text:
User:Egm6321.f10.team5/hw5

= Problem 1 - Higher order deriv.= From Meeting 26, p. 26-2

Find
Find $$y_{xxxxx}$$ in terms of derivative of $$y$$ with respect to $$t$$.

Author and Proof-Reader
[Author] Jo

[Proof-Reader] Oh, Sang Min

= Problem 2 - Solve using Method 2= From Meeting 26, p. 26-3

Find
Solve and plot $$x^2 y'' - 2 x y' + 2 y = 0$$ using the trial solution method where $$y = x^r$$ and boundary conditions are $$y(1) = 3$$ & $$y(2) = 4$$ or $$y(1) = -2$$ & $$y(2) = 5$$.

Solve
For $$y(1) = 3$$ & $$y(2) = 4$$ case, we get the following solution. For $$y(1) = -2$$ & $$y(2) = 5$$ case, we get the following solution.

Author and Proof-Reader
[Author] Jo

[Proof-Reader] Oh, Sang Min

= Problem 3 - Consider Char. eq.= From Meeting 27, p. 27-1

Given
The Euler L2-ODE-VC: $$ a_2x^2y''+a_1xy'+a_0y=0 \,$$ $$\displaystyle (Eq. 3.1) $$

The Euler L2-ODE-CC: $$ b_2y''+b_1y'+b_0y=0 \,$$ $$\displaystyle (Eq. 3.2) $$

The characteristic equation: $$ (r-\lambda)^2 = r^2-2\lambda r + \lambda^2 = 0 \,$$ $$\displaystyle (Eq. 3.3) $$

Find
1.1) Find $$a_2$$, $$a_1$$ , $$a_0$$ such that Eqn. 3.3 is the characteristic equation of Eqn. 3.1

1.2) Find the first homogeneous solution, $$y_1$$

1.3) Find the second homogeneous solution, $$y_2$$

1.4) Find the general homogeneous solution

2.1) Find $$b_2$$, $$b_1$$ , $$b_0$$ such that Eqn. 3.3 is the characteristic equation of Eqn. 3.2

2.2) Find the first homogeneous solution, $$y_1$$

2.3) Find the second homogeneous solution, $$y_2$$

2.4) Find the general homogeneous solution

Part 1.1
The coefficients of Eqn. 3.1 (a terms) can be found if we use the trial solution method. The trial solution for an Euler L2-ODE-VC is: $$ y=x^r \,$$ $$\displaystyle (Eq. 3.1.1) $$

Therefore we can calculate the first and second derivatives of the trial solution:

$$ y'=rx^(r-1) \,$$ $$\displaystyle (Eq. 3.1.2) $$

$$ y''=r(r-1)x^(r-2) \,$$ $$\displaystyle (Eq. 3.1.3) $$

Then substitute back into Egn. 3.1 so that it becomes,

$$ a_2x^2r(r-1)x^(r-2)+a_1xrx^(r-1)+a_0x^r=0 \,$$ $$\displaystyle (Eq. 3.1.4) $$

Then simplify,

$$ x^r[a_2r(r-1)+a_1r+a_0]=0 \,$$ $$\displaystyle (Eq. 3.1.5) $$

$$ a_2r^2-a_2r+a_1r+a_0=0 \,$$ $$\displaystyle (Eq. 3.1.6) $$

$$ a_2r^2+(a_1-a_2)r+a_0=0 \,$$ $$\displaystyle (Eq. 3.1.7) $$

We then set Eqn. 3.1.7 equal to the expanded out version of the characteristic equation (Eqn. 3.3) to obtain the values of the coefficients,

$$ a_2r^2+(a_1-a_2)r+a_0=r^2-2\lambda r+\lambda^2 \,$$ $$\displaystyle (Eq. 3.1.8) $$

It is then clear that,

$$ a_2 = 1 \,$$ $$\displaystyle (Eq. 3.1.9) $$

$$ (a_1-a_2)= -2\lambda ; a_1 = a_2-2\lambda = 1-2\lambda \,$$ $$\displaystyle (Eq. 3.1.10) $$

$$ a_0=\lambda^2 \,$$ $$\displaystyle (Eq. 3.1.11) $$

Part 1.2
To find the first homogeneous solution use the trial solution (Eqn. 3.1.1) and the characteristic equation (Eqn. 3.3). We find the roots of Eqn. 3.3 and substitute the first (real) one in for $$r$$ in the trial solution (Egn. 3.1.1). In this particular case the two roots are the same so $$r_1=r_2=\lambda$$. Substituting this into Eqn. 3.1.1 we get,

$$ y_1=x^\lambda \,$$ $$\displaystyle (Eq. 3.1.12) $$

Part 1.3
Since $$\lambda$$ has two identical roots we can find the second homogeneous solution using the form,

$$ y_2(x)=U(x)y_1(x) \,$$ $$\displaystyle (Eq. 3.1.13) $$

We then calculate the first and second derivatives of Eqn 3.1.13. The $$(x)$$ argument will be ignored to simplify the equations.

$$ y_2'=U'y_1+Uy_1' \,$$
$$\displaystyle (Eq. 3.1.14) $$

$$ y_2=Uy_1+2U'y_1'+Uy_1'' \,$$
$$\displaystyle (Eq. 3.1.15) $$

Plug Eqns. 3.1.13, 3.1.14, 3.1.15 into Eqn 3.1 and simplify,

$$ a_2x^2(Uy_1+2U'y_1'+Uy_1)+a_1x(U'y_1+Uy_1')+a_0Uy_1=0 \,$$
$$\displaystyle (Eq. 3.1.16) $$

$$ U(a_2x^2y_1+a_1xy_1'+a_0y_1)+U'(2a_2x^2y_1'+a_1xy_1)+Ua_2x^2y_1=0 \,$$
$$\displaystyle (Eq. 3.1.17) $$

This can be simplified further noting that $$U$$ is multiplied by Eqn 3.1, which equals $$0$$. So that completely eliminates one term in Egn. 3.1.17. We then substitute in the calculated values of the $$a$$ coefficients and the values for $$y_1$$ from part 1.1 and simplify,

$$ U'(2a_2x^2y_1'+a_1xy_1)+U''a_2x^2y_1=0 \,$$
$$\displaystyle (Eq. 3.1.18) $$

$$ U'[2x^2\lambda x^{\lambda -1}+(1-2\lambda )xx^\lambda ]+U''x^2x^\lambda =0 \,$$
$$\displaystyle (Eq. 3.1.19) $$

$$ U'[2x^2\lambda x^{\lambda -1}+(1-2\lambda )xx^\lambda ]+U''x^2x^\lambda =0 \,$$
$$\displaystyle (Eq. 3.1.19) $$

$$ U'[2\lambda x^{\lambda +1}+x^{\lambda +1}-2\lambda x^{\lambda +1}]+U''x^{\lambda +2} =0 \,$$
$$\displaystyle (Eq. 3.1.20) $$

$$ U'x^{\lambda +1}+U''x^{\lambda +2} =0 \,$$
$$\displaystyle (Eq. 3.1.21) $$

$$ (U'+ U''x)x^{\lambda +1} =0 \,$$
$$\displaystyle (Eq. 3.1.22) $$

$$ U'+ U''x =0 \,$$
$$\displaystyle (Eq. 3.1.23) $$

We then use variation of parameters to reduce the order of Eqn. 3.1.23,

$$ m=U' \,$$
$$\displaystyle (Eq. 3.1.24) $$

So we substitute in the change of parameters into Egn. 3.1.24 and use separation of variables to solve for $$m$$,

$$ m+ m'x =0 \,$$
$$\displaystyle (Eq. 3.1.25) $$

$$ m + x \frac {dm} {dx}=0 \,$$
$$\displaystyle (Eq. 3.1.26) $$

$$ m = -x \frac {dm} {dx} \,$$
$$\displaystyle (Eq. 3.1.27) $$

$$ \frac {m} {dm} = - \frac {x} {dx} \,$$
$$\displaystyle (Eq. 3.1.28) $$

Then invert to get,

$$ \frac {dm} {m} = - \frac {dx} {x} \,$$
$$\displaystyle (Eq. 3.1.29) $$

Then integrate each side and simplify,

$$ ln (m) = - (ln (x) + k_1) \,$$
$$\displaystyle (Eq. 3.1.30) $$

$$ m = \frac 1 x e^{- k_1} \,$$
$$\displaystyle (Eq. 3.1.31) $$

$$ k_2 = e^{-k_1} \,$$
$$\displaystyle (Eq. 3.1.32) $$

$$ m = \frac 1 x k_2 \,$$
$$\displaystyle (Eq. 3.1.33) $$

Then substitute the identity $$m = U'$$ into Eqn. 3.1.33 and integrate to solve for $$U$$,

$$ U = \int U' dx = \int \frac 1 x k_2 dx \,$$
$$\displaystyle (Eq. 3.1.34) $$

$$ U = k_2 [ln(x) + k_3] \,$$
$$\displaystyle (Eq. 3.1.35) $$

Substitute $$U$$ back into Eqn. 3.1.13 along with the known value of $$y_1$$ to obtain the 2nd homogeneous solution,

$$ y_2 = k_2 [ln(x) + k_3]x^{\lambda} \,$$
$$\displaystyle (Eq. 3.1.36) $$

Part 1.4
The general homogeneous solution of the Euler L2-ODE-VC will have the form,

$$ y(x) = C_1 y_1(x)+C_2 y_2(x) \,$$
$$\displaystyle (Eq. 3.1.37) $$

Plugging Eqns. 3.1.36 and 3.1.12 into Eqn. 3.1.37 and simplifying results in the general homogeneous solution,

$$ y(x) = C_1 x^{\lambda} + C_2 [k_2 [ln(x) + k_3]x^{\lambda}] \,$$
$$\displaystyle (Eq. 3.1.38) $$

$$ y(x) = x^{\lambda}[C_1 + C_2 k_2 ln(x) + C_2 k_3] \,$$
$$\displaystyle (Eq. 3.1.39) $$

$$ C_3 = C_1 + C_2 k_3 \,$$
$$\displaystyle (Eq. 3.1.40) $$

$$ C_4 = C_2 k_2 \,$$
$$\displaystyle (Eq. 3.1.41) $$

Therefore the general homogeneous solution simplifies to,

$$ y(x) = x^{\lambda}[C_4 + C_5 ln(x)] \,$$
$$\displaystyle (Eq. 3.1.42) $$

Part 2.1
The coefficients of Eqn. 3.2 (b terms) can be found if we use the trial solution method. The trial solution for an Euler L2-ODE-CC is: $$ y=e^{rx} \,$$
$$\displaystyle (Eq. 3.2.1) $$

