2 e^{rx} - 4r e^{rx} - 12e^{rx} = 0. \]

Step 5: Factor Out \( e^{rx} \)

Factor \( e^{rx} \) out of the equation:\[ e^{rx}(r^3 + 3r^2 - 4r - 12) = 0. \]Since \( e^{rx} eq 0 \), solve for \( r \) in\[ r^3 + 3r^2 - 4r - 12 = 0. \]

Step 6: Solve the Characteristic Equation

Find the roots of the characteristic equation:\[ r^3 + 3r^2 - 4r - 12 = 0. \]Use the rational root theorem or factor by trial and error to rewrite as \((r+3)(r-2)(r+2) = 0\).

Step 7: Find the Roots

Solve \((r+3)(r-2)(r+2) = 0\) by setting each factor equal to zero.- \(r+3=0\) gives \(r=-3\).- \(r-2=0\) gives \(r=2\).- \(r+2=0\) gives \(r=-2\).

Step 8: Write the General Solution

The roots are \( r = -3, 2, -2 \). Hence, the general solution of the differential equation is:\[ y(x) = C_1 e^{-3x} + C_2 e^{2x} + C_3 e^{-2x}, \]where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Key Concepts

Linear Homogeneous Differential Equations

A linear homogeneous differential equation is a specific type of differential equation where all terms are dependent on the function itself and its derivatives, without any additional constant or function on the right side of the equation.
For instance, the expression \( y''' + 3y'' - 4y' - 12y = 0 \) is linear because it consists solely of linear combinations of the derivatives of \( y \). It is homogeneous since it is equal to zero.
These equations are fundamental in studying various physical phenomena as they often describe systems in equilibrium.

• The order of the differential equation denotes the highest derivative, which in this case is three (a third-order equation).
• Being homogeneous implies the absence of any independent term—meaning the system it describes reacts only to its internal changes.

This kind of sophistication allows such equations to model natural laws effectively, such as principles governing wave functions in physics.

Characteristic Equation

The characteristic equation is a vital concept in solving linear homogeneous differential equations.
To determine the form of the general solution, we suppose the solution takes on the exponential form \( y = e^{rx} \), where \( r \) is a constant. Differentiating this assumption results in several expressions involving powers of \( r \).
By substituting these expressions back into the original differential equation, we factor out \( e^{rx} \), leading to the so-called characteristic equation, \( r^3 + 3r^2 - 4r - 12 = 0 \).

• Solving this polynomial characteristic equation allows us to find the specific values \( r \) that satisfy the equation.
• Techniques such as the Rational Root Theorem or synthetic division can assist in finding these roots efficiently.

The roots of the characteristic equation directly inform us about the nature of the solution to the differential equation.

General Solution

Once we've identified the roots of the characteristic equation, forming the general solution of the linear homogeneous differential equation becomes straightforward.
The general solution incorporates terms like \( e^{rx} \), where each \( r \) is a distinct root. In our case, the roots are \( r = -3, 2, -2 \).
The general solution thus becomes:

• \( y(x) = C_1 e^{-3x} + C_2 e^{2x} + C_3 e^{-2x} \)

Here, \( C_1, C_2, \) and \( C_3 \) are arbitrary constants that can be found if initial conditions are provided.
This general solution effectively captures all possible behaviors that the system described by the differential equation can exhibit.

Exponential Function Solution

The idea of an exponential function solution is central to finding solutions of linear homogeneous differential equations.
Employing the assumption \( y = e^{rx} \) forms the backbone of the characteristic equation approach. This form simplifies the problem as it turns the differentiation into multiplication by \( r \), allowing us to explore the dynamics of the equation as a polynomial.
The exponential function is crucial due to its unique property that the derivative of an exponential function is a scalar multiple of itself, making it fitting for linear equations.

• Each \( r \) in \( e^{rx} \) captures a distinct mode or pattern of behavior in the system being modeled.
• Alternative forms—like sinusoidal functions—may appear if the roots are complex, showcasing oscillatory solutions.

By using exponential functions, we efficiently generalize the types of solutions that apply to many physical and mathematical systems.