User:Egm6321.f09.team7/HW4

 Do remember to log in with your wiki username before editing so that your IP address is not recorded; see the history page. It would be helpful to give a title to each problem; see Team 3. See also the links to the lecture transparencies where HW problems were assigned in Team 1. Do create an internal link to each HW report subpage in your team wiki page to allow easy access to your HW reports.

See further comments below.

Egm6321.f09 13:18, 24 October 2009 (UTC)

See further comments in google doc for Team 7. Egm6321.f09 21:03, 23 November 2009 (UTC)

HW4: Problem 1
Problem Statment and Given:  p.19-1 For the Legendre equation 1 on p.14-2 with $$n=0$$, the homogeneous solution $$u_1(x)=1$$. Use reduction of order method 2 (undetermined factor) to find $$u_2(x)$$, the second homogeneous solution.

Solution:  In general for any given L2-ODE-VC Non Homogenous: $$y'' + a(x)y' + b(x)y = f(x)$$ Apply the following transformation where u1(x) is a homogenous solution: $$y(x) = U(x)u_1(x)$$ $$y(x) = U(x)u_1(x)$$ $$y' = U'u_1 + Uu_1'$$ $$y = Uu_1 + 2U'u_1' + Uu_1''$$ $$Uu_1 + U'(2u_1'+ u_1 a(x)) + U(u_1+ u_1' a(x) + u_1 b(x)) = f(x)$$
 * {| style="width:100%" border="0" align="left"

U''u_1 + U'(2u_1'+ u_1 a(x)) = f(x)$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

For this problem: $$u_1(x) = 1$$ $$u_1'(x) = 0$$ $$f(x) = 0$$ $$a(x) = -2x/(1-x^2)$$

We obtain the following ODE: $$U''-2x/(1-x^2) U' = 0$$

Multiply by $$(1-x^2)$$ and let $$U' = Z$$: $$(1-x^2) Z'-2xZ = 0$$ $$\frac{d((1-x^2) Z)}{dx} = 0$$ $$(1-x^2) Z= C_1$$ $$U'= \frac{C_1}{1-x^2}$$ $$U= \int \frac{C_1}{(1+x)(1-x)} dx$$ $$U= \frac{C_1}{2}[ \int ( \frac{1}{1+x} + \frac{1}{1-x} ) dx]$$

$$U= \frac{C_1}{2}[ ln( (1+x)(1-x) ) + k]$$ We obtain the following Solution:


 * {| style="width:100%" border="0" align="left"

y(x) = U(x)= ln( (1-x^2)^\frac{C_1}{2} ) + C_2$$ Apply  $$y(x) = U(x)u_1(x)$$ You get $$y(x) = U(x)$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

 But you have not identified the second homogeneous solution $$\displaystyle u_2(x)$$ yet. See Team 1. Egm6321.f09 16:59, 10 November 2009 (UTC)

HW4: Problem 2
Problem Statement and Given: 

Solve problem 1.1b page 28 from King book, see slide 19-1 in the notes.

Show that the function $$ u_{1}$$ is a solution of the differential equation given below. Use the reduction of order method to find the second independent solution, $$ u_{2}$$.

$$ xy''+2y'+xy=0$$

$$ u_{1}=x^{-1}sin(x)$$

Solution:

Find the first and second derivative of $$ u_{1}$$ and plug them into the equation to check if the equation is equal to zero.

$$ u_{1}'=x^{-1}cos(x)-x^{-2}sin(x)$$

$$ u_{1}''=-x^{-1}sin(x)-x^{-2}cos(x)-x^{-2}cos(x)+2x^{-3}sin(x)$$

$$ xu_{1}''+2u_{1}'+xu_{1}=0$$

$$ 2x^{-2}sin(x)-sin(x)-2x^{-1}cos(x)+2x^{-1}-2x^{-2}sin(x)+sin(x)=0$$

$$0=0$$

The function $$ u_{1}$$ is a solution of the differential equation given.

Now, we use the reduction of order method to find the second independent solution, $$ u_{2}$$.