Therefore we can calculate the first and second derivatives of the trial solution:

$$ y'=re^{rx} \,$$
$$\displaystyle (Eq. 3.2.2) $$

$$ y''=r^2e^{rx} \,$$
$$\displaystyle (Eq. 3.2.3) $$

Then substitute back into Egn. 3.2 so that it becomes,

$$ b_2r^2e^{rx} + b_1re^{rx} + b_0e^{rx}=0 \,$$
$$\displaystyle (Eq. 3.2.4) $$

Then simplify,

$$ e^{rx}[b_2r^2+b_1r+b_0]=0 \,$$
$$\displaystyle (Eq. 3.2.5) $$

$$ b_2r^2+b_1r+b_0=0 \,$$
$$\displaystyle (Eq. 3.2.6) $$

We then set Eqn. 3.2.6 equal to the expanded out version of the characteristic equation (Eqn. 3.3) to obtain the values of the coefficients,

$$ b_2r^2+b_1r+b_0=r^2-2\lambda r+\lambda^2 \,$$
$$\displaystyle (Eq. 3.2.7) $$

It is then clear that,

$$ b_2 = 1 \,$$
$$\displaystyle (Eq. 3.2.8) $$

$$ b_1= -2\lambda \,$$
$$\displaystyle (Eq. 3.2.9) $$

$$ b_0=\lambda^2 \,$$
$$\displaystyle (Eq. 3.2.10) $$

Part 2.2
To find the first homogeneous solution use the trial solution (Eqn. 3.2.1) and the characteristic equation (Eqn. 3.3). We find the roots of Eqn. 3.3 and substitute the first (real) one in for $$r$$ in the trial solution (Egn. 3.2.1). In this particular case the two roots are the same so $$r_1=r_2=\lambda$$. Substituting this into Eqn. 3.2.1 we get,

$$ y_1=e^{\lambda x} \,$$
$$\displaystyle (Eq. 3.2.11) $$

Part 2.3
Since $$\lambda$$ has two identical roots we can find the second homogeneous solution using the form,

$$ y_2(x)=U(x)y_1(x) \,$$
$$\displaystyle (Eq. 3.2.12) $$

We then calculate the first and second derivatives of Eqn 3.2.12. The $$(x)$$ argument will be ignored to simplify the equations.

$$ y_2'=U'y_1+Uy_1' \,$$
$$\displaystyle (Eq. 3.2.13) $$

$$ y_2=Uy_1+2U'y_1'+Uy_1'' \,$$
$$\displaystyle (Eq. 3.2.14) $$

Plug Eqns. 3.2.12, 3.2.13, 3.2.14 into Eqn 3.2 and simplify,

$$ b_2(Uy_1+2U'y_1'+Uy_1)+b_1(U'y_1+Uy_1')+b_0Uy_1=0 \,$$
$$\displaystyle (Eq. 3.2.15) $$

$$ U(b_2y_1+b_1y_1'+b_0y_1)+U'(2b_2y_1'+b_1y_1)+Ub_2y_1=0 \,$$
$$\displaystyle (Eq. 3.2.16) $$

This can be simplified further noting that $$U$$ is multiplied by Eqn 3.2, which equals $$0$$. So that completely eliminates one term in Egn. 3.2.16. We then substitute in the calculated values of the $$a$$ coefficients and the values for $$y_1$$ from part 2.1 and simplify,

$$ U'(2b_2y_1'+b_1y_1)+U''b_2y_1=0 \,$$
$$\displaystyle (Eq. 3.2.17) $$

$$ U'[2\lambda e^{\lambda x}+(-2\lambda )e^{\lambda x}]+U''e^{\lambda x} =0 \,$$
$$\displaystyle (Eq. 3.2.18) $$

$$ U''e^{\lambda x} =0 \,$$
$$\displaystyle (Eq. 3.2.19) $$

Therefore

$$ U'' =0 \,$$
$$\displaystyle (Eq. 3.2.20) $$

Then Integrate Eqn. 3.2.20 twice with respect to $$x$$

$$ U' = k_1 \,$$
$$\displaystyle (Eq. 3.2.21) $$

$$ U = x k_1 + k_2 \,$$
$$\displaystyle (Eq. 3.2.22) $$

Then plug Eqn. 3.2.22 and 3.2.1 into Eqn. 3.2.12 to get,

$$ y_2(x) = e^{\lambda x} (x k_1 + k_2) \,$$
$$\displaystyle (Eq. 3.2.23) $$

Part 2.4
The general homogeneous solution of the Euler L2-ODE-CC will have the form,

$$ y(x) = C_1 y_1(x)+C_2 y_2(x) \,$$
$$\displaystyle (Eq. 3.2.24) $$

Plugging Eqns. 3.2.23 and 3.2.12 into Eqn. 3.2.24 and simplifying results in the general homogeneous solution,

$$ y(x) = C_1 e^{\lambda x} + C_2 e^{\lambda x} (x k_1 + k_2) \,$$
$$\displaystyle (Eq. 3.2.25) $$

$$ y(x) = e^{\lambda x}[C_1 + C_2 + k_2] + C_2 k_1 x e^{\lambda x} \,$$
$$\displaystyle (Eq. 3.2.26) $$

$$ C_3 = C_1 + C_2 + k_3 \,$$
$$\displaystyle (Eq. 3.2.27) $$

$$ C_4 = C_2 k_1 \,$$
$$\displaystyle (Eq. 3.2.28) $$

Therefore the general homogeneous solution simplifies to,

$$ y(x) = C_3 e^{\lambda x} + C_4 x e^{\lambda x} \,$$
$$\displaystyle (Eq. 3.2.29) $$

Author and Proof-Reader
[Author] Rob Carroll

[Proof-Reader] Oh, Sang Min

= Problem 4 - Find particular solution= From Meeting 27, p 27-2.

Given
We are given the differential equation,
 * $$\displaystyle y'+P(x)y=Q(x)$$

where the homogeneous solution has been previously found to be,
 * $$\displaystyle y_h=Ae^{-\int{P(x)dx}}$$

Find
We are to find a particular solution, $$y_p$$, by variation of parameters.

Solution
We first let
 * $$\displaystyle y(x)=u(x)y_h$$

By taking the derivative we find,
 * $$\displaystyle y'(x)=e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)]$$

We substitution this into the given equation,
 * $$\displaystyle e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)]+P(x)u(x)e^{-\int{P(x)dx}}=Q(x)$$

Rearranging yields,
 * $$\displaystyle e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)+u(x)P(x)]=Q(x)$$

Which then simplifies to,
 * $$\displaystyle e^{-\int{P(x)dx}}u'(x)=Q(x)$$

We solve for $$u'(x)$$,
 * $$\displaystyle u'(x)=e^{\int{P(x)dx}}Q(x)$$

Integration yields,
 * $$\displaystyle u(x)=\int{e^{\int{P(x)dx}}Q(x)dx}$$

Keeping in mind that $$y(x)=u(x)y_h$$ and $$y(x)=y_p(x)+y_h(x)$$, we use $$u(x)$$ to find the particular solution,
 * $$\displaystyle y_p(x)=e^{-\int{P(x)dx}}\int{e^{\int{P(x)dx}}Q(x)dx}$$

Author and Proof-Reader
[Author]
 * Raul Riveros

[Proof-Reader]

= Problem 5 - Non-homog. L2-ODE-CC = From Meeting 28, p 28-2 to Meeting 29, p 29-1.

Given
Non-homogeneous L2-ODE-CC

Find
1)  Find PDEs for integrating factor $$ h(t,y) \! $$

2.1) Find $$ \bar{a}_1 $$ and $$ \bar{a}_0 $$ in terms of $$ a_2, a_1, a_0 \! $$

2.2) Find quadratic equation for $$ \alpha \!$$

2.3) Show reduced order equation using integrating factor

2.4) Use IFM to find $$ y \! $$ for general excitation $$ f(t) \! $$

2.5) Show that $$ \alpha \beta = \frac{a_0}{a_2} $$ and $$ \alpha + \beta = \frac{a_1}{a_2} $$

2.6) Deduce expression for particular solution for general excitation $$ f(t) \!$$

2.7) Verify with table of particular solutions considering $$ f(t) = t^2 \! $$

2.8) Solve with $$ f(t) = exp(-t^2) \! $$. Find coefficients $$ a_2, a_1, a_0 \! $$ such that (5.1) has the following characteristic equations:

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

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

Part 1
We can rearrange Eqn. 5.1 into the following form:

When in this form, it is apparent that the equation is exact since it can be represented as

where

The 2nd condition of exactness involves two relations which are as follows:

Computing the partial derivatives for (5.1.5) and (5.1.6),

$$ f_{tt} = f_{ty} = f_{yy} = f_{yp} = f_{y} = 0 \! $$

$$ g_{tp} = g_{yp} = g_{pp} = 0 \! $$

$$ g_{y} = a_0 \! $$

In order for (5.1.5) to be satisfied, $$ a_0 \! $$ must be zero.

$$ (5.1.3) \Rightarrow \phi(t,y,p) = h(t,y) + \int a_2 dp \! $$

$$ \phi_t = h_t \! $$

$$ \phi_y = h_y \! $$

Substituting these results into (5.1.4) yields

From (5.1.7) we can solve for $$ h \! $$ to arrive at

Part 2
Let $$ h(t) = exp(\alpha t), \alpha \in \Re \! $$.

