Josh Roberts

Report Problems

Report 1

Problem

Find a general solution. Check your answer by substitution.

$\displaystyle y''+4y'+(\pi ^{2}+4)y=0$

(1)

Solution

Note This is a homogenous linear 2nd order ordinary differential equation (ODE) with constant coefficients.

Trial Solution:

$\displaystyle y=e^{\lambda x}$

(2)

$\displaystyle y'=\lambda e^{\lambda x}$

(3)

$\displaystyle y''=\lambda ^{2}e^{\lambda x}$

(4)

Plugging (2),(3), and (4) into (1)

$\displaystyle (\lambda ^{2}+4\lambda +(\pi ^{2}+4))e^{\lambda x}=0$

(5)

The characteristic equation is

$\displaystyle \lambda ^{2}+4\lambda +(\pi ^{2}+4)=0$

(6)

Note The coefficients of this quadratic equation are

$\displaystyle a=1$

$\displaystyle b=4$

$\displaystyle c=\pi ^{2}+4$

The discriminant can be found as follows

$\displaystyle \Delta =b^{2}-4ac$

(7)

$\displaystyle \Delta =4^{2}-4(1)(\pi ^{2}+4)$

Note The discriminant is negative. Therefore, Eq (6) has complex roots of the form

$\displaystyle \lambda =-1/2a\pm i\omega$

(8)

where

$\displaystyle \omega ={\sqrt {c-b^{2}/4}}$

(9)

Now The general solution to the ODE is

$\displaystyle y(x)=e^{-bx/2}(Acos\omega x+Bsin\omega x)$

(10)

Plugging known quantities into (9) and (10)

$\displaystyle y(x)=e^{-2x}(Acos{\sqrt {(\pi ^{2}+4)-4}}\,\,x+Bsin{\sqrt {(\pi ^{2}+4)-4}}\,\,x)$

$\displaystyle y(x)=e^{-2x}(Acos\pi \,x+Bsin\pi \,x)$

Report 2

Problem

Given the two roots and the initial conditions:

$\displaystyle \lambda _{1}=-2\qquad \qquad \lambda _{2}=+5$

$\displaystyle y(0)=1\qquad \qquad y'(0)=0$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $r(x)=0$ .

Consider no excitation:

$\displaystyle r(x)=0$

Plot the solution.

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.

Solution

$\displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})$

(2.1.1)

Now Insert the given roots into (1)

$\displaystyle (\lambda -(-2))(\lambda -5)$

(2.1.2)

By the FOIL method, (2) gives the characteristic equation for the unknown L2-ODE-CC.

$\displaystyle \lambda ^{2}-3\lambda -10=0$

(2.1.3)

From this characteristic equation, (3), we can find the non-homogeneous L2-ODE-CC.

$\displaystyle y''-3y'-10=r(x)$

(2.1.4)

Now It is known from the given roots that the homogeneous L2-ODE-CC has the solution

$\displaystyle y_{H}(x)=C_{1}e^{-2x}+C_{2}e^{5x}$

(2.1.5)

Therefore, the overall solution is

$\displaystyle y(x)=y_{H}(x)+y_{P}(x)$

(2.1.6)

$\displaystyle y(x)=C_{1}e^{-2x}+C_{2}e^{5x}+y_{P}(x)$

(2.1.7)

where $y_{P}(x)$ is the particular solution.

Now Using the given initial conditions

$\displaystyle y(0)=1\qquad y'(0)=0$

$\displaystyle y(0)=C_{1}+C_{2}+y_{P}(0)=1$

(2.1.8)

Differentiating $y(x)$ and then inserting the initial condition.

$\displaystyle y'(x)=-2C_{1}e^{-2x}+5C_{2}e^{5x}+y_{P}'(x)$

(2.1.9)

$\displaystyle y'(0)=-2C_{1}+5C_{2}+y_{P}'(0)=0$

(2.1.10)

Now Multiplying (2.1.8) by 2 and adding it to (2.1.10)

$\displaystyle (2C_{1}+2_{C}2+2y_{P}(0)=2)+(-2C_{1}+5C_{2}+y_{P}'(0)=0)$

$\displaystyle C_{2}={\frac {2}{7}}-{\frac {2}{7}}y_{P}(0)-{\frac {1}{7}}y_{P}'(0)$

(2.1.11)

By inserting (2.1.11) into (2.1.8) and combining like terms $C_{1}$ is also found.

$\displaystyle C_{1}={\frac {5}{7}}-{\frac {5}{7}}y_{P}(0)+{\frac {1}{7}}y_{P}'(0)$

(2.1.12)

Now The overall solution $y(x)$ is given in terms of the initial conditions.

$\displaystyle y(x)={\frac {1}{7}}[(5-5y_{P}(0)+y_{P}'(0))e^{-2x}+(2-2y_{P}(0)-y_{P}'(0))e^{5x}]$

(2.1.13)

Now Considering no excitation

$\displaystyle r(x)=0$

Which causes

$\displaystyle y_{P}(x)=0\,\,,\,\,y_{P}'(x)=0$

So

$\displaystyle C_{1}={\frac {5}{7}}$

(2.1.14)

$\displaystyle C_{2}={\frac {2}{7}}$

(2.1.15)

The homogeneous solution to the L2-ODE-CC is therefore

$\displaystyle y_{H}(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}$

(2.1.16)

This solution will now be plotted using MATLAB.

File:HomogeneousL2ODE.jpg

Now Generate 3 non-standard and non-homogeneous L2-ODE-CC that satisfy the roots given in the problem statement.

$\displaystyle 2y''-6y'-20=r(x)$

(2.1.17)

$\displaystyle {\frac {3}{2}}y''-{\frac {9}{2}}y'-15y=r(x)$

(2.1.18)

$\displaystyle 3y''-9y'-30y=r(x)$

(2.1.19)

Exam 1 Material

Date: Tue, Feb 28, 2012

Study Guide

Sec 2.1 - Homogeneous L2-ODEs

A 2nd order ODE can be called linear if it can be written as

$\displaystyle y''+p(x)y'+q(x)y=r(x)$

(1)

Linear in y and its derivatives.
This equation is in standard form.

If $r(x)\equiv 0$ , then (1) reduces to

$\displaystyle y''+p(x)y'+q(x)y=0$

(2)

which is called a homogeneous L2-ODE.

If $r(x)\not \equiv 0$ , then (1) is called non homogeneous.

Superposition Principle (Linearity Principle)

For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution

of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

Initial Value Problems

Sec 2.2 - Homogeneous Linear ODEs with Constant Coefficients

These equations take the form

$\displaystyle y''+ay'+by=0$

(3)

Homogeneous solutions to these equations take the form

$\displaystyle y_{h}''+ay_{h}'+by_{h}=0$

(4)

and can be solved by the method of trial solution, also known as the method of undetermined coefficients.
The trial solution is

$\displaystyle y=e^{\lambda x}$

(5)

By inserting (5) and its 1st and 2nd derivatives into (4), one can obtain the characteristic equation

$\displaystyle \lambda ^{2}+a\lambda +b=0$

(6)

Finding the roots to the characteristic equation will allow for the homogeneous solution to be found. We know that for a quadratic equation
such as (6) that there are 3 kinds of roots, depending on the sign of the discriminant $a^{2}-4b$ ,namely.