/*! 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 13 Find a general solution. Check y... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 13

Find a general solution. Check your answer by substitution. $$y^{\prime \prime}-144 y=0$$

Short Answer

Expert verified
The general solution is \( y(x) = C_1 e^{12x} + C_2 e^{-12x} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \( y'' - 144y = 0 \).
02

Characteristic Equation

To solve this, we first find the characteristic equation by replacing \( y'' \) with \( r^2 \) and \( y \) with \( r \). This gives:\[ r^2 - 144 = 0 \] which can be rewritten as \( r^2 = 144 \).
03

Solve the Characteristic Equation

Solve \( r^2 = 144 \) to find the roots: \( r = \pm 12 \). So, the roots are \( r_1 = 12 \) and \( r_2 = -12 \). These are distinct real roots.
04

Write the General Solution

The general solution of a second-order linear homogeneous differential equation with real and distinct roots \( r_1 \) and \( r_2 \) is:\[ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]Substitute the found roots:\[ y(x) = C_1 e^{12x} + C_2 e^{-12x} \]
05

Substitute Back to Verify the Solution

To verify, calculate the first and second derivatives:\[ y'(x) = 12C_1 e^{12x} - 12C_2 e^{-12x} \]\[ y''(x) = 144C_1 e^{12x} + 144C_2 e^{-12x} \]Substitute \( y(x) \) and \( y''(x) \) back into the original equation:\[ y'' - 144y = (144C_1 e^{12x} + 144C_2 e^{-12x}) - 144(C_1 e^{12x} + C_2 e^{-12x}) = 0 \]The equation holds true, confirming the solution.

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 tackling a second-order linear homogeneous differential equation like \( y'' - 144y = 0 \), the first step in finding the solution is to form the characteristic equation. This equation effectively transforms the differential equation into an algebraic problem by substituting derivatives with powers of a variable, typically \( r \).
  • Replace \( y'' \) with \( r^2 \).
  • Replace \( y \) with \( r \).
The differential equation \( y'' - 144y = 0 \) becomes the characteristic equation \( r^2 - 144 = 0 \). Solving this characteristic equation simplifies the process of finding the roots which will determine the form of the solution to the differential equation.
Linear Homogeneous Differential Equation
A linear homogeneous differential equation is characterized by two key features: it is linear, and it is homogeneous.
  • Linear: Each term is a multiple of the function \( y \) or its derivatives. There are no products or powers of \( y \) or its derivatives.
  • Homogeneous: The equation equals zero when there are no forcing terms or functions independent of \( y \).
Using our differential equation \( y'' - 144y = 0 \) as an example, it fits these criteria perfectly. The terms involve only \( y \) and \( y'' \), and the equation is set to zero, indicating its homogeneity. Solving such equations typically involves finding the characteristic equation, as seen in the solution steps.
Distinct Real Roots
Finding roots is a critical part of solving a second-order linear homogeneous differential equation. In our equation's characteristic equation \( r^2 - 144 = 0 \), we are solving for \( r \). This results in roots \( r = 12 \) and \( r = -12 \), which are distinct and real.
  • Distinct: The roots are different from one another.
  • Real: The roots are not complex or imaginary numbers.
These properties significantly influence the general solution format. For distinct real roots like \( r_1 = 12 \) and \( r_2 = -12 \), the solution of the equation is built from exponential functions. The general solution takes the form: \[ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \] Here, \( C_1 \) and \( C_2 \) are constants determined by boundary or initial 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

Find a general solution. Check your answer by substitution. $$100 y^{\prime \prime}+20 y^{\prime}-99 y=0$$

Find a general solution. Check your answer by substitution. $$y^{\prime \prime}-2 y^{\prime}-5.25 y=0$$

Apply the given operator to the given functions. (Show all) steps in detail $$8 D^{2}+2 D-l ; \cosh \frac{1}{2} x, \quad \sinh ^{2} x, \quad e^{x / 2}$$

Find a (real) general solutlon. Which rule are you wing? (Show each step of your calcolation.) $$y^{\prime \prime}+9 y=\cos x+\frac{1}{3} \cos 3 x$$

Find a general solution. Check your answer by substitution. $$y^{\prime \prime}+2.6 y^{\prime}+1.69 y=0$$
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Turning Points in Physics

Read Explanation

Fields in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Famous Physicists

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.