Multiplying both sides of Eqn. 5.1 and integrating yields

Part 2.1
Assume that the LHS of (5.2.1) will be of the following form

We set $$ \bar{a}_2 = 0 $$ to reduce the order, differentiate (5.2.2) and set it equal to the integrand of (5.2.1) as follows

$$ \frac{d}{dt}\Big[exp(\alpha t)(\bar{a}_2y+\bar{a}_1y'+\bar{a}_0y) \Big] = exp(\alpha t)\Big(\bar{a}_1 y + (\alpha \bar{a}_1+\bar{a}_0)y' + \alpha \bar{a}_0 y \Big) = exp(\alpha t)\big[a_2y''+a_1y'+a_0y\big] \! $$

The remaining coefficients of (5.2.2) are thus given by

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

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

Part 2.2
With

$$ a_1 - \alpha a_2 = a_0/\alpha \! $$

we can rewrite it in quadratic form by multiplying both sides by $$ \alpha \! $$ to result in the following

Part 2.3
Returning now to (5.2.1), we substitute (5.2.2) as the result of the LHS integration yielding the following reduced order equation

Dividing through by $$ exp(\alpha t) \! $$ and then by $$ \bar{a}_1 $$ results in the following

Part 2.4
With the L1-ODE-CC found in (5.2.5), we choose the integrating factor as

$$ h(t) = exp \Big[ \int \frac{\bar{a}_0}{\bar{a}_1} dt \Big] = e^{\beta t} \! $$

And since we have (5.2.5) in the correct form (see (2) p. 10-3), we can solve for y using (6) on p. 10-3

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

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

Part 2.6
Eqn. (5.2.6) is actually the particular solution of the L2-ODE-CC given. Since $$ \alpha $$ and $$ \beta $$ are the roots of the characteristic equation, the complementary solution (or homogeneous solution) is found to be

Because the general solution is the linear combination of the homogeneous solution and particular solution, it is as follows

Part 2.7
If we use the trial solution lookup table for a particular solution if $$ f(t) = t^2 $$, we come up with a trial solution of

where $$ C_2 $$,$$ C_1 $$ and $$ C_0 $$ are undetermined constant coefficients.

To compare this particular solution to the one found in (5.2.6), we substitute $$ f(t) = t^2 $$ into (5.2.6) and find that

The result shown in (5.2.10) is clearly the same form as the trial solution in (5.2.9).

Part 2.8
We will now solve (5.1) for the forcing function $$ f(t) = exp(-t^2) $$.

Part 2.8.1
To begin with, since the characteristic equation is $$ (r+1)(r-2) $$, we know that the homogeneous solution is

and we know that the coefficients in (5.1) are

$$a_2 = 1 \!$$

$$a_1 = -1 \!$$

$$a_0 = -2 \!$$

Our integrating factor (p. 29-3) becomes

Now we turn to (2) p. 30-1 to find the particular solution

Part 2.8.2
To begin with, since the characteristic equation is $$ (r-4)^2 $$, we know that the homogeneous solution is

and we know that the coefficients in (5.1) are

$$a_2 = 1 \!$$

$$a_1 = -8 \!$$

$$a_0 = 16 \!$$

Our integrating factor (p. 29-3) becomes

Now using (2) p. 30-1 again to find the particular solution

Author and Proof-Reader
[Author] Michael Steinberg

[Proof-Reader]

= Problem 6 - Show that lecture agree with the book = From Meeting 30, p 30-1.

Given
Particular Solution from meeting 30.

$$\displaystyle y_p(x) = u_1(x)\int{\frac{1}{h(x)}\left(\int{h(x)f(x)\,dx}\right)\,dx} $$
$$\displaystyle (Eq. 6.1) $$

Particular solution from King p. 8 (1.6)

$$\displaystyle y_p(x) = \oint{f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds} $$
$$\displaystyle (Eq. 6.2) $$

Where $$\displaystyle W(x)$$ is the Wronskian defined as

$$\displaystyle W(x)=u_1(x)\,u_2(x)'-u_2(x)\,u_1(x)' $$
$$\displaystyle (Eq. 6.3) $$

Hint from meeting 30.

$$\displaystyle \frac{1}{h(x)}=\left( \frac{u_2(x)}{u_1(x)} \right)' $$
$$\displaystyle (Eq. 6.4) $$

Find
Show that 6.1 agrees with 6.2

Solve
Using quotient rule on equation 6.4 to solve for $$\displaystyle h(x)$$.

$$\displaystyle d\left(\frac{u}{v}\right)=\frac{v\,du-u\,dv}{v^2} $$

$$\displaystyle d\left(\frac{u_2(x)}{u_1(x)}\right)=\frac{u_1(x)\,u_2(x)'-u_2(x)\,u_1(x)'}{u_1(x)^2} $$
$$\displaystyle (Eq. 6.5) $$

Inverse of equation 6.5 solves for $$\displaystyle h(x)$$.

$$\displaystyle h(x)=\frac{u_1(x)^2}{u_1(x)\,u_2(x)'-u_2(x)\,u_1(x)'} $$
$$\displaystyle (Eq. 6.6) $$

Using integration by parts on equation 6.1

$$\displaystyle \int{u\,dv}=u\,v-\int{v\,du} $$

$$\displaystyle dv=\frac{1}{h(x)}=\left(\frac{u_2(x)}{u_1(x)}\right)' $$

$$\displaystyle u=\left(\int{h(x)f(x)\,dx}\right) $$

$$\displaystyle y_p(x) = u_1(x)\left\{\frac{u_2(x)}{u_1(x)} \left(\int{h(s)f(s)\,ds}\right)-\int{h(s)f(s)\frac{u_2(s)}{u_1(s)}\,ds}             \right\} $$
$$\displaystyle (Eq. 6.7) $$

Subsitute for h(s) as defined by equation 6.6

$$\displaystyle y_p(x) = u_1(x)\left\{\frac{u_2(x)}{u_1(x)} \left(\int{\frac{u_1(s)^2}{u_1(s)\,u_2(s)'-u_2(s)\,u_1(s)'}f(s)\,ds}\right)-\int{\frac{u_1(s)^2}{u_1(s)\,u_2(s)'-u_2(s)\,u_1(s)'}f(s)\frac{u_2(s)}{u_1(s)}\,ds}             \right\} $$
$$\displaystyle (Eq. 6.8) $$

Simplify equation 6.8

$$\displaystyle y_p(x) = \int{f(x)\frac{u_2(x)u_1(s)^2}{W(s)} \,ds}-\int{f(x)\frac{u_1(x)u_1(s)u_2(s)}{W(s)} \,ds}=\int{\frac{u_2(x)u_1(s)^2-u_1(x)u_1(s)u_2(s)}{W(s)}\,ds} $$
$$\displaystyle (Eq. 6.9) $$

Equation 6.9 does not equal equation 6.2 from King. This suggest there is either misprint in King or course notes.

If equation 6.1 was $$\displaystyle  y_p(x) = u_1(x)\int{\frac{1}{h(x)}\left(\int{\frac{h(x)f(x)\,dx}{u_1(x)}}\right)\,dx}$$ and solved similarly as shown above would be equal to equation 6.2.

Author and Proof-Reader
[Author] Mike Faraone

[Proof-Reader]

= Problem 7 - Finding two homog. solutions = From Meeting 31, p 31-1.

Given

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

$$\left( x+1 \right)y''-\left( 2x+3 \right)y'+2y=0$$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.1)$$
 * }

Trial solution :
 * {| style="width:100%" border="0"

$$\displaystyle y= {e}^{rx} $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.2)$$
 * }

Find
$$ u_1\ $$ and $$ u_2\ $$ of (7.1) using trial solution (7.2)

Solve
(7.2) ->
 * {| style="width:100%" border="0"

$$\displaystyle y'=r{{e}^{rx}}$$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.3)$$
 * }

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

$$\displaystyle y''={{r}^{2}}{{e}^{rx}}$$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.4)$$
 * }

(7.1) substitute with (7.3) and (7.4)
 * {| style="width:100%" border="0"

$$\displaystyle (x+1)r^2{e}^{rx} - (2x+3)r{e}^{rx} + 2{e}^{rx} = 0$$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.5)$$
 * }

->
 * {| style="width:100%" border="0"

$$\displaystyle {e}^{rx} ((x+1)r^2 - (2x+3)r + 2) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.6)$$
 * }

Using basic knowledge of Quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.7)$$
 * }

We can get two solutions about r

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

$$\displaystyle {r}_{1} = 2 \, \  {r}_{2}=  \frac{1}{ \left( x+1\right)} $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.8)$$
 * }

We choose $$r_1 $$ because it is constant
 * {| style="width:100%" border="0"

$$\displaystyle {u}_{1}(x) = {e}^{2x} $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.9)$$
 * }

Using (1) p.30-1

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

$$\displaystyle {u}_{2}(x)= \underbrace{{u}_{1}(x)}_{{e}^{2x}} \int \frac{1}{ u_1^2(x)}exp[-\int \underbrace{a_1(x)}_{\frac{-(2x+3)}{x+1}} \ dx] \ dx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$\displaystyle \ = {e}^{2x} \int \frac{1}{ {e}^{4x}}exp[\underbrace{\int \frac{2x+3}{x+1} \ dx}_{2x+log(x+1)}] \ dx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$\displaystyle \ = {e}^{2x} \int \frac{x+1}{{e}^{2x}} \ dx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Using wolfram mathematica

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

$$\displaystyle {u}_{2}(x) = - \frac{1}{4} (2x+3) $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (7.10)$$
 * }

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

$$ u_1 = {e}^{2x}, \ u_2 = - \frac{1}{4} (2x+3) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * }
 * }
 * }

Author and Proof-Reader
[Author] Oh, Sang Min

[Proof-Reader] Mike Faraone

= Problem 8 - Finding L2-ODE-VC(homog.) = From Meeting 31, p 31-3.

