User:Egm6321.f10.team4/HW5

= Problem 1 -  Solving multiple y derivative wrt x =

From Meeting 26, p 26-2.

Given
The multiple y derivatives wrt x that is given from F&9
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{x}=e^{-t}y_{t} $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }


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

$$\displaystyle y_{xx}=e^{-2t}(y_{tt}-y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-4)
 * }
 * }


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

$$\displaystyle y_{xx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-5)
 * }
 * }


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

$$\displaystyle y_{xxxx}=e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-6)
 * }
 * }

Two approaches to obtain y derivatives
method 1
 * $$\displaystyle \underbrace{\frac{d}{dt} y(x(t))}_{y_t=} = \underbrace{\frac{dx}{dt}}_{e^t} \cdot \underbrace{\frac{dy}{dx}}_{y_x} $$

rewriting the equation wrt y derivate with respect to x,
 * $$\displaystyle y_x = y_{t} \cdot e^{-t}$$

method 2
 * $$\displaystyle y_{x}= \underbrace{\frac{dy}{dx}}_{y_x=} = \underbrace{\frac{dy}{dt}}_{y_t} \cdot \underbrace{\frac{dt}{dx}}_{\frac{dx}{dt}=e^{-t}}$$


 * $$\displaystyle y_x = y_{t} \cdot e^{-t}$$

Find
Part a find $$\displaystyle y_{xxxxx}$$ in terms of derivative of y with respect to t

Part b Plug (eq 1.3) and (eq 1.4) into (eq 1.1) to obtain $$\displaystyle y_{tt} - 3y_{t} + 2y = 0$$

Solution for the part a
Solving for the 5th Derivative


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

$$\displaystyle y_{xxxxx} = \frac{d}{dx} (y_{xxxx}) $$ $$
 * $$\displaystyle (Eq. 1-7)
 * }
 * }


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

$$\displaystyle = \frac{dt}{dx} \cdot \frac{d}{dt} (y_{xxxx}) $$ $$
 * $$\displaystyle (Eq. 1-8)
 * }
 * }

Substituting (Eq 1-6) into the previous (Eq 1-8) yields,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{xxxxx} = \underbrace {\frac{dt}{dx}}_{e^{-t}} \cdot \frac{d}{dt} \left[e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) \right] $$ $$
 * $$\displaystyle (Eq. 1-9)
 * }
 * }


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

$$\displaystyle = e^{-t} \left[-4e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) + e^{-4t}(y_{ttttt} - 6y_{tttt} + 11y_{ttt} - 6y_{tt}) \right] $$ $$
 * $$\displaystyle (Eq. 1-10)
 * }
 * }


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

$$\displaystyle = e^{-5t} \left[-4y_{tttt} + 24y_{ttt} -44y_{tt} + 24y_{t} + y_{ttttt} - 6y_{tttt} + 11y_{ttt} - 6y_{tt}\right] $$ $$
 * $$\displaystyle (Eq. 1-11)
 * }
 * }

Collecting terms with reduction order of y derivate wrt t yields,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{xxxxx} = e^{-5t}(y_{ttttt} - 10y_{tttt} + 35y_{ttt} - 50y_{tt} + 24y_{t}) $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * $$\displaystyle (Eq. 1-12)
 * }
 * }

Solution for the part b
Plugging the obtained y derivatives into the given equation to obtain the following:

First, taking a look at what we have to use,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle x=e^t $$ $$
 * $$\displaystyle
 * }
 * }


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

$$\displaystyle y_{x}=e^{-t}y_{t} $$ $$
 * $$\displaystyle (Eq. 1-14)
 * }
 * }


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

$$\displaystyle y_{xx}=e^{-2t}(y_{tt}-y_{t}) $$ $$
 * $$\displaystyle
 * }
 * }

Substituting what we have into the coresponding variables yields,

reorganizing the equation by reduction order, we can solve the desired equation


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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle y_{tt} -2y_{t} + 2y = 0 $$ $$
 * $$\displaystyle (Eq. 1-17)
 * }
 * }

Egm6321.f10.team4.Yoon 00:00, 30 October 2010 (UTC) - Primary Author 68.17.105.125 01:01, 3 November 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 2: Method of Trial Solutions = From Meeting 26, p 26-3.

