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Chapter 31: Problem 10

Solve the given differential equations. $$\frac{d^{2} y}{d x^{2}}+y=0$$

Short Answer

Expert verified
The general solution is \( y(x) = C_1 \cos(x) + C_2 \sin(x) \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation \( \frac{d^{2} y}{d x^{2}} + y = 0 \) is a linear homogeneous second-order differential equation with constant coefficients.
02

Write the Characteristic Equation

To solve the differential equation, we start by writing the characteristic equation. For the equation \( \frac{d^{2} y}{d x^{2}} + y = 0 \), assume \( y = e^{rx} \). Then \( \frac{d^{2} y}{d x^{2}} = r^2 e^{rx} \), so \( r^2 e^{rx} + e^{rx} = 0 \). Dividing through by \( e^{rx} \), we get the characteristic equation: \( r^2 + 1 = 0 \).
03

Solve the Characteristic Equation

Solve the characteristic equation \( r^2 + 1 = 0 \). This equation can be rewritten as \( r^2 = -1 \). Solving for \( r \), we find the roots \( r = i \) and \( r = -i \), where \( i \) is the imaginary unit.
04

Write the General Solution

Since the roots \( r = i \) and \( r = -i \) are complex, the general solution of the differential equation is given by: \[ y(x) = C_1 \cos(x) + C_2 \sin(x) \]where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

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Key Concepts

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

Characteristic Equation
When faced with a second-order differential equation like \( \frac{d^{2} y}{d x^{2}} + y = 0 \), it's essential to translate it into a simpler algebra form called the characteristic equation. The primary goal is to transform the differential equation into a more manageable polynomial equation. Here's how it works:
  • Assume a solution of the form \( y = e^{rx} \). This assumption is useful because the exponential function retains its form under differentiation, making calculations straightforward.
  • When you substitute \( y = e^{rx} \) into the differential equation, you get a new equation: \( r^2 e^{rx} + e^{rx} = 0 \).
  • Cancel out \( e^{rx} \) (as long as \( e^{rx} eq 0 \)), leading to the characteristic equation \( r^2 + 1 = 0 \).
This process simplifies the problem by converting it into an equation involving only \( r \), making it easier to find the solution.
Complex Roots
When we solve the characteristic equation \( r^2 + 1 = 0 \), we end up with the equation \( r^2 = -1 \). This presents us with complex numbers, which have a real and an imaginary part:
  • The solutions are the complex roots \( r = i \) and \( r = -i \).
  • The imaginary unit \( i \) is defined such that \( i^2 = -1 \).
Complex roots bring exciting properties to the solutions of differential equations. When the characteristic equation yields complex roots of the form \( r = a + bi \) and \( r = a - bi \), the solutions take on a trigonometric form:\[y(x) = e^{ax} (C_1 \cos(bx) + C_2 \sin(bx))\]In our specific case, since the real part \( a \) is zero, the solutions are purely sinusoidal, showing up as cosine and sine functions.
General Solution
The final step is composing the general solution of the differential equation by leveraging the complex roots derived from the characteristic equation. When complex roots \( r = a + bi \) and \( r = a - bi \) are found, as in \( r = i \) and \( r = -i \), they give a solution shaped by the functions \( \cos \) and \( \sin \):\[y(x) = C_1 \cos(x) + C_2 \sin(x)\]Here:
  • \( C_1 \) and \( C_2 \) are arbitrary constants that can be determined if initial conditions or boundary conditions are provided.
  • The cosine and sine functions reflect the behavior of systems that exhibit oscillatory patterns, like springs or pendulums, under specific conditions.
This general solution covers all possible solutions to the equation, providing flexibility to match a wide range of physical situations. Initial conditions allow us to solve for \( C_1 \) and \( C_2 \), tailoring the solution to specific circumstances.

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