Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 22

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \((D-2)^{2}\left(D^{2}+9\right) y=0\)

Short Answer

Expert verified
Linear fourth-order solution: \( y = \left(C1 + C2x\right)e^{2x} + C3 \cos(3x) + C4 \sin(3x) \).

Step by step solution

01

- Identify the Type of Differential Equation

First, recognize that \( (D-2)^{2}\big(D^{2}+9\big) y=0 \) is a linear homogeneous differential equation with constant coefficients. Here, \( D \) represents the differentiation operator \( \frac{d}{dx} \).
02

- Find the Characteristic Equation

Form the characteristic equation by replacing \( D \) with \( r \) in the given differential operator. The characteristic equation is: \i.e., \( (r-2)^{2}(r^{2}+9) = 0 \).
03

- Solve the Characteristic Equation

Solve \( (r-2)^{2}(r^{2}+9) = 0 \). This gives the repeated root \( r=2 \) with multiplicity 2, and the complex roots \( r= \pm 3i \).
04

- Write the General Solution

Use the roots to write the general solution of the differential equation. Since \( r=2 \) is a repeated root, and \( r= \pm 3i \) are complex roots, the general solution is: \y = \( (C1 + C2x)e^{2x} + C3 \cos(3x) + C4 \sin(3x) \) where \( C1, C2, C3, \, \, \, and \, C4 \) are arbitrary constants.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

headline of the respective core concept
In the exercise, we encountered a **linear homogeneous differential equation**. This is a differential equation in which every term depends on the unknown function and its derivatives, and importantly, the equation equals zero. Such equations are crucial in understanding many physical systems where the principle of superposition applies. Another key feature is that they have constant coefficients, meaning coefficients are numbers rather than functions of the independent variable. This simplifies solving them, as the differential operator can be treated algebraically using characteristic equations.
headline of the respective core concept
A related concept is the **characteristic equation**. To form this equation, you simply replace the differentiation operator \(\frac{d}{dx}\) with an algebraic variable, usually represented by \(r\). For our problem, the differential equation \((D-2)^{2}\big(D^{2}+9\big) y=0\) becomes the characteristic equation \((r-2)^{2}( r^2+9)=0\). This equation helps us find the roots, which are essential in identifying the general solution of the differential equation.
headline of the respective core concept
Next, we explored **roots and the general solution**. Solving the characteristic equation \((r-2)^{2}(r^2+9)=0\) gives us the roots \r=2\ (a repeated root), and \(\pm 3i\) (complex roots). These roots are crucial in constructing the general solution. For real repeated roots like \r=2\, we use forms such as \(C_1 e^{2x}+C_2xe^{2x}\). For complex roots like \pm 3i\, we use combinations of sine and cosine: \(C_3 \cos(3x)+ C_4 \sin(3x)\). The general solution combines these forms, resulting in \(y = (C1 + C2x)e^{2x} + C3 \cos(3x) + C4 \sin( 3x)\).
headline of the respective core concept
Finally, **constant coefficients** play a significant role. These coefficients in the differential equation do not vary with the independent variable, making the characteristic equation straightforward to handle. Mathematically, they ensure that the solution techniques are algebraic and facilitate the use of exponentials and trigonometric functions for solutions. Constant coefficients are foundational to these types of problems because they simplify finding and interpreting the solution.

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

Show that the thickness of the ice on a lake increases with the square root of the time in cold weather, making the following simplifying assumptions. Let the water temperature be a constant \(10^{\circ} \mathrm{C}\), the air temperature a constant \(-10^{\circ} \mathrm{C}_{7}\) and assume that at any given time the ice forms a slab of uniform thickness \(x\). The rate of formation of ice is proportional to the rate at which heat is transferred from the water to the air. Let \(t=0\) when \(x=0\).

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \((1+y) y^{\prime}=y\) \(y=1\) when \(x=1\)

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}+4 y^{\prime}+5 y=2 e^{-2 x} \cos x\)

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(x\left(y y^{\prime \prime}+y^{\prime 2}\right)=y y^{\prime}\) Hint: Iet \(u=1 x\).

Solve the following differential equations. \(D(D+5) y=0\)
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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.

Sign-up for free