Given
Trial solution

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

$$\displaystyle y = \frac{{e}^{rx}}{x^2} $$
$$\displaystyle (8-1) $$
 * }
 * }

Character equation

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

$$\displaystyle r^2 + 3 = 0 $$
$$\displaystyle (8-2) $$
 * }
 * }

Find
L2-ODE-VC

Solve

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

$$\displaystyle y = \frac{{e}^{rx}}{x^2} $$
$$\displaystyle (8-3) $$
 * }
 * }

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

$$\displaystyle y' = \frac{r{e}^{rx}}{x^2} - \frac{2{e}^{rx}}{x^3} = \frac{{e}^{rx}}{x^2} (r - \frac{2}{x}) $$
$$\displaystyle (8-4) $$
 * }
 * }

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

$$\displaystyle y'' = \frac{{e}^{rx}}{x^2} {(r - \frac{2}{x})(r - \frac{2}{x}) + \frac{2}{x^2}} = \frac{{e}^{rx}}{x^2} (r^2 - \frac{4}{x}r + \frac{6}{x^2}) $$
$$\displaystyle (8-5) $$
 * }
 * }

 Using Reverse Eng.

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

$$\displaystyle \frac{{e}^{rx}}{x^2} [a_2(r^2 - \frac{4}{x}r + \frac{6}{x^2}) + a_1(r - \frac{2}{x}) + a_0] = \frac{{e}^{rx}}{x^2} (r^2 + 3) = 0 $$
$$\displaystyle (8-6) $$
 * }
 * }

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

$$\displaystyle a_2 r^2 + (a_1 - \frac{4a_2}{x})r + (\frac{6a_2}{x^2} - \frac{2a_1}{x} + a_0) = r^2 + 3 $$
$$\displaystyle (8-7) $$
 * }
 * }

Therefore

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

$$\displaystyle a_2 = 1, \ a_1 = \frac{4}{x}, \ a_0 = 3 + \frac{2}{x^2} $$
$$\displaystyle (8-8) $$
 * }
 * }

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

$$ y'' + \frac{4}{x} y' + (3 + \frac{2}{x^2})y = 0 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * }
 * }
 * }

Author and Proof-Reader
[Author] Oh, Sang min

[Proof-Reader] Mike Faraone

= References =

= Contributing members =

Jo : Authored Problem 1 and Problem 2.

Rob Carroll : Authored Problem 3

Oh, Sang Min : : Authored Problem 7 and Problem 8, proof-read problem 1, 2 and 3

Michael Faraone : Authored Problem 6, Proofed problems 7 and 8.

Michael Steinberg : Authored Problem 5

title: User:Egm6321.f10.team5/hw5


text:
User:Accelerometer/Portal mathematics

DRAFT

 * Not to be confused with Portal:Mathematics, the Wikiversity mathematics portal.

Portal mathematics is a recreational form of mathematics and geometry that uses "portals," imaginary objects that cause moving objects to be "teleported" on entry. This allows two points in space to be closer together than in "portal-less" space.

In fact, this leads to the First Axiom of Portals:

 * The shortest distance between two points is a pair of portals.

Lesson 1: Arithmetic on the Number Line
Most of us are already familiar with the concepts of addition and subtraction. In portal mathematics, it's essential to redefine these two operations on the number line to make portal mathematics easier to understand.

(Currently, I'm not sure how to implement multiplication and division. Help would be appreciated.)

Addition
In number line addition, we take the following steps to perform the operation $$x + y$$:
 * We place an ant on number $$x$$ on the number line.
 * The ant walks $$y$$ steps in the positive direction on the number line (the ant may actually walk the opposite direction if $$y$$ is negative.)
 * After the ant is done walking, the number on which he stands is $$x + y$$.

For example, to calculate 2 + 3, we start by placing an ant on number 2: ant <--+++++++++-->  0    1    2    3    4    5    6    7    8

The ant walks three spaces to the right (in the positive direction):

Step 1: ant---| <--+++++++++-->  0    1    2    3    4    5    6    7    8

Step 2: ant---| <--+++++++++-->  0    1    2    3    4    5    6    7    8

Step 3: ant---| <--+++++++++-->  0    1    2    3    4    5    6    7    8

And we end up like this:

ant <--+++++++++-->  0    1    2    3    4    5    6    7    8

The ant is on number 5, so 2 + 3 = 5.

But what about 3 + 2? We won't go through all the steps above again to figure out that 3 + 2 = 5. Although both 2 + 3 = 5 and 3 + 2 = 5 they use the same set of arguments, they are different as far as the ant is concerned:

 * In 2 + 3, 2 indicates a position on the number line—specifically, where the ant starts.
 * In 3 + 2, 2 indicates a distance and direction on the number line that the ant must walk. Direction is indicated by the sign of the number.

Since the first and second numbers of an arithmetic problem represent data with different purposes, we'll use different notation to indicate position and distance/direction:

 * $$\{x\}$$ indicates position $$x$$, and
 * $$[x]$$ indicates $$x$$ steps in the positive direction.

($$\{x\}$$ and $$[x]$$ are in fact one-dimensional coordinates and vectors, respectively.)

To denote addition, we use the $$\rightarrow$$ sign between the position and the distance/direction indicators, in that order. For example, $$\{2\}\rightarrow[3] = \{5\}$$.

Subtraction
In number line subtraction, we take the following steps to perform the operation $$x - y$$:
 * We place an ant on number $$x$$ on the number line.
 * The ant walks $$y$$ steps in the negative direction on the number line (the ant may actually walk the opposite direction if $$y$$ is negative.)
 * After the ant is done walking, the number on which he stands is $$x - y$$.

Seeing that $$x + y$$ = $$x + -y$$, we can extrapolate our $$\{x\}\rightarrow[y]$$ notation to allow subtraction:

 * $$\{x\}\leftarrow[y] = \{x\}\rightarrow[-y]$$

Lesson 2: Portals in One Dimension
In the above section, you probably thought I was just pulling your leg. Perhaps I was just re-enacting first-grade mathematics while introducing complicated and useless notation just to sound and look smart.

Portals 101
Suppose our ant is at the 0 mark and wishes to advance to the 7 mark as shown, but wants to walk a shorter distance in order to get there.

ant                                | <--+++++++++-->  0    1    2    3    4    5    6    7    8

We can add two points called portals. Portals are extremely difficult to define without examples, so we'll begin with one.

ant                                | <--++]+[++++-->  0    1    2    3    4    5    6    7    8

Here, ] indicates a left-facing portal, and [ right-facing. I'll explain what this means in a moment.

Now our ant walks to the 2 mark:

ant---|                            | <--++]+[++++-->  0    1    2    3    4    5    6    7    8

ant---|                       | <--++]+[++++-->  0    1    2    3    4    5    6    7    8

ant      ant---|         | <--++]+[++++-->  0    1    2    3    4    5    6    7    8

Notice that when the ant is on the 2 mark and attempts to move to the right, it emerges from the portal on the 4 mark.

Thus the ant only has to move a distance of 5, not 7, to go from 0 to 7. In other words, $$\{0\}\rightarrow[5] = \{7\}$$ in this portal system. Note that $$\{0\}\rightarrow[5] = \{7\}$$ clearly does not hold on the ordinary number line, since 0 + 5 is certainly not 7!

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Accelerometer/Portal mathematics


text:
Talk:Ideas in Geometry/Instructive examples/Section 2.4:

Hey, I think you did a really good job at explaining both the picture and the steps needed to figure out how to solve the problem. I think it was very helpful that you talked about two different methods of finding the solution so people who do not understand will be able to pick the way that works best for them. Also your explanations were easy to follow. Nice job!

I think you did a great job explaining yourself in terms of the picture and made it clear what the question was asking for. I understood completely how to do the problem just from your explanation.

The picture was very neat and practical, but is there a reason why the formula works? I understand that it does work, but how and why? Is there a mathematical rule or theorum that can be used? Rehus1 01:58, 3 November 2010 (UTC)


text:
Talk:Ideas in Geometry/Instructive examples/Section 1.1

Great job explaining this problem. It was very easy to follow. I actually drew it out as you instructed to ensure you worded the problem clearly enough to follow. The only mistake I think you made was calling the new segment to draw in "AB" when you meant to put in "BC." Good Work!

I agree with the person above me. I must say you did a very good job at explaining the material through your writing

I thought you did a very nice job explaining how to make a equilateral triangle with the given conditions. The explanation was pretty clear and your steps easily led to the solution. It would be even better if you included a picture/diagram along with your explanation like you said! And yes, the only mistake was the confusion on putting "AB" when it should be "BC." But great job!

Michelle Hur


text: Template:NumBlk

title: Template:NumBlk


text:
Template:NumBlk/doc

This template is copied from w:Template:NumBlk and is not meant for independent development.

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 *  : The body or content of the block.
 *  : Specify the block number.
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 *  : Specify the line style.
 *  : If set, put a box around the equation. (Experimental.)

