/*! 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 4 \(y^{\prime \prime}+3 y^{\prime}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 4

\(y^{\prime \prime}+3 y^{\prime}-4 y=0\)

Short Answer

Expert verified
The general solution is \(y(t) = C_1 e^{t} + C_2 e^{-4t}\).

Step by step solution

01

Identify the Type of Differential Equation

Determine that the given equation is a second-order linear homogeneous differential equation with constant coefficients: \(y^{\text{''}} + 3y^{\text{'}} - 4y = 0\).
02

Formulate the Characteristic Equation

Convert the differential equation to its characteristic equation by substituting \(y = e^{rt}\). This leads to the characteristic equation: \(r^{2} + 3r - 4 = 0\).
03

Solve the Characteristic Equation

Solve the quadratic characteristic equation \(r^{2} + 3r - 4 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), and \(c = -4\).
04

Find the Roots of the Characteristic Equation

Calculate the roots using the quadratic formula: \(r = \frac{-3 \pm \sqrt{9 + 16}}{2}\), which simplifies to \(r = \frac{-3 \pm 5}{2}\). Thus, the roots are \(r = 1\) and \(r = -4\).
05

Write the General Solution

Since the roots \(r_1 = 1\) and \(r_2 = -4\) are real and distinct, the general solution to the differential equation is \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). Substituting the roots, the solution is \(y(t) = C_1 e^{t} + C_2 e^{-4t}\).

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 this type of differential equation, our goal is to determine a characteristic equation. This is done by recognizing that the given equation is a second-order linear homogeneous differential equation with constant coefficients: \(y'' + 3y' - 4y = 0\). To convert this into its characteristic equation, we substitute y with an exponential function: \(y = e^{rt}\). Differentiating this function gives us: \(y' = re^{rt}\) and \(y'' = r^2e^{rt}\). By substituting these into the original differential equation, we get:
\[r^2e^{rt} + 3re^{rt} - 4e^{rt} = 0\]
Factoring out \(e^{rt}\) (which is never zero), we then simplify to get the characteristic equation:
\[r^2 + 3r - 4 = 0\].
Quadratic Formula
To solve the characteristic equation \(r^2 + 3r - 4 = 0\), we use the quadratic formula. The quadratic formula is a fundamental method for solving equations of the form \(ax^2 + bx + c = 0\). It is given by:
\[r = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a}\]
Here, we identify coefficients: \(a = 1\), \(b = 3\), and \(c = -4\). Plugging these values into the quadratic formula, we get:
\[r = \frac{-3 \, \pm \, \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\]
which simplifies to:
\[r = \frac{-3 \, \pm \, \sqrt{9 + 16}}{2}\]
Then:
\[r = \frac{-3 \, \pm \, \sqrt{25}}{2}\]
Hence, we find the roots:
\[r_1 = \frac{-3 + 5}{2} = 1\]
\[r_2 = \frac{-3 - 5}{2} = -4\].
General Solution
After finding the roots of the characteristic equation, the next step is to construct the general solution of the differential equation. Since the roots \(r_1 = 1\) and \(r_2 = -4\) are real and distinct, the general solution is formed by two exponential functions:
\[y(t) = C_1 \cdot e^{r_1 t} + C_2 \cdot e^{r_2 t}\]
Substituting the values of \(r_1\) and \(r_2\), we get:
\[y(t) = C_1 \cdot e^{t} + C_2 \cdot e^{-4t}\]
where \(C_1\) and \(C_2\) are constants determined by initial conditions. This general solution helps to describe all possible behaviors of the system modeled by the differential equation.

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 initial value problem \(x^{3} y^{\prime \prime \prime}+\) \(9 x^{2} y^{\prime \prime}+44 x y^{\prime}+58 y=0, y(1)=2, y^{\prime}(1)=10\), \(y^{\prime \prime}(1)=-2\). Graph the solution on the interval \([0.2,1.8]\) and approximate all local minima and maxima of the solution on this interval.

\(3 x^{2} y^{\prime \prime}-4 x y^{\prime}+2 y=0\)

(a) If \(y=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots\), what are the first three nonzero terms of the power series for \(y^{2}\) ? (b) Use this series to find the first three nonzero terms in the power series solution about \(x=0\) to Vander-Pol's equation, $$ y^{\prime \prime}+\left(y^{2}-1\right) y^{\prime}+y=0 $$ if \(y(0)=0\) and \(y^{\prime}(0)=1\).

\(x^{2} y^{\prime \prime}+x y^{\prime}+(x-1) y=0\)

\(x(1-x) y^{\prime \prime}+\left(\frac{1}{2}-3 x\right) y^{\prime}-y=0\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

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.