/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1083 Solve the equation: \(\left(d^{2... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 41: Problem 1083

Solve the equation: \(\left(d^{2} y / d x^{2}\right)+2(d y / d x)+y=0\)

Short Answer

Expert verified
The general solution to the given second-order linear homogeneous ordinary differential equation \(\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0\) is \(y(x) = (C_1 + C_2x)e^{-x}\), where \(C_1\) and \(C_2\) are arbitrary constants determined by the initial conditions of the problem.

Step by step solution

01

Write down the given equation

The given second-order linear ordinary differential equation is: \(\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0\)
02

Determine the characteristic equation

The characteristic equation of this differential equation is obtained by substituting \(y = e^{rx}\) and its derivatives into the given equation. The characteristic equation is of the form: \(ar^2 + br + c = 0\) Substituting the coefficients from our problem, we get: \(r^2 + 2r + 1 = 0\)
03

Solve the characteristic equation

Next, we need to find the roots of the characteristic equation. The above equation is a quadratic equation on r: \(r^2 + 2r + 1 = 0\) This equation has a discriminant given by \(\Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 1 = 0\). Since the discriminant is zero, the quadratic equation has only one real root (a repeated root). We can use the quadratic formula to solve for the root: \(r = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-2 \pm \sqrt{0}}{2 \cdot 1} = -1\) Now we know that \(r = -1\) is a repeated root.
04

Write the general solution

For a second-order linear homogeneous ordinary differential equation with constant coefficients and a repeated root, the general solution is given by: \(y(x) = (C_1 + C_2x)e^{rx}\) Where \(C_1\) and \(C_2\) are arbitrary constants determined by the initial conditions of the problem. For our problem, the repeated root is \(r = -1\), so the general solution is: \(y(x) = (C_1 + C_2x)e^{-x}\) This is the general solution to the given differential equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Characteristic Equation
When solving a second-order differential equation, the characteristic equation is a vital tool. It's formulated by assuming solutions of the form \(y = e^{rx}\) and substituting back into the differential equation. This leads to a polynomial equation, typically a quadratic:
  • Standard form: \(ar^2 + br + c = 0\)
For our specific equation, \( \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0\), substituting leads to:
  • Characteristic equation: \(r^2 + 2r + 1 = 0\)
This equation is crucial as its roots determine the form of the general solution to the differential equation. The roots can be real and distinct, real and repeated, or complex conjugates.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is dependent on the function or its derivatives. This forms a key part of second-order differential equations, especially when dealing with linear forms.

For example, our equation \( \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0\) is homogeneous because:
  • There are no isolated terms 鈥� each involves \(y\) or its derivatives.
  • The right side of the equation is zero, implying that the equation is set to zero.
These characteristics mean the solutions can include arbitrary constants, and the superposition principle applies 鈥� any linear combination of solutions is also a solution.
Second-Order Differential Equation
A second-order differential equation involves the second derivative of the function. These types of equations model physical processes like oscillations and waves.

For our exercise:
  • It is linear, as variables are not multiplied or raised to powers other than 1.
  • The general form is \(ay'' + by' + cy = 0\).
To solve, find the characteristic equation and solve for its roots. If roots are repeated as in our case with \(r = -1\), the general solution becomes:
  • \(y(x) = (C_1 + C_2x)e^{-x}\)
The inclusion of \(C_2x\) accounts for the repeated root, ensuring the solution remains general and valid for all conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the equation: \(y^{n}-2 y^{\prime}-3 y=4 e^{3 x}\).

Solve the differential equation \((\mathrm{dy} / \mathrm{dx})=2 \mathrm{xy}\) to obtain a general solution. Also find the particular solution if \(\mathrm{y}=5\) when \(\mathrm{x}=0\).

Solve the differential equation: \(\left(\mathrm{xy}^{2}-\mathrm{x}\right) \mathrm{dx}+\left(\mathrm{x}^{2} \mathrm{y}+\mathrm{y}\right) \mathrm{dy}=0\)

Find the general solution for: \((\mathrm{dy} / \mathrm{dx})=[(\mathrm{x}-2 \mathrm{y}) /(2 \mathrm{x}-\mathrm{y})]\).

Resolve the given operator \(\varphi(\mathrm{D})\) into linear factors, and arrange these factors in all possible orders. Operate upon the given function \(u\) with \(\varphi(D)\), and verify the fact that the results obtained are the same. \(\varphi(D)=2 D^{2}+D-6 ; u=\sin x+3 e^{x}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.