Problem 10 - Differential Equations and Linear Algebra

Question

Determine the general solution to the given differential equation. $$y^{\prime \prime \prime}+10 y^{\prime \prime}+25 y^{\prime}=0.$$

Step-by-Step Solution

Answer

The general solution to the given differential equation \(y^{\prime \prime \prime} + 10 y^{\prime \prime} + 25 y^{\prime} = 0\) is: \[ y(x) = C + D_1e^{-5x} + D_2xe^{-5x}, \] where \(C\), \(D_1\), and \(D_2\) are constants.

Step 1: Write down the characteristic equation

The given differential equation is: \[ y^{\prime \prime \prime}+10 y^{\prime \prime}+25 y^{\prime}=0. \] The corresponding characteristic equation for this differential equation is found by replacing \(y^{\prime \prime \prime}\) with \(r^3\), \(y^{\prime \prime}\) with \(r^2\), and \(y^{\prime}\) with \(r\): \[ r^3 + 10r^2 + 25r = 0 \]

Step 2: Factor the characteristic equation

Factor out the common term r from the characteristic equation: \[ r(r^2 + 10r + 25) = 0 \]

Step 3: Find the roots of the characteristic equation

We already have one root from the factored equation: r = 0. Next, we need to find the roots of the quadratic equation \(r^2 + 10r + 25\). To do this, we can use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = 10\), and \(c = 25\). Plug in these values and obtain the roots: \[ r = \frac{-10 \pm \sqrt{10^2 - 4(1)(25)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 100}}{2} = -5 \] So the roots of the characteristic equation are \(r = 0\) (with multiplicity 1) and \(r = -5\) (with multiplicity 2).

Step 4: Write the general solution based on the roots

Since there are three roots, the general solution will have three parts: - For the root \(r = 0\), the corresponding solution function is \(y_1 = Ce^{0x} = C\), where \(C\) is a constant. - For the root \(r = -5\), the corresponding solution functions are \(y_2 = D_1e^{-5x}\) and \(y_3 = D_2xe^{-5x}\), where \(D_1\) and \(D_2\) are constants since -5 is a root with multiplicity 2. Now, combine these solution functions to give the general solution of the given differential equation: \[ y(x) = Ce^0 + D_1e^{-5x} + D_2xe^{-5x} \] Or simply, \[ y(x) = C + D_1e^{-5x} + D_2xe^{-5x} \] This is the general solution to the given differential equation.

Key Concepts

Characteristic Equation

Understanding the characteristic equation is essential for solving higher-order linear homogeneous differential equations. This special equation, which emerges from assuming a solution of the form \(e^{rx}\), transforms the complex task of solving the differential equation into an algebraic problem. Our goal is to find the value of \(r\) that satisfies this equation.

For the given differential equation \(y''' + 10y'' + 25y' = 0\), the characteristic equation is \(r^3 + 10r^2 + 25r = 0\). Notice how each derivative of \(y\) corresponds to a power of \(r\) when we switch from calculus to algebra. This step simplifies finding the solution by focusing on algebraic roots rather than differential calculus.

Once we identify the characteristic equation, the roots are within reach through either factoring, as in this case, or other methods such as using the quadratic formula for quadratic characteristic equations. By solving the characteristic equation, we reveal the behavior of the differential equation's solutions in terms of exponential functions.

Roots of Differential Equations

In any differential equation's journey toward resolution, the roots of the characteristic equation represent critical milestones. In our example, after factoring, we identify one root directly: \(r = 0\). We then use the quadratic formula to find the other roots and determine that \(r = -5\) is a double root.

The significance of these roots can't be overstated—they guide us to the general form of the differential equation's solution. A single real root corresponds to an exponential function part of the solution, whereas repeated roots, like the case with \(r = -5\), introduce polynomial factors motivated by the need for linear independence among solutions. This concept, commonly overlooked, is a cornerstone for understanding the structure of solutions to linear differential equations and foreseeing the behavior of the solutions based on the nature of these roots.

General Solution of Differential Equations

Compiling the knowledge of our characteristic roots, we're equipped to draft the general solution to our differential equation. Each root translates to a term in the solution, with constants reflecting initial conditions or particular situations. In this scenario, \(C\), \(D_1\), and \(D_2\) are those placeholders, and the exponential functions, as well as \(x\) multiplied by an exponential for the repeated root, construct the solution.

A crucial detail to remember is the inclusion of \(xe^{-5x}\) for the double root at \(r = -5\). This nuanced touch ensures our final expression \(y(x) = C + D_1e^{-5x} + D_2xe^{-5x}\) accounts for all possible solution variations our differential equation might exhibit. By piecing together these exponential terms, reflective of the inherent behavior given by the roots, we articulate the full expanse of scenarios this differential equation can represent.