Problem 20 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given higher order differential equation. $$\frac{d^{3} x}{d t^{3}}-\frac{d^{2} x}{d t^{2}}-4 x=0$$

Step-by-Step Solution

Answer

The general solution is \( x(t) = C_1 e^{2t} + C_2 e^{-\frac{t}{2}} \cos \left( \frac{\sqrt{7}t}{2} \right) + C_3 e^{-\frac{t}{2}} \sin \left( \frac{\sqrt{7}t}{2} \right) \).

Step 1: Identify Type of Differential Equation

The given differential equation \( \frac{d^{3} x}{d t^{3}} - \frac{d^{2} x}{d t^{2}} - 4 x = 0 \) is a third-order linear homogeneous differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

For a differential equation of the form \( a \frac{d^{3}x}{dt^{3}} + b \frac{d^{2}x}{dt^{2}} + c x = 0 \), the characteristic equation is \( a r^{3} + b r^{2} + c = 0 \). Here, the characteristic equation is \( r^{3} - r^{2} - 4 = 0 \).

Step 3: Solve the Characteristic Equation

Factor the characteristic equation \( r^{3} - r^{2} - 4 = 0 \). First, check for rational roots using the Rational Root Theorem, which suggests trying \( r = 2 \), as it satisfies the equation. Use synthetic division to factor the polynomial: \((r - 2)(r^2 + r + 2) = 0\).

Step 4: Solve for Remaining Roots

Solve the quadratic \( r^2 + r + 2 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = 1, c = 2 \). This gives roots \( r = \frac{-1 \pm i\sqrt{7}}{2} \).

Step 5: Write the General Solution

The general solution of the differential equation is obtained by combining the solutions from the characteristic roots. It is \( x(t) = C_1 e^{2t} + C_2 e^{-\frac{t}{2}} \cos \left( \frac{\sqrt{7}t}{2} \right) + C_3 e^{-\frac{t}{2}} \sin \left( \frac{\sqrt{7}t}{2} \right) \), where \( C_1, C_2, \text{ and } C_3 \) are constants.

Key Concepts

Characteristic Equation

In the field of differential equations, a **characteristic equation** plays a crucial role in finding solutions to higher-order linear homogeneous differential equations. When you have a differential equation like the one given in the problem, transforming it into a characteristic equation simplifies the process.

For any differential equation of the form \(a \frac{d^{n}x}{dt^{n}} + b \frac{d^{n-1}x}{dt^{n-1}} + \ldots + c = 0\), the characteristic equation is determined by replacing each derivative with a power of \(r\). This way, an easier to handle algebraic expression results, which in this particular example is \(r^{3} - r^{2} - 4 = 0\).

Solving this equation allows us to find the values of \(r\), which directly help in forming the general solution of the differential equation. This process helps in transforming a differential problem into a more manageable algebraic one.

Linear Homogeneous Differential Equation

A **linear homogeneous differential equation** is a specific type of differential equation characterized by its linearity and homogeneity. It means the equation may consist of derivatives and dependent variables, but none of their products or powers beyond the first degree. In our example, the equation \(\frac{d^{3}x}{dt^{3}} - \frac{d^{2}x}{dt^{2}} - 4x = 0\) is linear because each term with \(x\) or its derivatives appears to the power of one.

The term "homogeneous" implies that all terms are either derivatives or multiples of the function by a constant, and all are equal to zero.

• This makes physics problems easier to handle, as systems without external influences, such as homogeneous equations, often describe them.
• Understanding this concept helps solve the equation systematically.

The absence of any additional forcing function or non-zero terms allows us to formulate a straightforward characteristic equation and solve for the general solution.

Constant Coefficients

The **constant coefficients** imply that the values multiplying each term in the differential equation remain the same and do not depend on the independent variable. In the given equation, all terms \(\frac{d^{3}x}{dt^{3}}, \frac{d^{2}x}{dt^{2}}, \text{and } x\) have coefficients that do not vary with \(t\).

This consistency means you need not worry about changes in coefficients, simplifying the work of solving the equation and forming the characteristic equation.

• Constant coefficients make it simpler to solve the polynomial characteristic equation using methods like the Rational Root Theorem or synthetic division.
• This process is a staple part of solving higher-order differential equations efficiently.

Using these principles, we reduce the complexity of dealing with time-variant equations significantly.

General Solution

The **general solution** of a differential equation represents the family of solutions accounting for all possible initial conditions. In the case of our solved equation, the general solution reflects the solutions derived from each of the characteristic roots.

Here, the roots are \( r = 2 \) from a real root and \( r = \frac{-1 \pm i\sqrt{7}}{2} \) from complex conjugate pairs. These yield:

• For the real root \(r = 2\): the solution \(C_1 e^{2t}\).
• For the complex roots: solutions based on Euler's formula, combining sine and cosine components: \(C_2 e^{-\frac{t}{2}} \cos\left(\frac{\sqrt{7}t}{2}\right) + C_3 e^{-\frac{t}{2}} \sin\left(\frac{\sqrt{7}t}{2}\right)\).

The constants \(C_1, C_2, \text{and } C_3\) are determined by the initial or boundary conditions specific to each problem instance. This framework gives a complete description of the dynamic system modeled by the differential equation.