Given
Solve the given L2_ODE_VC((4)P25.3 using Method 2 i.e.) Trial Solution $$\displaystyle y=x^r$$ with the boundary condition as followed


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

x^2y'' - 2xy' + 2y = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

The given boundary conditions
 * {| style="width:100%" border="0" align="left"

y(_{x=}1) = -2, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y(_{x=}2) = 5. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-3)
 * }
 * }

required

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

y(x) = \sum^{n}_{i=1} C_ix^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Find
Part a Solve (Eq. 2-1) using method2(trial solution) to make it have the shown format below.


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

y(x) = C_1x^{r_1} + C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-4)
 * }
 * }

Part b Plot the homogeneous equation

Solution for Part a
First, taking the given trial solution y's 1st and 2nd deriviates,
 * {| style="width:100%" border="0" align="left"

y=x^r $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-5)
 * }
 * }


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

y'=rx^{r-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-6)
 * }
 * }


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

y''=r(r-1)x^{r-2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-7)
 * }
 * }

Substiuting (Eq 2-5), (Eq 2-6), and (Eq 2-7) into the given L2_ODE(Eq 2-1) yields
 * {| style="width:100%" border="0" align="left"

\cancelto{_1}{x^2} r \cdot x^{r- \cancelto{1}{2}}(r-1) - 2 \cancelto{_1}{x} \cdot x^{r-\cancelto{1}{1}}r + 2x^r = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-8)
 * }
 * }

Simplifying and eliminating non-zero constant ahead of polynomial,
 * {| style="width:100%" border="0" align="left"

x^r \left( r^2 -r - 2r + 2 \right) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-9)
 * }
 * }

The charactoristic Equation of the ODE is
 * {| style="width:100%" border="0" align="left"

r^2- 3r + 2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-10)
 * }
 * }

Solving for the unknown 'r's yields
 * {| style="width:100%" border="0" align="left"

r_1 = 1, r_2 = 2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-11)
 * }
 * }

The solution takes the form of a homogeneous solution, so substituting the rs into the equation,
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{r_1}+C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-12)
 * }
 * }

Substituting the roots of the polynomial into the homogeneous format,
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{1}+C_2x^{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-13)
 * }
 * }

plugging the given boundary conditions yields
 * {| style="width:100%" border="0" align="left"

y(_{x1=}1)= C_1 \cdot (_{x1=} 1)^{1}+C_2 \cdot (_{x1=}1)^{2} = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y(_{x2=}2)= C_1 \cdot (_{x2=}2)^{1}+C_2 \cdot (_{x2=}2)^{2} = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-15)
 * }
 * }

Reorganizing the equation wrt $$\displaystyle C_1$$ and $$\displaystyle C_2$$,
 * {| style="width:100%" border="0" align="left"

C_1 + C_2 = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-16)
 * }
 * }


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

2C_1 + 4C_2 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-17)
 * }
 * }

Solving the above equations, we obtain $$\displaystyle C_1$$ and $$\displaystyle C_2$$,
 * {| style="width:100%" border="0" align="left"

C_1 = - \frac{13}{2}; C_2= \frac{9}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-18)
 * }
 * }

Therefore substiting the results from (Eq. 2-18) into the homogeneous solution format(Eq. 2-13)
 * {| style="width:100%" border="0" align="left"


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle y(x) = - \frac{13}{2}x + \frac{9}{2}x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2-19)
 * }
 * }

Solution for Part b
Plot the obtained equation,
 * {| style="width:100%" border="0" align="left"

y(x) = - \frac{13}{2}x + \frac{9}{2}x^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

</BR>

Egm6321.f10.team4.Yoon 00:00, 30 October 2010 (UTC) - Primary Author 68.17.105.125 01:00, 3 November 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 3 - Characteristic Equations for Euler L2-ODEs = From Meeting 27, p 27-1.