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title: Template:NumBlk/doc


text:
User talk:Accelerometer

Welcome
 Hello Accelerometer, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. Cirt (talk) 02:04, 29 October 2010 (UTC)

-- Cirt (talk) 02:04, 29 October 2010 (UTC)

title: User talk:Accelerometer


text:
Ideas in Geometry/Instructive examples/Section 1.2 problem

y=x^20-23x+4 at point x=0 f(x)=x^20-23x+4 f(x) is the slope of tangent line of this equation f(x)=20x^19-23 through calculus we that f(x)=x^a+....... (f^/)(x)=(a(x)^a-1)+...... f^/(x)is the slope of f(x) that is proven

f(x)=20x^19-23 at x=0 f(x)=(20(0)^19)-23 f(x)=-23 means the slope is -23 at x=0 y=0^20-23(0)+4   y=4

y=mx+b tangent line equation y=4 m=-23 x=0 4=-23(0)+b  4=b

so the equation is...... y=-23x+4 of the tangent line


text:
User:Osilion

This is Osilion's (my) user page.

title: User:Osilion


text:
User talk:K thakkar

 Hello K thakkar, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:17, 29 October 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 29 October 2010 (UTC)

title: User talk:K thakkar


text:
Template:EquationRef

'''This template is copied from w:Template:EquationRef and is not intended for independent development. '''

Use with Template:EquationNote.

title: Template:EquationRef


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text: Template:Str len

title: Template:Str len


text:
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title: Template:Str len/core


text:
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Use with Template:EquationRef.

PLEASE READ Template talk:EquationNote before modifying this template or importing the latest version from Wikipedia.

title: Template:EquationNote


text:
User:Egm6321.f10.team4.Yoon/HW5

= Problem 1 -  Solving multiple y derivative wrt x =

From

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

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 deribatives 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

= Problem 10: Method of Trial Solutions =

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
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

required

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

y(x) = \sum^{n}_{i=1} C_ix^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * }

Find
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
 *
$$\displaystyle (Eq. 2-4)
 * }
 * }

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

y=x^r $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-5)
 * }
 * }

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

y'=rx^{r-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-6)
 * }
 * }

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

y''=r(r-1)x^{r-2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\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
 *
$$\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
 *
$$\displaystyle (Eq. 2-9)
 * }
 * }

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

r^2- 3r + 2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\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
 *
$$\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
 *
$$\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
 *
$$\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
 *
$$\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
 *
$$\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
 *
$$\displaystyle (Eq. 2-16)
 * }
 * }

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

2C_1 + 4C_2 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\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
 *
$$\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" |

$$ \displaystyle y(x) = - \frac{13}{2}x + \frac{9}{2}x^2 $$ $$
 *
$$\displaystyle (Eq. 2-19)
 * }
 * }

Given
The particular solution for a L2-ODE-VC developed in class during Mtg 30 is given as:
 * {| style="width:100%" border="0"

$${{y}_{p}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}$$ Where $$h(x)$$ is equal to:
 *
$$\,(Eq. 6.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h(x)=\frac{{{u}_{1}}^{2}(x)}{{{u}_{2}}^{\prime }(x)\cdot {{u}_{1}}(x)-{{u}_{2}}(x)\cdot {{u}_{1}}^{\prime }(x)}$$
 *
$$\,(Eq. 6.2) $$
 * style= |
 * }

title: User:Egm6321.f10.team4.Yoon/HW5


text:
User talk:Henoch~enwikiversity

psychology
psychology

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title: User talk:Henoch~enwikiversity


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User talk:Paulmnguyen

Welcome to my talk page!

title: User talk:Paulmnguyen


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User talk:Simpliciodesilva

 Hello Simpliciodesilva, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User:Ha98574/Railroad/Korea/Connecting


text:
Talk:Ideas in Geometry/Instructive examples/Section 3.2 Problem 7-Using Brahmagupta's Formula

I like how you explained the problem in a very clear and easy to understand way. By explaining it in this way, you made it easy for everyone to follow. One suggestion that I might recommend would be to possibly switch up the format of the explanation, into possibly a step-by-step arrangement so that it might be a little easier to read. Other than that small suggestion though, the explanation is great and straightforward, which makes it very understandable to the average reader or student. Good Job!

TMcCormick31 21:18, 29 October 2010 (UTC)


text:
Talk:Ideas in Geometry/Instructive examples/9 from 2.1, Proof by Picture using Venn Diagrams

This is really good. This was a section that I actually had some trouble with and your explanations really helped. The only suggestions I really have for you are to try and downsize your picture and figure out how to insert the actual U and n symbols. U and N can get confusing when you are trying to memorize the symbols, but that is truly just a minor (really minor) detail. Downsizing the picture would help your readers see the entire project as a whole and allow them to better understand your explanations because they would be able to see it as they read the explanations. As I said these are really just minor adjustments, you did a great job! -Katelin Barker-

2.1.9
I think you did a really good job of explaining your answer well. The pictures definitely make everything clear as well. You have good step by step explanations. I really don't think there is much more you can add to this. All of your explanations were clear and concise. Overall, great job! --Christine Yu 06:44, 30 October 2010 (UTC)


title: File:Homework5a.png


title: File:Homework5b.png


text:
AP European History at GRCHS/The First Quarter

2A
I.	Age of Exploration:  A time when Europeans were trying to find a new route to Asia

 * a.	Reasons to go… God, Gold, & Glory

 * i.	God - people wanted to convert the newly discovered peoples to Christianity
 * ii.	Gold
 * iii.	Glory
 * iv.	The Realities

 * v.	Early Leaders in Exploration Potugal + Spain
 * b.	Technology of Exploration
 * i.	Cannon
 * ii.	Caravels over Galleys
 * iii.	Sails
 * iv.	Compass
 * v.	Astrolabe
 * c.	European Sponsorship

 * i.	In return for sponsorship

II. Exploration & Discovery:
 * a.	Portugal & Prince Henry the Navigator (1394-1460)

 * i.	Bartholomew Dias (1487)
 * ii.	Vasco da Gama (1497-99)
 * iii.	Pedro Cabral (1500-01)
 * b.	Spain & Ferdinand and Isabella
 * i.	Christopher Columbus (1492)

 * ii.	Ferdinand Magellan (1519-1521)

 * iii.	Hernando Cortes (1519-1522)
 * iv.	Francisco Pizarro (1531-1536)
 * c.	England & the British East/West India Company

 * i.	John Cabot (1497)
 * ii.	Sir Francis Drake (1577-1580)

 * iii.	Joint Stock Companies

 * a.	Virginia & London Companies
 * b.	Cash Crops
 * d.	France
 * i.	Jacques Cartier (1534-1541)
 * e.	Netherlands & the Dutch East/West India Company

 * f.	Golden Century of Spain was a time when the Spanish controlled much of the New World and was a powerful force in Europe.

 * g.	Price (Commercial) Revolution (16th Century)

III. Columbian Exchange:

 * a.	Definition
 * b.	From the New World
 * c.	From the Old World
 * d.	Triangle Trade & the Atlantic Economy
 * i.	From EUR
 * ii.	From AFR
 * a.	Bartolommeo de las Casas (1515)

 * b.	Atlantic Slave Trade

 * iii.	From AME
 * iv.	From EUR

IV. Roots of the Reformation (1400-1517):

 * a.	Secularism

 * b.	Simony
 * c.	Pluralism
 * d.	Indulgences

 * e.	Immorality

 * i.	Alexander VI (1492-1503)
 * ii.	Julius II (1503-1513)

 * iii.	Leo X (1513-1521)
 * iv.	Clement VII (1523-1534)
 * f.	Early Reformers

 * i.	Heretics, Excommunication, & Execution

V.	The Reformation:

VI. The Protestant Reformation:

 * a.	Martin Luther (1483-1546)

 * b.	Ninety-Five Theses (1517)
 * i.	It Said

 * ii.	His Intentions

 * c.	Luther & his Doctrine

 * i.	Faith Alone
 * ii.	Interpret the Bible
 * iii.	Two Sacraments
 * iv.	Scripture Alone (sola scriptura)
 * v.	Consubstantiation
 * a.	Transubstantiation
 * d.	Leo X & the Summons to Augsburg (1519)
 * i.	Luther thinks
 * ii.	The Cardinals think
 * iii.	In the end

 * e.	Prince Frederick III of Saxony (1520)

 * i.	German Princes, Luther, & the Politics of the Reformation

 * f.	Charles V, Luther, & the Pope

 * g.	Diet of Worms (1521)

 * i.	The problem

VII. Reaction to Luther’s Ideas in Germany:
 * a.	Clergy

 * b.	Nobles
 * i.	Diet of Speyer (1526)

 * ii.	Diet of Speyer & Protestatio (1529)

 * c.	Peasants & “Christian Freedom”

 * i.	On Christian Liberty (1520) & An Admonition to Peace (1525)

 * ii.	German Peasant Revolt/Swabian Peasant Revolt (1524-1525)

 * iii.	Their Demands in the Twelve Articles
 * iv.	Luther’s Response in the Against the Murdering, Thieving Hordes of Peasants

 * a.	Why did Luther side with the German Princes?

