Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 6

Find the general solution of the given equation by the method given. a. \(y^{\prime \prime}-3 y^{\prime}+2 y=10\). Method of Undetermined Coefficients. b. \(y^{\prime \prime}+y^{\prime}=3 x^{2}\). Variation of Parameters.

Short Answer

Expert verified
For part a, the general solution is \(y = c_{1} e^{x} + c_{2} e^{2x} + 5\). For part b, the general solution is \(y = c_{1} + c_{2} e^{-x} - x^2 e^{x} + 2x e^{x} - 2 e^{x}\).

Step by step solution

01

Part a: Find Homogeneous Solution

First, find the homogeneous solution \(y_{h}\) of the equation \(y^{\prime \prime}-3 y^{\prime}+2 y=0\). This is done by solving its characteristic equation \(r^2 - 3r + 2 = 0\) yielding \(r = 1, 2\) and hence \(y_{h} = c_{1} e^{x} + c_{2} e^{2x}\).
02

Part a: Find Particular Solution

Next step is to find the particular solution \(y_{p}\) using the method of Undetermined Coefficients. By inspection of the RHS of the given equation, we select \(y_p = A\) as our guess since RHS is a constant. Substituting \(y_p = A\) into the given equation gives \(A = 5\), so \(y_p = 5\).
03

Part a: Combine Solutions to Find General Solution

The general solution is the sum of the homogeneous solution and the particular solution, hence \(y = y_h + y_p = c_{1} e^{x} + c_{2} e^{2x} + 5\).
04

Part b: Find Homogeneous Solution

First, find the homogeneous solution of the equation \(y^{\prime \prime}+y^{\prime}= 0\). The characteristic equation for the homogeneous equation \(r^{2} + r = 0\) yields \(r = -1, 0\), which gives \(y_{h} = c_{1} + c_{2} e^{-x}\).
05

Part b: Use Variation of Parameters to Find Particular Solution

Use the method of variation of parameters to find the particular solution \(y_p\). The Wronskian of the homogeneous equations is given by \(W = -e^{-x}\) and the particular solution will be \(y_p = - \int x^2 e^x dx = -x^2 e^{x} + 2x e^{x} - 2 e^{x}\).
06

Part b: Combine Solutions to Find General Solution

Combine the homogeneous solution and the particular solution to get the general solution, hence \(y = y_h + y_p = c_{1} + c_{2} e^{-x} - x^2 e^{x} + 2x e^{x} - 2 e^{x}\).

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!

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

Consider the differential equation $$ \frac{d y}{d x}=\frac{x}{y}-\frac{x}{1+y} $$ a. Find the 1-parameter family of solutions (general solution) of this equation. b. Find the solution of this equation satisfying the initial condition \(y(0)=1 .\) Is this a member of the 1 -parameter family?

Numerically solve the nonlinear pendulum problem using the EulerCromer Method for a pendulum with length \(L=0.5 \mathrm{~m}\) using initial angles of \(\theta_{0}=10^{\circ}\), and \(\theta_{0}=70^{\circ} .\) In each case, run the routines long enough and with an appropriate \(h\) such that you can determine the period in each case. Compare your results with the linear pendulum period.

Find the solution of each initial value problem using the appropriate initial value Green's function. a. \(y^{\prime \prime}-3 y^{\prime}+2 y=20 e^{-2 x}, \quad y(0)=0, \quad y^{\prime}(0)=6\). b. \(y^{\prime \prime}+y=2 \sin 3 x, \quad y(0)=5, \quad y^{\prime}(0)=0\). c. \(y^{\prime \prime}+y=1+2 \cos x, \quad y(0)=2, \quad y^{\prime}(0)=0\). d. \(x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=3 x^{2}-x, \quad y(1)=\pi, \quad y^{\prime}(1)=0\).

Instead of assuming that \(c_{1}^{\prime} y_{1}+c_{2}^{\prime} y_{2}=0\) in the derivation of the solution using Variation of Parameters, assume that \(c_{1}^{\prime} y_{1}+c_{2}^{\prime} y_{2}=h(x)\) for an arbitrary function \(h(x)\) and show that one gets the same particular solution.

Consider an LRC circuit with \(L=1.00 \mathrm{H}, R=1.00 \times 10^{2} \Omega, C=\) \(1.00 \times 10^{-4} \mathrm{~F}\), and \(V=1.00 \times 10^{3} \mathrm{~V}\). Suppose that no charge is present and no current is flowing at time \(t=0\) when a battery of voltage \(V\) is inserted. Find the current and the charge on the capacitor as functions of time. Describe how the system behaves over time.
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