Part 1
Consider the Euler L2-ODE-VC:

Part 2
Consider the Euler L2-ODE-CC:

Part 1

 * 1) For Eq. 3-1, find $$\displaystyle a_2 $$, $$\displaystyle a_1 $$, and $$\displaystyle a_0 $$ such that $$\displaystyle \left( r-\lambda \right)^2=0 $$.
 * 2) Find the first homogeneous solution: $$\displaystyle y_1(x)=x^{\lambda} $$
 * 3) Find the second homogeneous solution: $$\displaystyle y_2(x)=u(x)x^{\lambda}=u(x)y_1(x) $$, where $$\displaystyle u(x) $$ is an unknown function of $$\displaystyle x $$. Also find the differential equation governing $$\displaystyle u(x) $$ and solve for $$\displaystyle u(x) $$, then $$\displaystyle y_2(x) $$.
 * 4) Find the general homogeneous solution.

Part 2

 * 1) For Eq. 3-2, repeat steps 1-4 in Part 1.

Part 1.1
Beginning with Method 2 (trial solution) from [http://upload.wikimedia.org/wikiversity/en/9/9d/2010_10_14_14_56_10.djvu Mtg. 25] p. 25-3, we will assume:

Thus the first and second derivatives of $$\displaystyle y $$ w.r.t. $$\displaystyle x $$ are:

Substituting into Eq. 3-1, we get:

Divide by $$\displaystyle x^r $$ to get:

We are trying to find the coefficients $$\displaystyle a_2 $$, $$\displaystyle a_1 $$, and $$\displaystyle a_0 $$ such that $$\displaystyle \left( r-\lambda \right)^2=0 $$, so we will set Eq. 3-6 equal to $$\displaystyle \left( r-\lambda \right)^2=0 $$:

We can see by inspection that this condition (i.e. Eq. 3-7) is satisfied if:

Part 1.2
The first homogeneous solution is $$\displaystyle y_1(x)=x^{\lambda} $$. We already know that $$\displaystyle (r-\lambda)^2=0 $$ (by definition), so we can find the roots immediately:

Thus we can write the first homogeneous solution as:

Part 1.3
The second homogeneous solution and its first two derivatives are:

Substituting Eqs. 3-8 into Eq. 3-1, we get:

Now, substitute Eqs. 3-11 into Eq. 3-12:

Divide through by $$\displaystyle x^{\lambda}$$: Distribute the terms outside the brackets: Now, rearrange and factor like terms: This reduces to the simple expression: Next, divide through by $$\displaystyle x $$: We can further simplify Eq. 3-18 by reducing its order if we define: Thus the differential equation governing $$\displaystyle u(x) $$ in the second homogeneous equation is: where $$\displaystyle v := u'(x) $$. In order to solve for $$\displaystyle v(x) $$, we will once again assume the form of: Substituting these into Eq. 3-20, we get: If we divide through by $$\displaystyle x^s $$, we obtain: Now, let's substitute this back into the expression for $$\displaystyle v(x)$$ in Eq. 3-21: We can now find $$\displaystyle u(x) $$ from:

Finally, substituting Eq. 3-25 into Eq. 3-11, we obtain:

Part 1.4
We can now write the general form of $$\displaystyle y(x) $$ as: Using Eqs. 3-10 and 3-26, and replacing the coefficient in Eq. 3-10, Eq. 3-27 becomes: The general homogeneous solution is:


 * Note: It is interesting that the form of the general homogeneous solution (Eq. 3-29) is identical to that of the second homogeneous solution (Eq. 3-26), despite having added the first homogeneous solution (Eq. 3-10). The difference is only in the magnitude of the constant.

