Problem 27 Solve the given initial-value pr... [FREE SOLUTION]

Chapter 3: Problem 27

Solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$ x^{2} y^{\prime \prime}+x y^{\prime}+y=0, y(1)=1, y^{\prime}(1)=2 $$

Short Answer

Expert verified

The solution is $ y(x) = x^{-1/2} ( \cos(\frac{\sqrt{3}}{2} \ln x) + 2\sqrt{3} \sin(\frac{\sqrt{3}}{2} \ln x) ) $.

Step by step solution

Identify the Type of Differential Equation

The given differential equation is $ x^2 y'' + x y' + y = 0 $. This is a second-order linear differential equation with variable coefficients. Recognizing this form, we can explore solution methods such as the Cauchy-Euler equation form.

Apply a Suitable Method

Given the structure $ x^2 y'' + x y' + y = 0 $, recognize it as a Cauchy-Euler equation, often solved using a substitution such as $ y = x^m $, leading to a characteristic equation.

Find the Characteristic Equation

Substitute $ y = x^m $ into the differential equation: $ m(m-1)x^m + mx^m + x^m = 0 $. Simplify to $ m^2 x^m = 0 $, giving the characteristic equation $ m^2 + m + 1 = 0 $.

Solve the Characteristic Equation

The characteristic equation is $ m^2 + m + 1 = 0 $. Solving this, use the quadratic formula: $ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 1 $, and $ c = 1 $. Calculate the discriminant: $ 1^2 - 4 \times 1 \times 1 = -3 $. Thus, roots are $ m = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i \sqrt{3}}{2} $.

Form the General Solution

For complex roots $ a \pm bi $, the general solution is $ y(x) = x^a (C_1 \cos(b \ln x) + C_2 \sin(b \ln x)) $. Here, $ a = -\frac{1}{2} $ and $ b = \frac{\sqrt{3}}{2} $, so the general solution is $ y(x) = x^{-1/2}(C_1 \cos(\frac{\sqrt{3}}{2} \ln x) + C_2 \sin(\frac{\sqrt{3}}{2} \ln x)) $.

Apply Initial Conditions

Use initial conditions $ y(1) = 1 $ and $ y'(1) = 2 $ to find $ C_1 $ and $ C_2 $. First, apply $ y(1) = 1 $: $ 1^{-1/2}(C_1 \cos(0) + C_2 \sin(0)) = 1 $ leading to $ C_1 = 1 $.

Use Derivative for Second Initial Condition

Differentiate the general solution for $ y' $, then use $ y'(1) = 2 $. Differentiating gives a general formula, evaluated at $ x = 1 $; solve for $ C_2 $ to satisfy the second initial condition, resulting in $ C_2 = 2 \sqrt{3} $.

Final Solution with Constants

Substitute $ C_1 = 1 $ and $ C_2 = 2\sqrt{3} $ back into the general solution: $ y(x) = x^{-1/2} ( \cos(\frac{\sqrt{3}}{2} \ln x) + 2\sqrt{3} \sin(\frac{\sqrt{3}}{2} \ln x) ) $.

Graph the Solution

Use a graphing utility like Desmos or Python's matplotlib to visually confirm the behavior of the solution curve. Input the derived equation and plot $ y(x) = x^{-1/2} ( \cos(\frac{\sqrt{3}}{2} \ln x) + 2\sqrt{3} \sin(\frac{\sqrt{3}}{2} \ln x) ) $ within a suitable range of $ x $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Second-Order Linear Differential Equations

Second-order linear differential equations are essential in various fields, including physics and engineering. These equations involve an unknown function and its derivatives, specifically the second derivative. The standard form of a second-order linear differential equation is $a(x)y'' + b(x)y' + c(x)y = 0$. In our exercise, the coefficients are functions of $x$, making it a Cauchy-Euler equation.

These equations have important properties that simplify finding their solutions:

Linearity: The equation is linear in terms of the function and its derivatives.
Order: The highest derivative is the second derivative.
Variable Coefficients: The coefficients $a(x), b(x), c(x)$ often follow specific patterns, simplifying analysis.

The solution of such equations typically involves finding a general solution that satisfies these conditions, before applying any initial values.

Defining the Initial Value Problem

An initial value problem (IVP) is a differential equation coupled with specified values at a particular point. The purpose of initial conditions is to ensure a unique solution among the infinitely many possible solutions of a differential equation.

In our case, the initial value problem is given by the equation $ x^{2} y^{\prime \prime}+x y^{\prime}+y=0 $ with conditions $ y(1)=1 $ and $ y^{\prime}(1)=2 $. Here鈥檚 why initial conditions matter:

They provide specific values for the solution and its derivative at $x = 1$.
They help determine the constants in the solution that appear after solving the differential equation itself.

Without initial conditions, it would be impossible to hon the infinitesimal spectrum of possible solutions to a specific one.

Exploring the Characteristic Equation

The characteristic equation is a crucial step in solving second-order linear differential equations. For Cauchy-Euler equations, this is derived by assuming a solution of the form $y = x^m$. Substituting this into the differential equation allows us to derive what is called the characteristic equation.

In this exercise, we derived the characteristic equation as $m^2 + m + 1 = 0$ by substituting $y = x^m$. Key points include:

These equations are typically polynomial equations in terms of $m$.
The roots of the characteristic polynomial reveal the form of the general solution.

The nature of these roots greatly affects our solution, part of which involves identifying whether they are real or complex.

Handling Complex Roots Solution

Complex roots in the characteristic equation signal a particular behavior in the solution of differential equations. When solving $m^2 + m + 1 = 0$, we found the roots $m = \frac{-1 \pm i\sqrt{3}}{2}$. These complex roots yield solutions that involve trigonometric functions along with the powers of $x$.

For this scenario, the general solution is structured using:

Real part $a$ translates into exponentiation: $x^a$.
Imaginary part $b$ introduces oscillatory behavior: $C_1\cos(b\ln x) + C_2\sin(b\ln x)$.

This method takes advantage of the imaginary parts to capture oscillations, typically in the form $\cos$ and $\sin$ terms, combined with a power function of $x$. In essence, this results in a solution that captures transitional behavior across domains of $x$.

91影视

Solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$ x^{2} y^{\prime \prime}+x y^{\prime}+y=0, y(1)=1, y^{\prime}(1)=2 $$

Short Answer

Step by step solution

Identify the Type of Differential Equation

Apply a Suitable Method

Find the Characteristic Equation

Solve the Characteristic Equation

Form the General Solution

Apply Initial Conditions

Use Derivative for Second Initial Condition

Final Solution with Constants

Graph the Solution

Key Concepts

Understanding Second-Order Linear Differential Equations

Defining the Initial Value Problem

Exploring the Characteristic Equation

Handling Complex Roots Solution

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