 * d.	Believers, the Augsburg Confession, & the Lutheran Church (1530)

 * e.	Charles V & his Lackluster Response
 * i.	Habsburg-Valois Wars

 * ii.	Ottomans Turks & the 1530’s

 * iii.	Schmalkaldic League & War (1546-1547)

 * iv.	Peace of Augsburg (1555)

 * v.	Death of Charles V (1556)
 * a.	His Son, Phillip II (1556-1598) & the Spanish Habsburgs
 * b.	His Brother, Ferdinand I (1556-1564) & the Austrian Habsburgs

VIII. Reformation Throughout Europe:
 * a.	Switzerland
 * i.	Ulrich Zwingli (1484-1531) of Zurich
 * a.	His Reforms
 * ii.	Anabaptists (1525)
 * a.	Their Reforms

 * iii.	Jean Calvin (1509-1564)

 * a.	Institutions of the Christian Religion (1536)
 * i.	Predestination

 * ii.	Good Works
 * iii.	Protestant Work Ethnic

 * b.	Theocracy at Geneva & his Ecclesiastical Ordinances

 * i.	Presbyters

 * ii.	Michael Servetus
 * c.	French Calvinists (Huguenots)
 * b.	Low Countries
 * c.	Scotland, John Knox, & Presbyterianism

 * i.	Book of Common Order (1564)
 * d.	England & Henry VIII

 * i.	Henry VIII (1509-1547), Catherine of Aragon, & Succession

 * a.	Mary Tudor
 * ii.	Anne Boleyn, Separation, & Elizabeth I

 * a.	Elizabeth I
 * b.	Act in Restraint of Appeals (1533)
 * c.	Act of Succession (1534)

 * d.	Act of Supremacy (1534)

 * e.	Book of Common Prayer
 * iii.	Religious Reform for the Anglican Church

 * a.	Pilgrimage of Grace (1536)
 * iv.	Edward VI (1547-1553)
 * v.	Mary Tudor (1553-1558)
 * vi.	Elizabeth I (1558-1603)

IX. Catholic Reformation (Counter-Reformation):

 * a.	Reaction to Protestantism under Pope Paul III (1534-1549)

 * i.	Sacred Congregation of the Holy Office (1542)
 * b.	Reaction to Protestantism under Pope Paul IV (1555-1559)

 * i.	Index of Forbidden Books
 * c.	Council of Trent (1545-1563)

 * i.	Rejected Protestant ideas

 * ii.	Seminary Education
 * iii.	No Indulgences/Simony/Pluralism
 * d.	Ignatius of Loyola (1491-1556)

 * i.	Society of Jesus (Jesuit Order) & Spiritual Exercises (1548)

 * a.	Basic Tenets

 * b.	Results
 * e.	Baroque

 * i.	Basic Ideas

 * ii.	Gianlorenzo Bernini (1598-1680) & The Ecstasy of Saint Theresa
 * iii.	Peter Paul Rubens (1577-1640) & The Horrors of War, Massacre of the Innocents
 * f.	Results

X.	Culture during the Two Reformations:
 * a.	Role of Women in the Protestant Reformation

 * b.	Role of Women in Catholic Reformation
 * i.	Witch Hunts

I.Unit 2 I.D. Terms:

1)Age of Exploration:

2)Bartholomew Dias:

3)Vasco de Gama:

4)Christopher Columbus:

5)Ferdinand Magellan:

6)Hernando Cortez:

7)Francisco Pizarro:

8)Joint-Stock Companies:

9)Golden Century of Spain:

10)Price (Commercial) Revolution:

11)Columbian Exchange:

12)Triangle Trade:

13)Atlantic Economy:

14)Protestant Reformation:

15)Pluralism:

16)Simony:

17)Indulgences:

18)Martin Luther:

19)Charles V:

20)Diet of Worms:

21)German Peasant Revolt:

22)Schmalkaldic War:

23)Peace of Augsburg:

24)John Calvin:

25)Predestination:

26)Theocracy:

27)John Knox:

28)Henry VIII:

29)Act of Supremacy:

30)Catholic Reformation:

31)Baroque:

32)Council of Trent:

33)Ignatius Loyola:

2B
I.	Wars of Religion (1572-1648):

 * a.	Medieval Idea of Christendom

 * b.	Protestant Realities
 * c.	Catholic Realities
 * d.	Major Conflicts in Europe

II. France III. Low Countries

IV. Holy Roman Empire

V.	France, King Francis I (1515-1547), & the Huguenots:

 * a.	Problems w/ Taxation(taille [land] tax)

 * b.	Problems w/ the Nobility

 * a.	Huguenot Minority in France

 * c.	Problems w/ the People

 * d.	The Solution: the Concordat of Bologna (1516)
 * a.	It Said
 * b.	Significance on the Nobility

 * c.	Significance on Revenue
 * d.	Significance on the Clergy

 * e.	Significance on the Huguenots
 * e.	Henry II (1547-1559) [son of Francis I], Calvinism, & Persecution

 * f.	Catherine de’ Medici (1519-1589) [wife of Henry II] & Her Regency

 * a.	The Problem w/ Catherine
 * b.	The Problem w/ Regencies

 * c.	House of Guise [C], the House of Montmorency [C], & the House of Bourbon [P]

 * g.	Charles IX (1560-1574) [1st son of Henry II] & Marriage to Navarre (1572)

 * a.	Henry of Navarre (1553-1610) [Leader of House Bourbon] & His Sister

 * b.	Problems w/ the House of Guise
 * c.	Henry Guise [Leader of House Guise] & the St. Bartholomew’s Day Massacre (1572)

 * i.	Results
 * h.	Henry III (1574-1589) [2nd son of Henry II] & Financial Crisis

 * a.	Problems w/ the Nobles

 * i.	Catholic League (1576) & Henry Guise

 * b.	Henry of Navarre & Heir to the Throne (1584)

 * i.	Henry & Politiques

 * i.	War of the Three Henry’s (1584-1589)

 * a.	Henry of Guise’s Execution (1588)

 * b.	Henry III’s Assassination (1589)
 * c.	Henry of Navarre’s Win (1589)
 * j.	Henry of Navarre, Unity, & “Paris is worth a Mass” (1593)

 * k.	Henry IV of the Bourbon Dynasty & the Edict of Nantes (1598)
 * a.	Edict of Nantes
 * i.	La Rochelle
 * l.	Rebuilding France in the Aftermath
 * a.	Maximilien Bethune & the Duke of Sully

 * b.	Taxation & the Paulette (Office Inheritance)

 * c.	Colonization
 * d.	Infrastructure
 * e.	Bureaucracy

 * f.	Ignored Provincial Complaints, Parlements (royal law courts), & Estates General (Legislature)
 * i.	Charles Loyseau & Treatise on Orders and Plain Dignities

 * g.	Henry’s Life, marriage to Marie de’ Medici, & Death (1610)

 * m.	Louis XIII (1610-1643) [son of Henry VI]

 * a.	Called Estates General (1616)

 * b.	Ignored the Edict of Nantes (1627)

 * c.	Cardinal Richelieu (1585-1642) & Reasons of the State

 * i.	Intendant System & Generalities

 * a.	Political Purpose

 * ii.	Religious Views & Foreign Policy

VI. Low Countries & the Dutch Revolt:

 * a.	Charles V & Dutch Happiness
 * b.	Phillip II & Dutch Unhappiness

 * c.	Dutch Revolt (1556-1587)

 * d.	Duke of Alva & the Council of Blood

 * e.	William of Orange [the Silent] (1576) & Protestant Revival

 * f.	Alexander Farnese (Phillip’s nephew)
 * a.	Ten Southern Counties (Spanish Netherlands) [Belgium]

 * b.	Seven Northern Counties (Dutch United Provinces) [the Netherlands]

 * i.	Union of Utrecht (1579)
 * ii.	Dutch United Provinces (1581)
 * a.	Phillip II’s Reality

 * g.	Elizabeth I (1558-1603), Dutch Support, War w/ Spain

 * a.	English Textiles
 * b.	Fear of Dutch Conquest

 * c.	Assassination Attempt by Phillip & Mary

 * h.	Phillip’s Invasion & the Spanish Armada (1588)

 * a.	Why Spain Lost

 * b.	Why England Won
 * c.	Significance

 * i.	Phillip III (1598-1621) [son of Phillip II] & Peace (1609)

 * j.	Spain’s Decline
 * a.	Wars w/ England & the Dutch
 * b.	Gold Trade Slowed
 * c.	Population Increase & Lack of a Middle Class

 * d.	Rise of Northern European States

 * i.	Miguel de Cervantes (1547-1616) & Don Quixote

 * e.	Treaty of the Pyrenees (1659)

 * i.	Spanish/French Border
 * ii.	French Italian Possessions
 * iii.	Austrian Italian Possessions

VII. Holy Roman Empire & the Thirty Years’ War:

 * a.	Peace of Augsburg to 1618

 * b.	Prelude to War: Ferdinand II (1578-1637) [Habsburg King of Austria] & Holy Roman Emperorship

 * a.	Bohemian Diet (1618)
 * b.	Problem w/ Bohemia
 * c.	Defenestration of Prague (1618)

 * a.	Catholic League
 * b.	Protestant Union
 * d.	The Bohemian Phase of the Thirty Years’ War (1618-1625)

 * a.	Protestant Union led by Prince Frederick V of the Palatinate

 * b.	Catholic League led by Ferdinand II, Phillip III (Spain), & Johannes von Tilly

 * c.	Results of the Bohemian Phase

 * e.	The Danish Phase of the Thirty Years’ War (1625-1629)

 * a.	Protestant Union led by Christian IV (1588-1648)

 * b.	Catholic League led by Ferdinand II, Phillip III (Spain), Johannes von Tilly, & Albrecht Wallenstein

 * c.	Results of the Danish Phase

 * i.	Edict of Restitution (1629)
 * f.	The Swedish Phase of the Thirty Years’ War (1629-1635)

 * a.	Protestant Union led by Gustavus Adolphus (1594-1632)

 * b.	Catholic League led by Ferdinand II, Phillip III (Spain), Johannes von Tilly, & Albrecht Wallenstein

 * c.	Results of the Swedish Phase

 * g.	The French Phase of the Thirty Years’ War (1635-1648)

 * a.	Protestant Union led by Louis XIII

 * b.	Catholic League led by Ferdinand III (1637-1657) & Phillip IV (1621-1665)

 * c.	Results of the French Phase

 * h.	Peace of Westphalia (1648)

 * a.	Spain & the Dutch
 * b.	Switzerland
 * c.	France & Alsace & Lorraine

 * d.	Papal Limitations
 * e.	Bohemia & Austria
 * f.	Brandenburg-Prussia
 * g.	Peace of Augsburg
 * h.	German Princes
 * i.	Results of the Thirty Years’ War

 * a.	Michel de Montaigne (1533-1592) & Essays

V.	Unit 2 I.D. Terms:

1)	Huguenots:

2)	Concordat of Bologna:

3)	Henry III:

4)	Henry of Navarre:

5)	War of the Three Henry’s:

6)	Edict of Nantes:

7)	Louis XIII:

8)	Cardinal Richelieu:

9)	Intendant System:

10)	Dutch Revolt:

11)	Phillip II:

12)	Spanish Armada:

13)	Elizabeth I:

14)	Treaty of the Pyrenees:

15)	Thirty Years’ War:

16)	Ferdinand II:

17)	Defenestration of Prague:

18)	Bohemian Phase:

19)	Danish Phase:

20)	Swedish Phase:

21)	French Phase:

22)	Pease of Westphalia:

3A
I.	Absolutism:
 * a.	Absolutism is

A political philosophy, system of govenrment that says the King should be powerful.