Part 2
For the reader's convenience, Eq. 3-2 is reproduced below:

Part 2.1
Let us assume that $$\displaystyle y $$ has the form:

Its first and second derivatives are thus:

When we substitute these into Eq. 3-2, we get:

Divide through by $$\displaystyle e^{rx}b_2 $$:

Recall that we are trying to find the coefficients $$\displaystyle b_2 $$, $$\displaystyle b_1 $$, and $$\displaystyle b_0 $$ such that $$\displaystyle \left( r-\lambda \right)^2=0 $$, so we will set Eq. 3-32 equal to $$\displaystyle \left( r-\lambda \right)^2=0 $$:

By inspection, we see that Eq 3-33 is satisfied if the coefficients are:

Part 2.2
The first homogeneous solution is $$\displaystyle y_1(x)=e^{\lambda x} $$. We also know that, by definition, $$\displaystyle (r-\lambda)^2=0 $$, so we can find the roots immediately:

Thus we can write the first homogeneous solution as:

Part 2.3
In order to find the second homogeneous solution, we start with:

Its first two derivatives are :

Divide Eq. 3-2 by $$\displaystyle b_2 $$ to make it easier to work with:

Now we can substitute $$\displaystyle y_2(x) $$ (Eq. 3-37) and its derivatives, along with the b-coefficients (Eqs. 3-34) into Eq. 3-38:

Divide through by $$\displaystyle e^{\lambda x} $$ :

Consolidating the terms in Eq. 3-40, we obtain the differential equation governing $$\displaystyle u(x) $$:

Integrating Eq. 3-41 twice yields the expression for u(x):

Now we will substitute this back into Eq. 3-37 to obtain the second homogeneous equation:

Part 2.4
We can write the general form of $$\displaystyle y(x) $$ as:

Next, we substitute Eqs. 3-36 and 3-43, and replace the coefficient in 3-36 to get:

The general homogeneous solution is thus:


 * Note: Just as in Part 1, we see that the general solution (Eq. 3-45) has the same form as the second homogeneous equation (Eq. 3-43).

Egm6321.f10.team4.ejm 18:27, 1 November 2010 (UTC) Author

= Problem 4 - Particular solution using variation of parameters= From Meeting 27, p 27-2.

Find
The particular solution $$\displaystyle y_P $$ of $$\displaystyle y $$

Solution
In HW2 problem 8 we determined that the homogeneous solution to equation 4-1 was

Taking the derivative of equation 4-2,

where

substituting equation 4-5 into 4-4 yields

Now if we substitute equations 4-2 and 4-6 into equation 4-1

Which can be reduced to

Solving for $$\displaystyle {A(x)}' $$

Integrating both sides of equation 4-7

Substituting equation 4-8 back into equation 4-2

Since we know

$$\displaystyle y_P(x) $$ must be


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

$$\displaystyle y_P(x)= {C_1}{exp{\left [ - \int^{x} a_0(s) ds \right ]}} \int^{x}\left [b(s) exp \left [ \int^{s} a_0(r) dr \right ] \right ] $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4-12)
 * }
 * }

Egm6321.f10.team4.Auerbach 04:16, 3 November 2010 (UTC) - Primary Author

= Problem 5 - Equation of Motion: Mass Spring Damper System= From Meeting 28, p 28-2 to Meeting 29, p 29-1.

Given
A spring-mass-damper system with an applied force:



Consider the nonhomogeneous L2_ODE_CC:

$${{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y=f\left( t \right)$$ <p style="text-align:right;">$$\displaystyle (Eq. 5-1) $$

where $${a_2}$$ is the mass constant m, $${a_1}$$ is the damping constant c, $${a_0}$$ is the spring constant k, $$f\left ( t \right )$$ is the forcing function

Find
5.1 PDEs for the integrating factor $$\displaystyle h(t,y)$$

5.2 Trial solution for the integrating factor $$\displaystyle h(t,y)$$


 * 5.2.1 $$\displaystyle \bar{a}_1$$ and $$\displaystyle \bar{a}_2$$ in terms of $$\displaystyle a_2,a_1,a_0$$