Put the equation in the form:

$$ y''+a_{1}(x)y'+a_{0}y=0$$

$$ y''+\frac{2}{x}y'+y=0$$

From the notes in class, slides 17-1 to 17-4, we get the formula:

$$ u_{2}=u_{1}\int_{}^{x} \frac{1}{(u_{1}(t))^{2}}exp(-a_{1}(t))dt$$ Where: $$ -a_{1}(t)=-\int_{}^{t} a(s)ds $$

$$ -a_{1}(t)=-\int_{}^{t} \frac{2}{s}ds = -2 ln (t)=ln(t)^{-2}$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{(t^{-1}sin(t))^{2}}exp(ln t^{-2})dt$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{t^{-2}sin^{2}(t)}t^{-2}dt$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{ sin^{2}(t)} dt$$

$$ u_{2}=x^{-1}sin(x)(-cot(x)) $$

$$ u_{2}=-x^{-1}cos(x) $$

HW4: Problem 3
Problem Statement and Given:  $$ xy''+2y'+xy=x$$ from problem 2: $$ y''+\frac{2}{x}y'+1y=x$$ $$ a_1=\frac{2}{x}$$ $$ a_0=1$$ $$ f(x)=x$$ $$u_1(x)=\frac{sin(x)}{x}$$ $$u_2(x)=-\frac{cos(x)}{x}$$

Solution: The solution is in the form $$y(x)=c_1(x)u_1(x)+c_2(x)u_2(x)$$ The terms $$y(x)=c_1(x)$$ and $$y(x)=c_2(x)$$ can be found by using the Wronskian matrix.

$$ u'_2(x)=\frac{cos(x)}{x^2}+\frac{sin(x)}{x}$$ $$ u'_1(x)=\frac{cos(x)}{x}-\frac{sin(x)}{x^2}$$

$$ W={\left[\begin{matrix}u_1(x) & u_2(x) \\ u'_1(x) & u'_2(x)\end{matrix}\right]}$$ $$detW=u_1(x)u'_2(x)-u'_1(x)u_2(x)=(\frac{sin(x)}{x})(\frac{cos(x)}{x^2}+\frac{sin(x)}{x})-(\frac{cos(x)}{x}-\frac{sin(x)}{x^2})(-\frac{cos(x)}{x})$$

$$detW=\frac{cos(x)sin(x)}{x^3}+\frac{sin^2(x)}{x^2}+\frac{cos^2(x)}{x^2}-\frac{cos(x)sin(x)}{x^3}=\frac{1}{x^2}(sin^2(x)+cos^2(x))=\frac{1}{x^2}$$

$$ W^{-1}=\frac{1}{detW}{\left[\begin{matrix}u_1(x) & -u_2(x) \\ -u'_1(x) & u'_2(x)\end{matrix}\right]}$$

$$ W^{-1}=\frac{1}{\frac{1}{x^2}}{\left[\begin{matrix}\frac{sin(x)}{x} & \frac{cos(x)}{x} \\ -(\frac{cos(x)}{x}-\frac{sin(x)}{x^2}) & (\frac{cos(x)}{x^2}+\frac{sin(x)}{x})\end{matrix}\right]}$$

$$ {\left[\begin{matrix}c'_1 \\ c'_2 \end{matrix}\right]}= W^{-1}{\left[\begin{matrix}0 \\ f(x)\end{matrix}\right]}$$

$$ {\left[\begin{matrix}c'_1 \\ c'_2 \end{matrix}\right]}= x^2{\left[\begin{matrix}\frac{sin(x)}{x} & \frac{cos(x)}{x} \\ -(\frac{cos(x)}{x}-\frac{sin(x)}{x^2}) & (\frac{cos(x)}{x^2}+\frac{sin(x)}{x})\end{matrix}\right]} {\left[\begin{matrix}0 \\ f(x)\end{matrix}\right]}$$

$$ {\left[\begin{matrix}c'_1 \\ c'_2 \end{matrix}\right]}= {\left[\begin{matrix}x^2cos(x) \\ x^2sin(x) \end{matrix}\right]}$$

Integrate to find $$ c'_1$$ and $$c'_2 $$

$$c_1=\int_{}^{x} s^2cos(s)ds=(x^2-2)sin(x)+2xcos(x)+C_1$$ $$c_2=\int_{}^{x} s^2sin(s)ds=2xsin(x)-(x^2-2)cos(x)+C_2$$

