Problem 19 - Advanced Engineering Mathematics

Question

In Problems \(15-28\), find the general solution of the given higher-order differential equation. $$ \frac{d^{3} u}{d t^{3}}+\frac{d^{2} u}{d t^{2}}-2 u \quad 0 $$

Step-by-Step Solution

Answer

The general solution is \( u(t) = C_1 e^{t} + e^{-t}(C_2 \cos t + C_3 \sin t) \).

Step 1: Write the differential equation

The given differential equation is \( \frac{d^{3} u}{d t^{3}} + \frac{d^{2} u}{d t^{2}} - 2u = 0 \). This is a third-order linear homogeneous differential equation with constant coefficients.

Step 2: Find the characteristic equation

Assume the solution is of the form \( u(t) = e^{rt} \). Substitute \( u(t) = e^{rt} \) into the differential equation to form the characteristic equation: \( r^3 + r^2 - 2 = 0 \).

Step 3: Solve the characteristic equation

We need to solve \( r^3 + r^2 - 2 = 0 \). This can be done by trying possible rational roots using the Rational Root Theorem. Possible rational roots are \( \pm 1, \pm 2 \). Upon testing, we find that \( r = 1 \) is a root.

Step 4: Factor the characteristic polynomial

Since \( r = 1 \) is a root, we factor the polynomial as \( (r - 1)(r^2 + 2r + 2) = 0 \). To verify, divide \( r^3 + r^2 - 2 \) by \( r - 1 \) using synthetic division or polynomial long division and check that the quotient is \( r^2 + 2r + 2 \).

Step 5: Solve the quadratic factor

Solve \( r^2 + 2r + 2 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This gives \( r = -1 \pm i \) (where \( i \) is the imaginary unit representing \( \sqrt{-1} \)). These are complex roots.

Step 6: Write the general solution

The general solution based on the roots \( r = 1, -1 \pm i \) is given by: \( u(t) = C_1 e^{t} + e^{-t}(C_2 \cos t + C_3 \sin t) \), where \( C_1, C_2, \) and \( C_3 \) are constants determined by initial conditions.

Key Concepts

Characteristic Equation

To solve a higher-order linear homogeneous differential equation, one important step is finding the characteristic equation. For the given differential equation, \( \frac{d^{3} u}{d t^{3}} + \frac{d^{2} u}{d t^{2}} - 2u = 0 \), we start by assuming a solution of the form \( u(t) = e^{rt} \). This assumption helps to transform the differential equation into an algebraic equation in terms of \( r \). By substituting \( u(t) = e^{rt} \) into the equation, we obtain the characteristic equation \( r^3 + r^2 - 2 = 0 \). Solving this characteristic equation is crucial as it helps us find the roots, which subsequently allows us to construct the general solution to the differential equation.

Linear Homogeneous Differential Equation

The given differential equation, \( \frac{d^{3} u}{d t^{3}} + \frac{d^{2} u}{d t^{2}} - 2u = 0 \), is classified as a linear homogeneous differential equation. Here are a few key points about this type of equation:

• "Linear" indicates that the equation is a sum of derivatives of \(u(t)\) without any products or non-linear terms.
• "Homogeneous" means the equation is equal to zero. This absence of a non-zero term makes it simpler to solve compared to non-homogeneous equations.
• The equation involves derivatives up to the third order, making it a higher-order differential equation.

Understanding these properties helps us apply the correct methods to solve the equation and find all possible solutions.

Complex Roots

When solving the characteristic equation \( r^3 + r^2 - 2 = 0 \), one of the steps involves dealing with complex roots. After finding \( r = 1 \) as a root and factoring the polynomial, the quadratic factor \( r^2 + 2r + 2 = 0 \) remains. Using the quadratic formula, we solve for \( r \) and find two complex roots: \( r = -1 \pm i \). Complex roots in differential equations often lead to solutions involving sine and cosine functions. They are typically expressed in the form of \( e^{\alpha t} (C \cos(\beta t) + D \sin(\beta t)) \), where \( \alpha \) and \( \beta \) come from the real and imaginary parts of the roots respectively. For our equation, this results in a component of the solution that involves \( e^{-t}(C_2 \cos t + C_3 \sin t) \). Dealing with complex roots is essential in understanding the behavior of solutions to differential equations, especially in oscillatory systems.

General Solution

The general solution to a linear homogeneous differential equation is an expression that encompasses all possible solutions to the equation. For the given third-order differential equation, we use the roots \( r = 1, -1 \pm i \) found from the characteristic equation. The general solution is a combination of all terms derived from these roots:

• The real root \( r = 1 \) yields the exponential term \( C_1 e^{t} \).
• The complex roots \( -1 \pm i \) contribute the terms \( e^{-t}(C_2 \cos t + C_3 \sin t) \).

Thus, the general solution is \( u(t) = C_1 e^{t} + e^{-t}(C_2 \cos t + C_3 \sin t) \). In this expression, \( C_1, C_2, \) and \( C_3 \) are arbitrary constants, determined by initial conditions or boundary values specific to a given problem. The general solution provides a comprehensive picture of all potential solutions to the differential equation.