User:Egm6321.f09.Team6/HW4

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See further comments in google doc for Team 6. Egm6321.f09 21:03, 23 November 2009 (UTC)

= Problem 1 =

Problem Statement
P.19-1 For the Legendre equation 1 on p.14-2 with n = 0, the homogeneous solution u1(x) = 1. Use reduction of order method 2 (undetermined factor) to find u2(x), the second homogeneous solution.

Solution
Equation 1 on p.14-2 with n=0, reduces to
 * {| style="width:100%" border="0" align="left"

(1-x^2)y''-2x y' = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.1)
 * }
 * }

The above equation is rewritten as,


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

y'' - \frac{2x}{1-x^2} y' = 0 $$ $$ Where,
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

a_1(x)=\frac{-2x}{1-x^2} $$ $$ From Lecture 17-4, Equation 2:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.3)
 * }
 * }


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

u_2 (x) = u_1(x) \int_{}^x {\frac{1}} \exp \left[ { - \int_{}^t {a_1 (s)ds} } \right]dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\int {} a_1(s)ds=\int {} \frac{-2s}{(1-s^2)}ds= \log(1-s^2) $$ $$ Substituting $$\displaystyle u_1(t)=1$$ and $$\int {} a_1(s)ds $$ in Equ.2
 * \displaystyle
 * \displaystyle
 * $$\displaystyle (1.5)
 * }
 * }


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

u_2 (x) = (1) \int_{}^x {\frac{1}{(1)^2} \exp \left[ - \log(1-s^2) \right]}dt = \int_{}^x {\frac{1}{1-s^2}}dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.6)
 * }
 * }


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

\int_{}^x {\frac{1}{1-s^2}}dt = \int_{}^x \left[\frac{1/2}{1-s} +\frac{1/2}{1+s} \right]ds = -\frac{1}{2}\log(1-x)+\frac{1}{2}\log(1+x) $$ $$ Thus, the second homogeneous solution by reduction of order method 2 is given by
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.7)
 * }
 * }


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

u_2(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1.8)
 * }
 * }

= Problem 2 =

Problem Statement
From King et al., p.28, problem 1.1.b: Show that the functions $$u_1$$ are solutions to the differential equations given below. Use the reduction of order method to find the second independent solution, $$u_2$$.


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


 * $$\displaystyle
 * $$\displaystyle

u_1 = x^{-1}sinx

$$ $$
 * $$\displaystyle (2.1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle
 * $$\displaystyle

x\frac{d^2y}{dx^2}+2\frac{dy}{dx}+xy = 0

$$ $$
 * $$\displaystyle (2.2)
 * }
 * }

Solution
Part 1: Showing that $$u_1$$ is a solution to the differential equation.


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


 * $$\displaystyle
 * $$\displaystyle

y = x^{-1}sinx

$$ $$
 * $$\displaystyle (2.3)
 * }
 * }


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


 * $$\displaystyle
 * $$\displaystyle

\frac{dy}{dx}=\frac{d}{dx}\left(x^{-1}sinx\right)=-x^{-2}sinx+x^{-1}cosx

$$ $$
 * $$\displaystyle (2.4)
 * }
 * }


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


 * $$\displaystyle
 * $$\displaystyle

\frac{d^2y}{dx^2}=\frac{d^2}{dx^2}\left(x^{-1}sinx\right)=2x^{-3}sinx-2x^{-2}cosx-x^{-1}sinx

$$ $$
 * $$\displaystyle (2.5)
 * }
 * }

Substituting into the original equation, we have:


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


 * $$\displaystyle
 * $$\displaystyle

x\left(2x^{-3}sinx-2x^{-2}cosx-x^{-1}sinx\right)+2\left(-x^{-2}sinx+x^{-1}cosx\right)+x\left(x^{-1}sinx\right)=0

$$ $$
 * $$\displaystyle (2.6)
 * }
 * }

The left-hand side of the equation above reduces to zero, proving that $$u_1$$ is a solution to the differential equation (2.2).

Part 2: Solving for $$u_2$$. We will look for a solution of the form $$y(x)=U(x)u_1(x)$$. First, we represent the first and second derivative as follows:


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


 * $$\displaystyle
 * $$\displaystyle

\frac{dy}{dx}=\frac{dU}{dx}u_1+U\frac{du_1}{dx}

$$ $$
 * $$\displaystyle (2.7)
 * }
 * }


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


 * $$\displaystyle
 * $$\displaystyle

\frac{d^2y}{dx^2}=\frac{d^2U}{dx^2}u_1+2\frac{dU}{dx}\frac{du_1}{dx}+U\frac{d^2u_1}{dx^2}

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.8)
 * }
 * }

Next, we substitute Equations 2.7 and 2.8 into the differential equation (2.2).


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


 * $$\displaystyle
 * $$\displaystyle

x\left(\frac{d^2U}{dx^2}u_1+2\frac{dU}{dx}\frac{du_1}{dx}+U\frac{d^2u_1}{dx^2}\right)+2\left(\frac{dU}{dx}u_1+U\frac{du_1}{dx}\right)+x\left(U(x)u_1(x)\right)=0

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.9)
 * }
 * }

We can now collect the terms.