 * a.	Examples of Absolutist States

France, Russia, Austria, Prussia
 * b.	Philosophical Support of Absolutism
 * i.	Thomas Hobbes (1588-1679) & Leviathan

 * ii.	Jacques Bossuet (1627-1704)

Devine Right of Kings- God said they should have a powerful King
 * b.	Absolutism was a reaction to

 * c.	Absolutism in Reality
 * a.	Law & Tradition
 * b.	Nobles’ Role in GOV

 * c.	Economy of the State
 * d.	Breaking Point of the People
 * d.	Characteristics of an Absolutist State
 * a.	Centralized Monarchy
 * b.	Royal Dynasty

 * c.	Role of the Nobility

The King reduced their power but increased their status.

 * d.	Ignoring of Legislatures

King ignores town councils and other courts, they can only meet if they are called by the King.
 * e.	Oppression of the Peasants

Take away their rights.
 * f.	Strengthening of the Bureaucracy

 * i.	Problems

 * g.	Expansion of the Citizen Military

Wants a standing army.

 * i.	War as a “Reason of the State”

Everything you do should improve the state; War should increase territory.
 * h.	State Religion

State should control, King appoints state religious leaders.
 * i.	Architecture & Construction

Symbols of Power; victories
 * i.	Baroque
 * ii.	Neoclassicism

Artistic painting (classical themes) II. Absolutism in France: “One Law, One King, One Faith”: Louis XIV goal to a Absolute Monarchy; centralized government
 * a.	Louis XIV (1638-1715) [reigns 1643-1715], the Sun King

The King who put absolutism in place in France, only 5 years old when he took reign
 * a.	Anne of Austria (1601-1666) & Cardinal Jules Mazarin (1602-1661)

 * b.	Mazarin’s Rule of France

 * i.	Fronde (1648-1653)

 * ii.	Mazarin’s Response
 * iii.	Louis XIV’s Response (1653)

 * a.	Results

 * b.	Louis XIV, the Absolute & his Reign

 * a.	Louis as God’s Representative on Earth

 * b.	Louis, his Nobles, & Taxation (capitation [head tax] & vingtieme [5% tax on land])

 * i.	Status as a Control

 * a.	Everything Has A Place & Every Place Has A Thing

 * ii.	Levels of Status & Appointments

 * a.	Nobility of the Robe & Merchants

 * b.	Nobility of the Sword & Court Life

 * c.	Louis XIV, the Palace at Versailles, & the Supreme Symbol of Royal Power

 * i.	Palace at Versailles (construction 1669-1686)

 * ii.	Baroque Architecture at Versailles

 * iii.	Political Purposes
 * a.	Foreign Diplomats
 * b.	Peasants
 * c.	Nobility of the Sword

 * i.	Those Not Participating
 * iv.	Court Life

 * a.	Ceremonies
 * i.	Conduct

 * ii.	The French Playwright, Moliere

 * a.	The Bourgeois Gentlemen & Tartuffe
 * v.	Impact of Versailles on Europe

 * a.	St. Petersburg under Peter I & the Potsdam Palace under Frederick II
 * d.	Louis XIV & Mercantilism

 * i.	Mercantilism

 * a.	Mercantilist Principles
 * i.	Wealth in Finite (Limited)

 * ii.	Wealth as Power
 * iii.	Wealth & Competition

 * iv.	Wealth, Power, & Influence

 * b.	Mercantilist Economic Policies

 * ii.	Jean-Baptiste Colbert (1619-1683), Louis XIV’s Finance Minister (1661)

 * a.	His Domestic Policies

 * e.	Louis XIV’s Persecution of Minorities

 * i.	The Revocation of the Edict of Nantes (1685)

 * a.	Results

 * b.	Economic Significance

 * f.	Limits to French Absolutism
 * i.	No Nation-State

 * ii.	Provincial Estates

 * iii.	Parlements

 * a.	Parlement of Paris & the Jansenists

 * iv.	Uncontrollable Factors by Jean Bodin

 * v.	Reality for France

III. The Habsburg Monarchy & the Holy Roman Empire:

 * a.	The Limits of the Holy Roman Emperor

 * b.	The Rise of Austria

 * a.	Possessions
 * b.	The Metropolitan Monarchy

 * c.	Problems with the Nobles

 * d.	Problems with Hungary (Magyars)

 * i.	Hungarian Rebellion (1679)

 * ii.	The Siege of Vienna (1683) & Leopold I (1658-1705) [King of Austria & Holy Roman Emperor]
 * iii.	Poland, the Great Vanquisher of the Turks

 * iv.	Lifting of the Siege of Vienna

 * v.	Pope Innocent XI & Formation of the Holy League (1684)

 * vi.	War of the Holy League (1686-1687)

 * vii.	Peace of Karlowitz (1699)

 * a.	It Said

 * e.	Reality for Austria

IV. The Rise of Prussia & the Hohenzollern Dynasty:

 * a.	Brandenburg-Prussia
 * b.	Benefits from the Thirty Years’ War

 * a.	Junkers & the Peasants

 * c.	Problems from the Thirty Years’ War

 * d.	Frederick William, the Great Elector (1640-1688) & the Path to Centralization

 * a.	Agreement with the Junkers (1653)

 * b.	With the Money from Taxation

 * c.	His Goal

 * i.	Results

 * d.	Creation of State Officials

 * e.	General Directory over Finance, War, and Royal Domains

 * f.	Results Under Frederick William

 * e.	King Frederick I of Prussia (1688-1713) [son of Frederick William]
 * f.	Frederick William I (1713-1740) [son of Frederick I], the Sergeant King

 * a.	Creation of Fiscals

 * b.	Appointment of Commoners

 * c.	“Prussia is not an country with an army, it is an army with a country”

 * i.	Size of the Army
 * d.	Reality for Prussia

V.	The Expansion of Muscovy & the Creation of Russia:

 * a.	Russia & the Mongols (13th Century to 15th Century)

 * b.	Rise of the Muscovy & Ivan III (1462-1505)

 * a.	Poland to the West, Mongols to the East, Crimean Tartars to the South
 * c.	Ivan IV (1533-1584), the Terrible, & Expansion of the Muscovite State

 * a.	His Goals

 * b.	Territorial Expansion
 * c.	Control of the Boyars

 * d.	 The Terrible Years (1570’s)

 * i.	War with Poland

 * ii.	Oprinchnina

 * iii.	Murder of his Daughter-in-Law, unborn-Grandson, & Son (Ivan)

 * a.	Fyodor (Ivan IV’s other son)
 * d.	Time of Troubles (1584-1613)

 * a.	Rise of the Cossacks

 * b.	Multiple Czars

 * e.	Michael Romanov (1613-164), the Assembly of Nobles, & Stability
 * a.	Why Him?
 * f.	Romanov Russia
 * a.	Russian Orthodox Church & the Romanovs

 * i.	The Old Believers

 * b.	Peasant Life

 * c.	Foreign Enemies
 * d.	Westward Expansion & War with Poland (1640’s)
 * i.	Treaty of Andrussovo (1667)

 * g.	Sweden, Muscovite Russia’s Great Enemy

 * a.	Charles XII (1697-1718)

 * h.	Peter I [the Great] (1682-1725), Tsar of Russia, Vanquisher of the Swedes

 * a.	His Problems with Russian Culture

 * b.	Peter’s European Tour & Modernization (Westernization)

 * c.	Peter & His Reforms in Russia
 * i.	Bureaucracy

 * a.	Creation of a Senate

 * ii.	Table of Ranks (1722) & the Service State

 * a.	Use of Commoners

 * iii.	Cultural Shift

 * a.	St. Petersburg

 * iv.	Old Believers & Other Objectors
 * v.	Economy & Taxation

 * vi.	Military

 * vii.	Territorial Expansion & the Port of Azoz (1696)

 * d.	Great Northern War (1700-1721)

 * i.	Battle of Narva (1700)

 * ii.	Battle of Poltava (1709)

 * iii.	Treaty of Nystadt (1721)

 * a.	Territorial Gains
 * i.	Limitations to Russian Absolutism

 * j.	Reality for Russia

VI. European Balance of Power Politics:

 * a.	Renaissance Idea of Balance of Power

 * b.	International Law

 * a.	Samuel von Pufendorf (1632 – 1694) & Of the Law of Nature and Nations (1672)

