User:Egm6321.f11.team2.rho/HW6

=R*6.1- Solve Nonhomogeneous L2-ODE-CC=

Given
As an application of nonhomogeneous L2-ODE-CC:

2nd exactness condition for N2-ODES is:

For problme 2, integrating factor is

Find
1. Find the PDEs that governs the integrating factor $$\displaystyle h(x,y)$$ using 2nd exactness condition for N2-ODEs

2. Solve (6.1.1) using trial solution for integrating factor of $$\displaystyle h(t)={{e}^{\alpha t}}$$

2.1) Find $$ (\bar a_1,\bar a_0)$$ in terms of $$ (a_0,\ a_1,\ a_2) $$.

2.2) Find the quadratic equation for $$\displaystyle \alpha $$.

2.3) Reduced-order equation: lead to

$$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$$

$$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\int e^{\alpha t}f(t)dt$$

2.4) Use the IFM to solve the equation in part 2.3.

2.5) Show that

$$\displaystyle  \alpha \beta = \frac {a_0}{a_2}$$.

$$\displaystyle  \alpha+\beta=\frac{a_1}{a_2}$$

2.6) Deduce the particular solution $$\displaystyle y_P(t)$$ for general excitation $$\displaystyle f(t)$$

2.7) Verify result with table of particular solutions for

$$\displaystyle f(t) = t exp(bt) $$

2.8) Solve the nonhomogeneous L2-ODE-CC with the following excitation

Hyperbolic function: $$\displaystyle f(t) = tanh t $$

For the coefficients (a0,a1,a2),consider two characteristic equations:

2.8.1) $$\displaystyle (r+1)(r-2) = 0$$

2.8.2) $$\displaystyle (r-4)^2 = 0 $$

2.9) For each case in questions 2.8.1 and 2.8.2, determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

Part 1
Multipling $$\displaystyle H(t,y)$$ in the both sides, (6.1.1) can be rewritten as follows:

In the 2nd exactness condition,

From (6.1.5) and (6.1.6), We can get the following relation:

Assuming that H(t,y) and F(t,y) are only the functions of t in (6.1.2) and (6.1.3), and P=y', we can get the following relations in the 2nd exactness condition:

From (6.1.9) and (6.1.10), we can get the following result:

Part 2.1
where $$ \bar{a}_2 = 0 $$

$$ \frac{d}{dt}\Big[e^{\alpha t}(\bar{a}_1y'+\bar{a}_0y) \Big] = e^{\alpha t}\Big(\bar{a}_1 y + (\alpha \bar{a}_1+\bar{a}_0)y' + \alpha \bar{a}_0 y \Big) = e^{\alpha t}\big[a_2y+a_1y'+a_0y\big] \! $$

Thus, we can the the following relation.

$$ \bar{a}_1 = a_2 \! $$

$$ \bar{a}_0 = a_1 - \alpha a_2 = a_0/\alpha \! $$

Part 2.2
Multiplying $$\displaystyle \alpha $$ in the both second and third sides of (6.1.9) and rewriting it, we can get the following quadratic equation.

Part 2.3
From Part 2.1,

The above function can be rewriten as follows:

where $$ \beta=\frac{\bar a_0}{\bar a_1}$$

Part 2.4
The integrating factor to use IFM is $$ \bar{h}(t) = exp \Big[ \int \underbrace{\frac{\bar{a}_0}{\bar{a}_1}}_{\beta} dt \Big] = e^{\beta t} \! $$

The IFM would be used for the solution as follows: $$\displaystyle y(t)=\frac{1}{\bar h(t)}\int^t \bar h(s)B(s)ds $$ The result of Part 2.3 will be devided by $$\displaystyle \bar a_1 $$ to obtain B(s). $$\displaystyle y'+\beta y = \frac{e^{-\alpha t}}{\bar a_1} \int e^{\alpha t}F(t)dt = B(s) $$ Then we can combine $$\displaystyle h(t), \ B(s) $$ into $$\displaystyle y(t) $$ to find

Part 2.5
$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_0/\alpha}{a_2} $$

Multiplying $$\displaystyle \alpha $$ in the both first and last sides of the above equation, we can get the following:

$$ \alpha\beta = \frac{a_0}{a_2} $$

$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_1 - \alpha a_2}{a_2} = \frac{a_1}{a_2} - \alpha $$

Thus, $$ \alpha + \beta = \frac{a_1}{a_2} $$

Part 2.6
When $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are the roots of the characteristic equation, the homogeneous solution is

The particular solution $$\displaystyle y_P(t) $$ is

Part 2.7
Since $$\displaystyle F(t) = t e^{bt} $$,

Part 2.8.1
Thus, the L2-ODE is given by

coefficients $$\displaystyle {{a}_{0}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{2}}$$ are

From the root of the above characteristic equations, we can see the following:

$$\displaystyle \alpha=-1, \beta=2$$

The excitation is hyperbolic function as follows: $$\displaystyle f(t)=tanh(t)$$

With the website of "wolframalpha", the general solution y(t) is $$\displaystyle y(t)=\frac{1}{6}\left[1+2e^{-2t}log(e^{2t}+1)-4e^ttan^{-1}(e^{-t})\right]$$

Part 2.8.2
Thus, the L2-ODE is given by

The coefficients $$\displaystyle {{a}_{0}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{2}}$$ are

From the root of the above characteristic equations, we can see the following:

$$\displaystyle \alpha=\beta=4$$

The excitation is hyperbolic function as follows: $$\displaystyle f(t)=tanh(t)$$

With the website of "wolframalpha", the general solution y(t) is $$\displaystyle y(t)=-\frac{Li_2(-e^{2t}\cdot e^{-4t})}{2}-\frac{e^{-2t}}{2}+\frac{1}{16}$$

=R*6.7- Homogeneous Solution of Legendre Function Using Variation of parameters=

Given
The general form of the Legendre Equation is

$$\displaystyle P_2(x)$$ should be

for the following result of $$\displaystyle Q_2(x)$$

Find
Show

using the Method of Variation of Parameters.

Solution
When n=2, the given Legendre Equation is as

(6.7.4) cam be rewritten as follows:

Let $$\displaystyle y(x) = U(x)\cdot {u}_{1}(x)$$ and $$\displaystyle P_2(x) = {u}_{1}(x)$$.

where $$\displaystyle \begin{align} &y = U {u}_{1} \\ &y' = U {u}_{1}'+U' {u}_{1} \\ &y= U {u}_{1}+2U' {u}_{1}'+U'' {u}_{1} \\ \end{align} $$

To reduce the order of (6.7.6), $$\displaystyle Z:=U'$$ willb be used.

To use integrating factor method, we first need to find the integrating factor, h(x), as follows:

The soloution of Z is

From the definition of Z(x)= U',

Solving y(x) with U(x),

2nd homogeneous solution is

Thus,