 * 5.2.2 Quadratic equation for $$\displaystyle \alpha $$


 * 5.2.3 Reduced order equation using integrating factor $$e^{\alpha t}[\bar{a}_{1}y^{'}+\bar{a}_{0}y^{'}] = \int e^{\alpha t} f\left ( t \right )dt$$ $$\bar{a}_{1}y^{'}+\bar{a}_{0}y^{'} = e^{-\alpha t}\int e^{\alpha t} f\left ( t \right )dt$$


 * 5.2.4 By implementing IFM (Integrating Factor Method), find $$y\left ( t \right )$$ for general excitation $$f\left ( t \right )$$


 * 5.2.5 Show $$\displaystyle \alpha \beta =\frac{\alpha _{0}}{\alpha _{2}} $$ and $$\displaystyle \alpha + \beta =\frac{\alpha _{1}}{\alpha _{2}} $$


 * 5.2.6 Deduce an expression for the particular solution $$\displaystyle y_P $$ for a general excitation $$\displaystyle f(t) $$


 * 5.2.7 Verify with the table of particular solutions. Take into consideration that $$f\left ( t \right ) = t^{2}$$


 * 5.2.8 Solve the linear 2nd order ordinary differential equation with constant coefficients (L2_ODE_CC) with $$f(t)=exp(-t^2) $$. Find the coefficients $$\displaystyle a_2,a_1,a_0$$ such that the L2_ODE_CC accepts the following as characteristic equations:


 * 5.2.8.1 $$\displaystyle (r+1)(r-2)=0 $$


 * 5.2.8.2 $$\displaystyle (r-4)^2=0 $$

5.1
The equation of motion for a spring-mass-damper system can be written as:

Using an integrating factor $$\displaystyle h(t,y)$$, equation 5-1 becomes

If we then set $$\displaystyle p=y'$$, equation 5-2 can be written in a form that satisfies the first condition of exactness:

The second condition of exactness requires the following 2 conditions must be met:


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

$$\displaystyle {g_{pp}}=f_{tp}+pf_{yp}+2f_{y} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5-3)
 * }
 * }


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

$$\displaystyle {f_{tt}}+2pf_{ty}+p^2f_{yy}=g_{tp}+pg_{yp}-g_{y} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5-4)
 * }
 * }

Taking the partial derivatives in equation 5-3,


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

$$\displaystyle g_{pp}=0 $$
 * }
 * }


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

$$\displaystyle f_{tp}=0 $$
 * }
 * }


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

$$\displaystyle f f_{yp}=0 $$
 * }
 * }


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

$$\displaystyle f_{y}=h_ya_0 $$
 * }
 * }

Equation 5-3 reduces to


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

$$\displaystyle 0=0+p0+2h_ya_0 $$
 * }
 * }

so,


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

$$\displaystyle h_y=0 $$
 * }
 * }

Taking the partial derivatives in equation 5-4,


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

$$\displaystyle f_{tt}=h_{tt}a_2 $$
 * }
 * }


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

$$\displaystyle f_{ty}=h_{ty}a_2=0 $$
 * }
 * }


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

$$\displaystyle f_{yy}=h_{yy}a_2=0 $$
 * }
 * }


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

$$\displaystyle g_{tp}=h_{t}a_1 $$
 * }
 * }


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

$$\displaystyle g_{yp}=h_{y}a_1=0 $$
 * }
 * }


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

$$\displaystyle g_{y}=h_{y}a_1p+h_{y}a_0y+ha_0-h_yf(t)=ha_0 $$
 * }
 * }

Equation 5-4 reduces to


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

$$\displaystyle h_{tt}a_2+2p0+p^20=h_{t}a_1+p0-ha_0 $$
 * }
 * }

or


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

$$\displaystyle h_{tt}a_2-h_{t}a_1+ha_0=0 $$
 * }
 * }

5.2
Starting off with the trial solution for the intergration factor $$\displaystyle h(t)=exp(\alpha t)$$, equation 5-1 becomes

