/*! 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 31 Solve. \(2 y^{\prime \prime}+2... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 31

Solve. \(2 y^{\prime \prime}+2 y^{\prime}-5 y=0\)

Short Answer

Expert verified
\[y(t) = C_1 e^{\left(\frac{-1 + \sqrt{11}}{2} t\right)} + C_2 e^{\left(\frac{-1 - \sqrt{11}}{2} t\right)}\]

Step by step solution

01

Write the characteristic equation

Start by converting the differential equation into its characteristic equation. Given the differential equation is \[2y'' + 2y' - 5y = 0\], the characteristic equation is formed by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1:\[2r^2 + 2r - 5 = 0\]
02

Solve the characteristic equation using the quadratic formula

The quadratic formula is given by \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] For the characteristic equation \(2r^2 + 2r - 5 = 0\), identify where a = 2, b = 2, and c = -5. Substitute these values in to get: \[r = \frac{-2 \pm \sqrt{(2)^2 - 4(2)(-5)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 40}}{4} = \frac{-2 \pm \sqrt{44}}{4} = \frac{-2 \pm 2\sqrt{11}}{4} = \frac{-1 \pm \sqrt{11}}{2}\]
03

Write the general solution of the differential equation

Since the roots of the characteristic equation are real and distinct, the general solution is given by: \[y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\] where \(r_1 = \frac{-1 + \sqrt{11}}{2}\) and \(r_2 = \frac{-1 - \sqrt{11}}{2}\).Thus, the general solution is: \[y(t) = C_1 e^{\left(\frac{-1 + \sqrt{11}}{2} t\right)} + C_2 e^{\left(\frac{-1 - \sqrt{11}}{2} t\right)}\]

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
In the field of differential equations, a characteristic equation is a crucial concept. For any second-order linear differential equation like:
2y'' + 2y' - 5y = 0
We derive the characteristic equation by substituting:
  • y'' with r^2
  • y' with r
  • y with 1
This transforms the differential equation into a polynomial form. The resulting characteristic equation for our example is: 2r^2 + 2r - 5 = 0
This equation helps us find the roots, which in turn allows us to determine the general solution for the differential equation.
quadratic formula
To solve the characteristic equation, often we rely on the quadratic formula. For a quadratic equation of the form:
ax^2 + bx + c = 0
The quadratic formula is:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Applying this to our characteristic equation 2r^2 + 2r - 5 = 0, identify the coefficients: a = 2, b = 2, and c = -5, and substitute them into the quadratic formula. This results in:
\[ r = \frac{-2 \pm \sqrt{(2)^2 - 4(2)(-5)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 40}}{4} = \frac{-2 \pm \sqrt{44}}{4} = \frac{-1 \pm \sqrt{11}}{2} \]
This gives us two distinct roots:
  • \(r_1 = \frac{-1 + \sqrt{11}}{2}\)
  • \(r_2 = \frac{-1 - \sqrt{11}}{2}\)
general solution
With the roots from the characteristic equation, we can now form the general solution of the differential equation. For real and distinct roots (as we have here):
  • \( r_1 = \frac{-1 + \sqrt{11}}{2} \)
  • \( r_2 = \frac{-1 - \sqrt{11}}{2} \)
The general solution is written as: \[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]
Plugging in our roots, we get: \[ y(t) = C_1 e^{\left( \frac{-1 + \sqrt{11}}{2} t \right)} + C_2 e^{\left( \frac{-1 - \sqrt{11}}{2} t \right)} \]
Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions. This solution represents the complete answer to the original differential equation, encompassing all possible solutions based on different initial values.

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 the equilibrium points and assess the stability of each. \(x^{\prime}=x-e^{y}, y^{\prime}=2 \ln x+y-6\)

Let \(x\) and \(y\) represent the populations (in thousands) of prey and predators that share a habitat. For the given system of differential equations, find and classify the equilibrium points. \(x^{\prime}(t)=0.5 x-0.4 x y, y^{\prime}(t)=-0.4 y+0.2 x y\)

Find the equilibrium points and assess the stability of each. \(x^{\prime}=(2-x) \sqrt{y}, y^{\prime}=x y-8\)

Solve the initial-value problem. \(y^{\prime \prime \prime}+3 y^{\prime \prime}=18 x-18\) \(y(0)=3, y^{\prime}(0)=-1, y^{\prime \prime}(0)=10\)

Solve the initial-value problem. \(x^{\prime}=2 x-y, y^{\prime}=2 x+5 y, x(0)=3, y(0)=-5\)
See all solutions

Recommended explanations on Biology Textbooks

Genetic Information

Read Explanation

Biological Structures

Read Explanation

Chemistry of Life

Read Explanation

Bioenergetics

Read Explanation

Communicable Diseases

Read Explanation

Ecology

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.