Problem 26 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$x^{2} y^{\prime \prime}-5 x y^{\prime}+8 y=0, \quad y(2)=32, y^{\prime}(2)=0$$

Step-by-Step Solution

Answer

The solution is \(y(x) = x^4 - 8x^2\).

Step 1: Identify the Type of Differential Equation

The differential equation given is a second-order linear homogeneous ordinary differential equation of the form \(x^2 y'' - 5x y' + 8y = 0\).

Step 2: Assume a Solution Form

Since this is a Cauchy-Euler equation, we use the solution form \(y(x) = x^m\), where \(m\) is a constant to be determined.

Step 3: Compute Derivatives

Find the first and second derivatives: \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\).

Step 4: Substitute and Form the Characteristic Equation

Substitute \(y, y',\) and \(y''\) into the differential equation: \(x^2(m(m-1)x^{m-2}) - 5x(mx^{m-1}) + 8(x^m) = 0\). Simplify to get \(m(m-1)x^m - 5mx^m + 8x^m = 0\).

Step 5: Simplify and Solve Characteristic Equation

Factor out \(x^m\): \(x^m(m(m-1) - 5m + 8) = 0\). This leads to \(m^2 - 6m + 8 = 0\).

Step 6: Solve the Quadratic Equation

Solve \(m^2 - 6m + 8 = 0\) using the quadratic formula: \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=1, b=-6, c=8\). The solutions are \(m_1=4\) and \(m_2=2\).

Step 7: Write the General Solution

The general solution to the equation is \(y(x) = C_1x^4 + C_2x^2\), where \(C_1\) and \(C_2\) are constants.

Step 8: Apply Initial Conditions to Determine Constants

Use the initial conditions \(y(2) = 32\) and \(y'(2) = 0\). Substitute \(y = C_1(2^4) + C_2(2^2) = 32\). This gives \(16C_1 + 4C_2 = 32\).

Step 9: Derive the Expression for the First Derivative

The derivative \(y'(x) = 4C_1x^3 + 2C_2x\). Apply the initial condition \(y'(2) = 0\) to get \(32C_1 + 4C_2 = 0\).

Step 10: Solve the System of Equations

Solve the equations \(16C_1 + 4C_2 = 32\) and \(32C_1 + 4C_2 = 0\) simultaneously to find \(C_1 = 1\) and \(C_2 = -8\).

Step 11: Write the Particular Solution

Substitute the constants back to find the particular solution, \(y(x) = x^4 - 8x^2\).

Step 12: Graph the Solution Curve

Use a graphing utility to plot \(y(x) = x^4 - 8x^2\). The graph will pass through the point \((2, 32)\) and have a horizontal tangent at \((2, 32)\).

Key Concepts

Cauchy-Euler equation

The Cauchy-Euler equation is a special type of second-order linear differential equation. Its standard form is written as:

• \[ x^2y'' + axy' + by = 0 \]

This equation is characterized by having coefficients that are powers of the independent variable \(x\). The main feature of the Cauchy-Euler equation is that it allows for solutions of the form \(y(x) = x^m\), where \(m\) is a constant to be determined. Such a form is chosen because it simplifies the solution process by converting the equation into a polynomial form, making it easier to work with.
For example, in our given equation, \(x^2y'' - 5xy' + 8y = 0\), we identify it as a Cauchy-Euler equation because it fits this format, with coefficients based on powers of \(x\). To solve it, we substitute the assumed form into the equation and derive a simpler characteristic equation. This approach effectively reduces a differential problem to solving a basic algebraic polynomial equation.

Initial-value problem

An initial-value problem, often abbreviated as IVP, involves finding a function that satisfies a differential equation along with specified values at a particular point. In essence, it provides a snapshot initial condition that allows for solving for specific constants in a general solution.
For our problem, we were given the initial conditions:

• \(y(2) = 32\)
• \(y'(2) = 0\)

These initial conditions mean the particular solution must pass through the point \((2, 32)\) and that the slope of the tangent to the curve is zero at \(x = 2\). These conditions are instrumental in determining the unique solution from the general solution formed by our differential equation. Through substituting these conditions back into the equation, we are able to solve for the arbitrary constants \(C_1\) and \(C_2\) in the expression, providing us with a specific solution.

Second-order linear differential equation

A second-order linear differential equation is a type of differential equation that involves the unknown function and its derivatives up to the second derivative. Its general form is:

• \[ ay'' + by' + cy = f(x) \]

where \(a\), \(b\), and \(c\) are constants or functions of \(x\), and \(f(x)\) is a given function. These equations are crucial in modeling various physical phenomena, such as mechanical vibrations and electrical circuits, due to their ability to describe relationships involving accelerating systems.
In our example, the function is homogeneous, meaning \(f(x) = 0\). This simplifies our task, as solving homogeneous linear equations often involves determining a characteristic equation that allows us to ascertain the solution form based on its roots. Only linear expressions involved with \(y\), \(y'\), and \(y''\) classify it as linear, which impacts how we solve them using principles like superposition.

Characteristic equation

The characteristic equation is a vital step in solving linear differential equations, allowing for the transformation from a differential to an algebraic problem. It is formulated by substituting the assumed solution form into the differential equation.
For a second-order Cauchy-Euler equation in the assumed form \(y(x) = x^m\), the derivatives introduce terms where powers of \(x\) cancel out naturally, leading us to a polynomial in \(m\). For our problem, this was:

• \[ m^2 - 6m + 8 = 0 \]

Solving this quadratic equation gives the values of \(m\), which are 4 and 2 in our case, leading to two separate solutions based on these exponents.
This process transforms the differential equation task into finding roots of a polynomial, making solutions more approachable and enabling us to form a general solution in terms of arbitrary constants. Those solutions can later be adjusted using initial value conditions, as demonstrated, to find the specific functions describing our system.