Plug this into the general solution: $$y(x)=[(x^2-2)sin(x)+2xcos(x)+C_1][\frac{sin(x)}{x}]-[2xsin(x)-(x^2-2)cos(x)+C_2][\frac{cos(x)}{x}]$$

$$y(x)=\frac{(x^2-2)sin^2(x)}{x}+\frac{2xcos(x)sin(x)}{x}+\frac{C_1sin(x)}{x}-\frac{2xcos(x)sin(x)}{x}+\frac{(x^2-2)cos^2(x)}{x}-\frac{(C_2cos(x)}{x}$$

$$y(x)=\frac{x^2-2}{x}(cos^2(x)+sin^2(x))+\frac{C_1sin(x)}{x}-\frac{C_2cos(x)}{x}$$

$$y(x)=\frac{x^2-2}{x}+\frac{C_1sin(x)}{x}-\frac{C_2cos(x)}{x}$$

Method 2:  Assume the general solution is $$y(x)=U(x)u_1(x)$$ From the lecture notes let $$Z(x)=U'(x)$$ so $$u_1(x)Z'+[a_1(x)u_1(x)+2u'_1(x)]Z=f(x)$$ $$(\frac{sin(x)}{x})Z'+[(\frac{2sin(x)}{x^2})+2(\frac{cos(x)}{x}-\frac{sin(x)}{x^2})]Z=x$$ $$(\frac{sin(x)}{x})Z'+(\frac{2 cos(x)}{x})Z=x$$ $$Z'+(\frac{2 cos(x)}{sin(x)})Z=\frac{x^2}{sin(x)}$$

$$Z(x) = csc(x) (c_1 csc(x)-(x^2-2) cot(x)+2 x)$$

$$U(x)=\int_{}^{x}Z(s)ds$$ $$U(x)=csc(x) (-c_1 cos(x)+x^2-2)+c_2$$ Plug $$U(x)$$ into general solution to get: $$y(x)=(csc(x) (-c_1 cos(x)+x^2-2)+c_2)(\frac{sin(x)}{x})$$ $$y(x)=\frac{x^2-2}{x}+\frac{C_1sin(x)}{x}-\frac{C_2cos(x)}{x}$$

HW4: Problem 4
Problem Statement: 

Describe in words, step by step, the method of attack given only the homogeneous or non-homogeneous L2. ODE. VC.

Solution:

Answer: Try reduction of method order 0, then order 1, and later integral factor method. If they do not work, use trial solution method. If both of the roots cannot be determine we can use the undetermined factor method to find u2 knowing u1. For the non-homogenous case solve the homogeneous solution and then solve the particular solution using variations of parameters.

 See my comments on the same problem in Team 3. See also the nice flow chart in Team 1. Egm6321.f09 13:24, 24 October 2009 (UTC)

HW4: Problem 5
Problem Statement:  1.3a in King Book. Solution: '''See attached file contains the solution! '''

Problem Statement
Pg. 21-3 confirm King pg. 28 pr 1.1 a

given the H-L-2ODE-VC below verify the first homogeneous solution and find the second homogeneous solution

$$(x-1)y^{''}-xy^{'} + y =0 $$

Problem Solution
For this problem the first homogeneous solution is given as: $$u_1=e^{x}$$

but how was this solution obtained? This solution can be obtained using the method of trial solution and starting with the familiar

form of $$ y(x)= e^{rx} \qquad \qquad r \quad is \quad constant$$

Now using this trial solution we will find the different differentiations of y:

$$ y(x) = e^{rx} $$

$$ y^{'}(x) = re^{rx} $$

$$ y^{''}(x) = r^{2}e^{rx} $$

Substituting these back into the original problem statement gives:

$$ (x-1)r^{2}e^{rx} - x(re^{rx}) + e^{rx} = 0 $$

Taking out the common exponential term reduces the equation to:

$$ (r^{2}x - r^{2} - xr + 1) = 0 $$

This equation can be examined to find two solutions for r, the first and most obvious (and also given away by the original homogeneous solution) is that r=1. This leads us back to the problem statement and answer to our question:

$$ u_{1}(x) = e^{x} $$

unfortunately the second solution for r2 would be a function of x which violates the original trial solution condition that r be a constant. Therefore we will have to rely on other methods to find the second homogeneous solution.

To find a second solution knowing the first the method of undetermined factors can be utilized