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


 * $$\displaystyle
 * $$\displaystyle

U\left(\frac{d^2U}{dx^2}+2\frac{du_1}{dx}+xu_1\right)+xu_1\frac{d^2U}{dx^2}+\frac{dU}{dx}\left(2x\frac{du_1}{dx}+2u_1\right)=0

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.10)
 * }
 * }

Since $$u_1$$ is a solution of the differential equation (2.2), the term multiplying by $$U$$ is zero. Next, we define $$Z:=\frac{dU}{dx},$$ and $$\frac{dZ}{dx}=\frac{d^2U}{dx^2}$$. By removing the $$u_1$$ terms, which are equal to zero and substituting the new definitions, we arrive at the following first-order, linear differential equation:


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


 * $$\displaystyle
 * $$\displaystyle

xu_1\frac{dZ}{dx}+Z\left(2x\frac{du_1}{dx}+2u_1\right)=0

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.11)
 * }
 * }

Dividing by $$Zxu_1$$, we have


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


 * $$\displaystyle
 * $$\displaystyle

\frac{1}{Z}\frac{dZ}{dx}+\frac{2}{u_1}\frac{du_1}{dx}+\frac{2}{x}=0

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.12)
 * }
 * }

By integration, we have


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


 * $$\displaystyle
 * $$\displaystyle

log|Z|+2log|u_1|+\int^x\frac{2}{s}ds=C

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.13)
 * }
 * }

This can be used to solve for $$Z$$.


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


 * $$\displaystyle
 * $$\displaystyle

Z=\frac{k}{u_{1}^{2}}exp\left(-\int^x\frac{2}{s}ds\right)=\frac{dU}{dx}$$, where $$k=e^C

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.14)
 * }
 * }

Integrating 2.14 gives us an equation for $$U(x)$$.


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


 * $$\displaystyle
 * $$\displaystyle

U(x)=\int^x\frac{c}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt+\tilde{c}

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.15)
 * }
 * }

This gives us an equation for $$y(x)$$.


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


 * $$\displaystyle
 * $$\displaystyle

y(x)=u_1(x)\int^x\frac{c}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt+\tilde{c}u_1(x)

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.16)
 * }
 * }

Finally, we arrive at the equation for $$u_2(x)$$.


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


 * $$\displaystyle
 * $$\displaystyle

u_2(x)=u_1(x)\int^{x}\frac{1}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.17)
 * }
 * }

By solving Equation 2.17, we arrive at the solution for $$u_2$$.


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


 * $$\displaystyle
 * $$\displaystyle

u_2(x)=x^{-1}cosx

$$ $$
 * <p style="text-align:right;">$$\displaystyle (2.18)
 * }
 * }

= Make-up Problem 1.2 c =

= Make-up Problem 1.3 b =

Problem Statement
Solve $$ \frac{d^2y}{dx^2} + 4y = 2{\sec{2x}} $$

Solution
We know that,

$$ y_{Gen} = y_{H} + y_{P} $$

- To Find Homogeneous Solution

$$ y'' + 4y = 0 $$

$$ r^2 +4 = 0 $$

$$ r= +2i, -2i $$

$$ y_{H}= C_{1}{\cos{2x}} + C_{2}{\sin{2x}} $$

- To Find Particular Solution by using Variation of Paramaters

$$ {\cos(2x)}C'_{1} + {\sin(2x)}C'_{2} = 0 $$

$$ -2{\sin(2x)}C'_{1} + 2{\cos(2x)}C'_{2} = 2{\sec(2x)} $$

$$ C'_{1} = -tan(2x) $$

$$ C'_{2} = 1 $$

$$ \int C'_{1} = -\int {tan(2x)} = 2ln{\cos(2x)} = ln{cos^2(2x)} $$

$$ \int C'_{2} = \int 1 = x $$

Thus, we obtain the general solution as

$$ y_{Gen} = y_{H} + y_{P} $$

$$ Y_{Gen} = C_{1}{\cos(2x)} + C_{2}{\sin(2x)} + ln{\cos(2x)} + x{\sin(2x)} $$

Therefore, the particular solution is

$$y_p(t)=-y_1\int\frac{y_2f(t)}{W(y_1,y_2)}dt+y_2\int\frac{y_1f(t)}{W(y_1,y_2)}dt=-cos(t)\int sin(t)f(t)dt+sin(t)\int cos(t)f(t)dt$$

= Make-up Problem 1.3 c =



= Make-up Problem 1.3 d =

Problem Statement
Find the general solution of

$$\frac{d^2y}{dx^2}+y=f(x), y(0)=y'(0)=0$$

Solution
First, we solve for the homogeneous solution.

$$y_H(t)=c_1cos(t)+c_2sin(t)$$

So we have

$$y_1(t)=cos(t)$$ $$y_2(t)=sin(t)$$

The Wronskian of these two functions is

$$W=\left| \begin{array}{ c c } cos(t) & sin(t) \\ -sin(t) & cos(t) \end{array} \right|=cos^2(t)+sin^2(t)=1 $$

Therefore, the particular solution is

$$y_p(t)=-y_1\int\frac{y_2f(t)}{W(y_1,y_2)}dt+y_2\int\frac{y_1f(t)}{W(y_1,y_2)}dt=-cos(t)\int sin(t)f(t)dt+sin(t)\int cos(t)f(t)dt$$

Therefore, the general solution is

$$y=c_1cosx+c_2sinx-cosx\int^{x}sin(t)f(t)dt+sinx\int^{x}cos(t)f(t)dt$$

By substitution of the initial conditions, we find the $$c_1=c_2=0$$ which yields the following result for the general solution:

$$y=-cosx\int^{x}sin(t)f(t)dt+sinx\int^{x}cos(t)f(t)dt$$

= Contributing Members = Egm6321.f09.Team6.PaulMoore 20:54, 20 October 2009 (UTC)

Egm6321.f09.Team06.sada 23:46, 20 October 2009 (UTC)

Egm6321.f09.Team6.dianafoster 00:20, 21 October 2009 (UTC)

Egm6321.f09.team6.breaux 13:40, 21 October 2009 (UTC)