If we integrate both sides with respect to $$\displaystyle t$$

Assuming the result will be a reduction of order

5.2.1
Differentiating the LHS of equation 5-7 and the LHS of equation 5-6 and comparing:

Examining like $$\displaystyle y''$$ terms:

Thus

Examining like $$\displaystyle y'$$ terms:

Thus

Examining like $$\displaystyle y$$ terms:

Thus

5.2.2
Using equations 5-10 and 5-11

or

5.2.3
Starting with equation 5-7

Dividing both sides by $$\displaystyle \bar{a}_1$$

5.2.4
Using what was previously discussed in [http://upload.wikimedia.org/wikiversity/en/0/00/2010_09_14_15_00_52.djvu Mtg. 10],

where

Thus, using the equation presented in 10-3

5.2.5
From the above definitions of $$ \displaystyle \alpha $$ and $$ \displaystyle \beta $$, and since $$ \displaystyle \bar{a}_{1}=a_2 $$:

5.2.6
The solution to $$\displaystyle y $$ is made up of the particular solution, $$\displaystyle y_P $$, and the homogeneous solutions,$$\displaystyle y_{H1,2} $$. The solution of $$\displaystyle y $$ presented in equation 5-12 can be expanded to:

so

5.2.7
Let's start off by defining $$\displaystyle f\left ( t \right ) = t^{2}$$

substituting earlier definitions of $$\displaystyle \alpha$$ and $$\displaystyle \beta$$, we can reduce to

Using the table for particular solutions:

Substituting back in equation 5-1

Thus,

Which yields the same particular solution as before,

5.2.8
Let's start off by defining $$\displaystyle f\left ( t \right ) = exp(-t^{2})$$

5.2.8.1
Given the characteristic equation

or

so,

5.2.8.2
Given the characteristic equation

or

so,

Egm6321.f10.team4.Corella.RL 20:57, 5 November 2010 (UTC) - Co-author Egm6321.f10.team4.Auerbach 04:16, 3 November 2010 (UTC) - Co-author Egm6321.f10.team4.ejm 20:40, 5 November 2010 (UTC) Editor

= Problem 6 - Expressions for the Particular Solution= From Meeting 30, p 30-1.

Given
From our lecture notes:

$$\displaystyle y_p (x) =  u_1 (x) \int \frac {1}{h (x)} [ \int h (x) f (x) dx ]  dx $$<p style="text-align:right;">$$\displaystyle (Eq. 6-1) $$

From the text by King (pg.8, Eqn. 1.6)

$$\displaystyle y_p (x) =  \int^{x} f(s) \frac {u_1 (s) u_2 (x) - u_1 (x) u_2 (s)}{W(s)} ds $$<p style="text-align:right;">$$\displaystyle (Eq. 6-2) $$

where $$\displaystyle W(s) $$ is the Wronskian, defined as

$$\displaystyle W(s) =  u_1 u_2'  -  u_2 u_1' $$<p style="text-align:right;">$$\displaystyle (Eq. 6-3) $$

Show
that the given expressions (6-1) and (6-2) are equivalent.

Solution
We will begin by expressing $$\displaystyle h(x) $$ in terms of $$\displaystyle u_1 $$ and $$\displaystyle u_2 $$.

$$\displaystyle \frac {1}{h(x)} =  (\frac {u_2}{u_1})' $$<p style="text-align:right;">$$\displaystyle (Eq. 6-4) $$

By the quotient rule, this can be rewritten

$$\displaystyle \frac {1}{h(x)} =  \frac {u_1 u_2'  -  u_2 u_1'}{u_1^2} $$<p style="text-align:right;">$$\displaystyle (Eq. 6-5) $$

and it follows that

$$\displaystyle h(x) =  \frac {u_1^2}{u_1 u_2'  -  u_2 u_1'} $$<p style="text-align:right;">$$\displaystyle (Eq. 6-6) $$

By substituting the expressions from (6-5) and (6-6) into our given expression (6-1)

$$\displaystyle y_p(x) =  u_1 (x) \int \frac {u_1 u_2' - u_2 u_1'}{u_1^2} [\int^{x} f(s) \frac {u_1^2}{u_1 u_2' - u_2 u_1'} ds]  dx $$

where $$\displaystyle (s) $$ is a dummy variable.