The first step is to rearrange the original equation to give the form:

$$ y^{''} - (\frac{x}{x-1})y^{'} + (\frac{1}{x-1})y = 0 $$

making the identification:

$$ a_{1} = (\frac{-x}{(x-1)}) $$

$$ a_{0} = (\frac{1}{(x-1)} $$

Now assume and answer with the form:

$$ y(x) = U(x)u_{1}(x) $$

after taking the first and second derivatives of y and plugging them back in the following is found:

$$ 0= U^{'} \left[ a_{1}u_{1} + 2u_{1}^{'} \right ] + U^{''}u_{1} $$

After some rearranging and using the integrating factor method as described in the homework from page 17-4 it is found that the second homogeneous solution takes the following for:

$$ u_{2} = u_{1} \int_{}^{x} \frac{1}{(u_{1}(t))^{2}} e (-\bar{a}_{1}(t)dt) $$

with the following resulting from the derivation:

$$ \bar{a}_{1} = \int{}^{t} a_{1}(s)ds = - \left [x + ln(x-1) \right ] $$

Using this value into Equation 2 yields and analyzing the integral as a first step yields:

$$ \int{}^{x} \frac{1}{e^{2t}} [ e^{t} + (t-1) ] dt $$

$$ u_{2} = e^{x} [ \frac13e^{-x} + \frac{x}2e^{-2x} ] $$

$$ u_{2} =-\frac13e^{-x} + \frac{x}2$$

knowing u1 and u2 and adding in two constants of integration the full homogeneous solution is:

$$ y(x) = K_{1}e^{x} + K_{2} (-\frac13 e^{-x} + \frac{x}2) $$

Problem Statement
Pg. 23-1 []

Part A

Given the H-L-2ODE-VC below find the full solution using both the Variation of Parameters, and the Alternative Method:

$$(x-1)y^{''}-xy^{'} + y = x $$

Problem Solution
The first method to be used is the Variation of Parameters, to use this method it is necessary to have both homogeneous solutions. These solutions were found above in the problem from 21-3 and are as follows:

$$u_1=\exp^{x}$$

$$ u_{2} =-\frac13e^{-x} + \frac{x}2$$

The first step in our solution is to assume that the solution takes the following form:

$$ y(x) = c_1(x)u_1(x) + c_2(x) u_2(x)$$

from here the first and second derivatives are found and added back to the equation taking out the common terms gives us

$$ c_1({u_1}^{}+{a_1}{u_1}^{'}+{a_0}u_1)+c_2({u_2}^{}+{a_1}{u_2}^{'}+{a_0}u_2)+{c_1}^'{u_1}^'+{c_2}^'{u_2}^'=f $$

The first two terms in the above equation are homogeneous solutions are therefore = 0 so this leaves us with

$${c_1}^'{u_1}^'+{c_2}^'{u_2}^'=f $$

combining this with the "breakthrough assumption" of:

$${c_1}^'{u_1}+{c_2}^'{u_2}=0 $$

allows to construct a matrix known as the wronskion matrix which has the following form:

$$\left|  \begin{matrix} u_1 & u_2 \\ {u_1}^' & {u_2}^' \end{matrix}  \right|$$

Let the wronskion matrix be denoted as W and its determinent be known as the wronskion and be denoted by W

Then using the definition of the Wronskion and solving a linear first order ODE with constant coefficients as shown on page 19-3 we can find two new variables:

$$ d_1(x) =  -\int{}^{x} \frac{u_2(s)f(s)}{W(s)}  ds $$

$$ d_2(x) =  \int{}^{x} \frac{u_1(s)f(s)}{W(s)}  ds $$

for our application all of the above are known quantities except for W:

$$ W =  \frac13 + \frac12 e^x  -  \frac{x}2  e^x + \frac13 $$

$$ W  =  \frac23  +  \frac12 e^{x} - \frac{x}2  e^x  $$

so we can determine the values of d_1 and d_2

$$ d_1 = -\frac14 [5xe^{-x} - 8 e^{-x} + x^3]$$

$$ d_1 = 6x e^x - 6e^x  - 3e^{-x}  +  3xe^{-x}  $$

The final form of the solution is in the form:

$$ y(x) = Au_1 +  Bu_2  +d_1u_1  + d_2u_2$$

with A and B being constants of integration and coming from IC or BC's solving for our problem gives:

$$ y(x) = Ax + B(-\frac13e^{-x} + \frac{x}2) + \frac14[5x-8+x^3e^x] + [6x e^x - 6e^x  - 3e^{-x}  +  3xe^{-x}]*[-\frac13e^{-x} + \frac{x}2]$$

The next step is to work the the same problem but using the alternate method. This process is shown in the following papers:



Contributing Authors
Egm6321.f09.team7.hua 01:27, 21 October 2009 (UTC) Egm6321.f09.team7.mm 02:19, 21 October 2009 (UTC) --Egm6321.f09.team7.benedict 23:05, 21 October 2009 (UTC)

 Only 3 contributing team members ? Egm6321.f09 13:24, 24 October 2009 (UTC)