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Chapter 24: Problem 65

Solve the equation \(y^{\prime \prime}+4 y=0\) (a) near the ordinary point \(x=0\).

Short Answer

Expert verified
The short answer for the given differential equation \(y'' + 4y = 0\) near the ordinary point \(x = 0\) is \(y(x) = c_1 \cos(2x) + c_2 \sin(2x)\), where \(c_1\) and \(c_2\) are arbitrary constants.

Step by step solution

01

Identify the Differential Equation Type

We have a second-order linear differential equation with constant coefficients given by: \(y'' + 4y = 0\)
02

Use the Characteristic Equation Method

We will use the characteristic equation method to solve the given equation. To find the characteristic equation, we replace \(y''\) with \(m^2\) and y with 1. The characteristic equation is: \(m^2 + 4 = 0\)
03

Solve the Characteristic Equation

Now, we will solve the characteristic equation for the values of m: \(m^2 + 4 = 0\) Subtracting 4 from both sides: \(m^2 = -4\) Taking the square root of both sides: \(m = \pm 2i\)
04

Write the General Solution

As we have two imaginary roots for the characteristic equation (\(m = \pm 2i\)), the general solution for the given differential equation is of the form: \(y(x) = c_1 \cos(2x) + c_2 \sin(2x)\) Here, \(c_1\) and \(c_2\) are arbitrary constants. This is the solution for the given second-order linear differential equation near the ordinary point \(x=0\).

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Key Concepts

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

Characteristic Equation
A characteristic equation is derived from a differential equation and helps us identify the type of solutions we might expect. For second-order linear differential equations like \(y'' + 4y = 0\), the form \(y'' = m^2\) allows us to substitute and simplify. By replacing \(y''\) with \(m^2\) and \(y\) with 1, we reach the characteristic equation \(m^2 + 4 = 0\). This transformation is crucial because it turns a differential equation into an easier algebraic equation.
  • The characteristic equation is quadratic in nature.
  • Solving it helps determine the roots which influence the type of solution to the differential equation.

If you struggle with second-order equations, remember that forming a characteristic equation allows you to use algebraic methods to solve a problem involving derivatives.
Imaginary Roots
When we compute the characteristic equation \(m^2 + 4 = 0\) and solve for \(m\), we find that \(m^2 = -4\). Solving this gives us \(m = \pm 2i\). This indicates two imaginary roots, which affect the solution of the differential equation. An important point to note here is:
  • Imaginary roots typically appear in conjugate pairs.
  • They often lead to oscillatory solutions in differential equations.

These imaginary numbers represent complex solutions but can be nicely expressed in terms of real functions鈥攕pecifically, sine and cosine. Recognizing these patterns is critical as they shape the general form of solutions.
General Solution
For a second-order linear differential equation with constant coefficients yielding imaginary roots, like \(m = \pm 2i\), the general solution takes a specific form. In this case, it is \(y(x) = c_1 \cos(2x) + c_2 \sin(2x)\). This represents the harmonious coexistence of sine and cosine functions, as imaginary roots correspond to oscillatory behavior.
  • The constants \(c_1\) and \(c_2\) depend on the initial conditions or boundary values.
  • This method enables us to predict the behavior around the ordinary point \(x=0\).

The beauty of the general solution is that it describes all possible solutions under different conditions, driven by the nature of the differential equation's roots.
Ordinary Point
An ordinary point in the context of differential equations is a value of \(x\) where the equation behaves nicely, and solutions exist without issues, as with \(x=0\) in our exercise. It means that coefficients of derivatives in the equation are well-defined and finite at that point. This is crucial in both finding and applying solutions:
  • Only around ordinary points can every method smoothly apply, like the characteristic equation method.
  • The absence of singularities or irregularities makes computation simpler and intuitive.

Whenever tackling a differential equation, identifying ordinary points helps to confidently proceed with known techniques and ensures the reliability and validity of the result.

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Most popular questions from this chapter

Use the Taylor Series method to solve \(y^{\prime \prime}+x y^{\prime}+(2 x-1) y=0\) with \(\quad y(-1)=2, y^{\prime}(-1)=-2\)

Find as a power series the integral of the differential equation $$ \mathrm{y}^{\prime \prime}=\mathrm{e}^{\mathrm{x}} \mathrm{y}^{2}-\left(\mathrm{y}^{\prime}\right)^{2} $$ (a) which fulfills the conditions $$ \mathrm{y}(0)=0, \quad \mathrm{y}^{\prime}(0)=1 $$

Find, up to terms of the fourth degree, the integral of the differential equation, \(u^{\prime \prime} \cos x+u^{\prime} \sin x+(\cos x+\sin x) u=0\) for which \(\mathrm{u}(\pi / 4)=2, \mathrm{u}^{\prime}(\pi / 4)=0\)

Determine, in a Taylor series expansion, the solution of the initial value problem \(x^{\prime}=-x^{2}, \quad\) (a) \(x=-(1 / 2)\) when \(s=0\)

Solve Gauss' hypergeometric equation $$ x(1-x) y^{\prime \prime}+[c-(a+b+1) x] y^{\prime}-a b y=0 $$
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