Problem 16 - Advanced Engineering Mathematics

Question

In Problems 13-28, use the procedures developed in this chapter to find the general solution of each differential equation. $$ 2 y^{\prime \prime \prime}+9 y^{\prime \prime}+12 y^{\prime}+5 y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1e^{-\frac{1}{2}x} + e^{-2x}(C_2\cos(x) + C_3\sin(x))\).

Step 1: Identify the Type of Differential Equation

The given equation is a homogeneous linear differential equation of third order with constant coefficients. We'll need to find the characteristic equation next.

Step 2: Write the Characteristic Equation

For the differential equation \(2y''' + 9y'' + 12y' + 5y = 0\), the characteristic equation is obtained by replacing \(y\) with \(e^{rx}\) giving us: \[2r^3 + 9r^2 + 12r + 5 = 0\]

Step 3: Solve the Characteristic Equation

Use methods such as synthetic division or the rational root theorem to find the roots of the characteristic equation:* Guess solutions for \(r\) (possible rational roots are \(\pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}\)).* Upon testing, find that \(r = -\frac{1}{2}\) is a root. Perform polynomial division to reduce the polynomial.

Step 4: Factor the Polynomial

Divide \(2r^3 + 9r^2 + 12r + 5\) by \(r + \frac{1}{2}\) to get a quadratic polynomial. The quotient is:\(2r^2 + 8r + 10\).

Step 5: Solve the Quadratic Polynomial

Now solve the quadratic polynomial \(2r^2 + 8r + 10 = 0\) using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Where \(a = 2, b = 8, c = 10\).Calculate: \[r = \frac{-8 \pm \sqrt{64 - 80}}{4}\]\[r = \frac{-8 \pm \sqrt{-16}}{4}\]This results in complex roots: \(-2 \pm i\).

Step 6: Write General Solution

With one real root \(r_1 = -\frac{1}{2}\) and complex roots \(r_2 = -2 + i\) and \(r_3 = -2 - i\), the general solution is:\[y(x) = C_1e^{-\frac{1}{2}x} + e^{-2x}(C_2\cos(x) + C_3\sin(x))\] where \(C_1, C_2, C_3\) are arbitrary constants.

Key Concepts

Characteristic Equation

In the context of differential equations, a characteristic equation is a crucial tool. It helps convert a higher-order differential equation into something more manageable. For a homogeneous linear differential equation with constant coefficients, like \[2 y^{\prime \prime \prime}+9 y^{\prime \prime}+12 y^{\prime}+5 y=0,\]we use the characteristic equation to find potential solutions.

• The characteristic equation is formed by assuming a solution of the form \(y = e^{rx}\) where \(r\) is a constant.
• By substituting \(y = e^{rx}\) into the differential equation, we derive a polynomial equation in terms of \(r\).

In this example, the characteristic equation becomes:\[2r^3 + 9r^2 + 12r + 5 = 0.\]This polynomial equation is the first step towards finding the general solution of the differential equation.

Complex Roots

Complex roots often occur in the process of solving characteristic equations. After solving the characteristic equation, you may find some or all of the roots to be complex numbers.

• Complex roots arise when the discriminant—\(b^2 - 4ac\)—of a quadratic equation is negative.
• In this example, the quadratic polynomial \(2r^2 + 8r + 10 = 0\) yields complex roots: \(-2 \pm i\).

These complex roots can be represented as \(a + bi\) and \(a - bi\), where \(a\) is the real part, and \(b\) is the imaginary part.

• In general, complex roots lead to oscillatory solutions represented with functions like sine and cosine.

So here, the presence of complex roots informs us that part of the general solution involves trigonometric functions.

General Solution

The general solution of a homogeneous linear differential equation is a combination of solutions derived from its characteristic equation. For the equation provided, \[2 y^{\prime \prime \prime}+9 y^{\prime \prime}+12 y^{\prime}+5 y=0,\]we determined a real root and complex conjugate roots from the characteristic equation.

• The real root leads to the term \(C_1e^{-\frac{1}{2}x}\) in the solution.
• The complex roots \(-2 \pm i\) contribute the term \(e^{-2x}(C_2\cos(x) + C_3\sin(x))\).

Thus, the general solution becomes:\[y(x) = C_1e^{-\frac{1}{2}x} + e^{-2x}(C_2\cos(x) + C_3\sin(x)),\]where \(C_1, C_2, C_3\) are constants determined by initial conditions or boundary conditions.

Polynomial Division

Polynomial division is a method used to simplify the process of solving polynomial equations, especially when a root is known or guessed. In the context of finding the characteristic equation's roots, it becomes essential.

• We use polynomial division to simplify higher-degree polynomials by checking rational roots through trial and error or established roots.
• In our example, once we determined \(r = -\frac{1}{2}\) as a root, we performed polynomial division on \(2r^3 + 9r^2 + 12r + 5\).

The polynomial division gives us a simpler quadratic polynomial:\[2r^2 + 8r + 10.\]Solving this quadratic provides complex roots that are crucial for completing the general solution. Polynomial division thus helps reduce complexity and find the necessary roots more efficiently.