Substituting the Wronskian into the second integral gives

$$\displaystyle y_p(x) =  u_1 (x) \int \frac {u_1 u_2' - u_2 u_1'}{u_1^2} [\int^{x} f(s) \frac {u_1^2}{W(s)} ds]  dx $$

Combining the two integrals into one gives

$$\displaystyle y_p(x) =  u_1(x) \int^{x} f(s)  \frac {u_1(s)^2}{W(s)}  \frac {u_1(s) u_2(x)  -  u_2(s) u_(x)}{u_1(s)^2}  ds $$

The primed terms are now expressed in terms of x, and the unprimed terms are expressed in terms of the dummy variable s.

The $$\displaystyle u_1 (s)^2 $$ terms will cancel, leaving the expression

$$\displaystyle y_p(x) = u_1(x) \int^{x} f(s) \frac {u_1(s) u_2(x)  -  u_2(s) u_1(x)}{W(s)}  ds $$

which is our equation (6-2), multiplied by the term $$\displaystyle u_1(x) $$. Had we started with the general solution similar to that as expressed in Meeting 10

$$\displaystyle y(x) =  \frac {1}{h (x)}  \int h (x) f (x) dx  $$

then there would be no additional $$\displaystyle u_1(x) $$ term. In order to get perfect reconciliation between the lecture notes and the text, we would have to modify our problem statement to not include the additional $$\displaystyle u_1(x) $$ term.

128.227.69.48 20:18, 2 November 2010 (UTC) Egm6321.f10.team4.osentowski - Author

= Problem 7 Find two homogeneous solutions= From Meeting 31, p 31-1.

Solution
First start by computing the first and second integrals of $$\displaystyle y $$

Substitute $$\displaystyle y,y', y''$$ into Eq. 8-1 to obtain the following

Divide Eq. 8-4 by $$\displaystyle e^{rx}$$ to obtain the characteristic equation.

We will obtain two roots by solving Eq. 8-5

First root

Second root

The two roots are real and distinct and will generate the following two homogeneous solutions

Egm6321.f10.team4.petralanda.n 18:39, 1 November 2010 (UTC)Naiara Petralanda

= Problem 8 Find a L2_ODE_VC through reverse engineering = From Meeting 31, p 31-3.

Given
Given the following characteristic equation

and the trial solution method form

Find
Using reverse engineering find a L2_ODE_VC that will satisfy Eq. 8-1 using the trial solution method on Eq. 8-2.

Solution
Take the first derivative of Eq. 8-2

Take the second derivative of Eq. 8-2

Generate the equation using preliminary functions $$\displaystyle a_0, a_1, a_2 $$

Divide both sides by $$\displaystyle y $$ and equate it with the given characteristic equation Eq. 8-1

We will obtain three equations based on the power of $$\displaystyle r $$

The first equation will solve for $$\displaystyle a_2 $$

The second equation will yield the value of $$\displaystyle a_1 $$

If we substitute the solution on Eq. 8-8

The third and last equation will allow us to solve for the third variable coefficient

If we insert the solutions shown in Eq. 8-8 and Eq. 8-10 we will get the form of $$\displaystyle a_o $$

If we combine Eq. 8-8, Eq. 8-10 and Eq. 8-12 we can form the L2_ODE_VC

Egm6321.f10.team4.petralanda.n 18:40, 1 November 2010 (UTC)Naiara Petralanda