 * c.	Wars to Maintain Balance of Power

 * a.	Louis XIV & France’s “Natural Borders”

 * b.	War of the League of Augsburg (1688-1697)
 * i.	England/Dutch United Provinces/Austria/Spain vs. France
 * ii.	Treaty of Ryswick (1692)

 * c.	War of Spanish Succession (1701-1714)

 * i.	Spain, Charles II, & Succession (1700)

 * ii.	Philip of Anjou (Louis XIV’s grandson)

 * iii.	Philip V, King of Spain (1700)
 * iv.	Philip V, Heir to the French Throne (1701)

 * v.	England/Dutch United Provinces/Austria/Prussia vs. Spain/France

 * vi.	Peace of Utrecht (1713)
 * a.	It Said
 * b.	Territorial Exchanges for Austria
 * c.	Territorial Exchanges for England

 * vii.	 Louis XV (1715-1774)

 * d.	Creation of the Modern State

I.	Unit 3 I.D. Terms:

1)	Absolutism:

2)	Reasons of the State:

3)	Baroque:

4)	“One Law, One King, One Faith”:

5)	Louis XIV:

6)	Fronde:

7)	Parlement:

8)	Palace at Versailles:

9)	Mercantilism:

10)	Jean-Baptiste Colbert:

11)	Metropolitan Monarchy:

12)	Siege of Vienna:

13)	Leopold I:

14)	War of the Holy League:

15)	Junkers:

16)	Frederick William, the Great Elector:

17)	Frederick William I:

18)	Ivan IV:

19)	Time of Troubles:

20)	Michael Romanov:

21)	Peter I:

22)	Modernization:

23)	Great Northern War:

24)	Treaty of Nystadt:

25)	European Balance of Power:

26)	War of the League of Augsburg:

27)	Treaty of Ryswick:

28)	War of Spanish Succession:

29)	Philip of Anjou:

30)	Peace of Utrecht:

3B
I.	’’’Absolutism: ’’’

 * a.	Absolutism is

 * a.	Examples of Absolutist States

 * b.	Philosophical Support of Absolutism

 * i.	Thomas Hobbes (1588-1679) & Leviathan

 * ii.	Jacques Bossuet (1627-1704)

 * b.	Absolutism was a reaction to

 * c.	Absolutism in Reality

 * a.	Law & Tradition

 * b.	Nobles’ Role in GOV

 * c.	Economy of the State

 * d.	Breaking Point of the People

 * d.	Characteristics of an Absolutist State

 * a.	Centralized Monarchy

 * b.	Royal Dynasty

 * c.	Role of the Nobility

 * d.	Ignoring of Legislatures

 * e.	Oppression of the Peasants

 * f.	Strengthening of the Bureaucracy

 * i.	Problems

 * g.	Expansion of the Citizen Military

 * i.	War as a “Reason of the State”

 * h.	State Religion

 * i.	Architecture & Construction

 * i.	Baroque

 * ii.	Neoclassicism

II. ’’’Absolutism in France: “One Law, One King, One Faith”:’’’

 * a.	Louis XIV (1638-1715) [reigns 1643-1715], the Sun King

 * a.	Anne of Austria (1601-1666) & Cardinal Jules Mazarin (1602-1661)

 * b.	Mazarin’s Rule of France

 * i.	Fronde (1648-1653)

 * ii.	Mazarin’s Response

 * iii.	Louis XIV’s Response (1653)

 * a.	Results

 * b.	Louis XIV, the Absolute & his Reign

 * a.	Louis as God’s Representative on Earth

 * b.	Louis, his Nobles, & Taxation (capitation [head tax] & vingtieme [5% tax on land])

 * i.	Status as a Control

 * a.	Everything Has A Place & Every Place Has A Thing

 * ii.	Levels of Status & Appointments

 * a.	Nobility of the Robe & Merchants

 * b.	Nobility of the Sword & Court Life

 * c.	Louis XIV, the Palace at Versailles, & the Supreme Symbol of Royal Power

 * i.	Palace at Versailles (construction 1669-1686)

 * ii.	Baroque Architecture at Versailles

 * iii.	Political Purposes

 * a.	Foreign Diplomats

 * b.	Peasants

 * c.	Nobility of the Sword

 * i.	Those Not Participating

 * iv.	Court Life

 * a.	Ceremonies

 * i.	Conduct

 * ii.	The French Playwright, Moliere

 * a.	The Bourgeois Gentlemen & Tartuffe

 * v.	Impact of Versailles on Europe

 * a.	St. Petersburg under Peter I & the Potsdam Palace under Frederick II

 * d.	Louis XIV & Mercantilism

 * i.	Mercantilism

 * a.	Mercantilist Principles

 * i.	Wealth in Finite (Limited)

 * ii.	Wealth as Power

 * iii.	Wealth & Competition

 * iv.	Wealth, Power, & Influence

 * b.	Mercantilist Economic Policies

 * ii.	Jean-Baptiste Colbert (1619-1683), Louis XIV’s Finance Minister (1661)

 * a.	His Domestic Policies

 * e.	Louis XIV’s Persecution of Minorities

 * i.	The Revocation of the Edict of Nantes (1685)

 * a.	Results

 * b.	Economic Significance

 * f.	Limits to French Absolutism

 * i.	No Nation-State

 * ii.	Provincial Estates

 * iii.	Parlements

 * a.	Parlement of Paris & the Jansenists

 * iv.	Uncontrollable Factors by Jean Bodin

 * v.	Reality for France

III. ’’’The Habsburg Monarchy & the Holy Roman Empire: ’’’

 * a.	The Limits of the Holy Roman Emperor

 * b.	The Rise of Austria

 * a.	Possessions

 * b.	The Metropolitan Monarchy

 * c.	Problems with the Nobles

 * d.	Problems with Hungary (Magyars)

 * i.	Hungarian Rebellion (1679)

 * ii.	The Siege of Vienna (1683) & Leopold I (1658-1705) [King of Austria & Holy Roman Emperor]

 * iii.	Poland, the Great Vanquisher of the Turks

 * iv.	Lifting of the Siege of Vienna

 * v.	Pope Innocent XI & Formation of the Holy League (1684)

 * vi.	War of the Holy League (1686-1687)

 * vii.	Peace of Karlowitz (1699)

 * a.	It Said

 * e.	Reality for Austria

IV. ’’’The Rise of Prussia & the Hohenzollern Dynasty: ’’’

 * a.	Brandenburg-Prussia

 * b.	Benefits from the Thirty Years’ War

 * a.	Junkers & the Peasants

 * c.	Problems from the Thirty Years’ War

 * d.	Frederick William, the Great Elector (1640-1688), & the Path to Centralization

 * a.	Agreement with the Junkers (1653)

 * b.	With the Money from Taxation

 * c.	His Goal

 * i.	Results

 * d.	Creation of State Officials

 * e.	General Directory over Finance, War, and Royal Domains

 * f.	Results Under Frederick William

 * e.	King Frederick I of Prussia (1688-1713) [son of Frederick William]

 * f.	Frederick William I (1713-1740) [son of Frederick I], the Sergeant King

 * a.	Creation of Fiscals

 * b.	Appointment of Commoners

 * c.	“Prussia is not an country with an army, it is an army with a country”

 * i.	Size of the Army

 * d.	Reality for Prussia

V.	’’’The Expansion of Muscovy & the Creation of Russia: ’’’

 * a.	Russia & the Mongols (13th Century to 15th Century)

 * b.	Rise of the Muscovy & Ivan III (1462-1505)

 * a.	Poland to the West, Mongols to the East, Crimean Tartars to the South

 * c.	Ivan IV (1533-1584), the Terrible, & Expansion of the Muscovite State

 * a.	His Goals

 * b.	Territorial Expansion

 * c.	Control of the Boyars

 * d.	 The Terrible Years (1570’s)

 * i.	War with Poland

 * ii.	Oprinchnina

 * iii.	Murder of his Daughter-in-Law, unborn-Grandson, & Son (Ivan)

 * a.	Fyodor (Ivan IV’s other son)

 * d.	Time of Troubles (1584-1613)

 * a.	Rise of the Cossacks

 * b.	Multiple Czars

 * e.	Michael Romanov (1613-1645), the Assembly of Nobles, & Stability

 * a.	Why Him?

 * f.	Romanov Russia

 * a.	Russian Orthodox Church & the Romanovs

 * i.	The Old Believers

 * b.	Peasant Life

 * c.	Foreign Enemies

 * d.	Westward Expansion & War with Poland (1640’s)

 * i.	Treaty of Andrussovo (1667)

 * g.	Sweden, Muscovite Russia’s Great Enemy

 * a.	Charles XII (1697-1718)

 * h.	Peter I [the Great] (1682-1725), Tsar of Russia, Vanquisher of the Swedes

 * a.	His Problems with Russian Culture

 * b.	Peter’s European Tour & Modernization (Westernization)

 * c.	Peter & His Reforms in Russia

 * i.	Bureaucracy

 * a.	Creation of a Senate

 * ii.	Table of Ranks (1722) & the Service State

 * a.	Use of Commoners

 * iii.	Cultural Shift

 * a.	St. Petersburg

 * iv.	Old Believers & Other Objectors

 * v.	Economy & Taxation

 * vi.	Military

 * vii.	Territorial Expansion & the Port of Azoz (1696)

 * d.	Great Northern War (1700-1721)

 * i.	Battle of Narva (1700)

 * ii.	Battle of Poltava (1709)

 * iii.	Treaty of Nystadt (1721)

 * a.	Territorial Gains

 * i.	Limitations to Russian Absolutism

 * j.	Reality for Russia

VI. ’’’European Balance of Power Politics: ’’’

 * a.	Renaissance Idea of Balance of Power

 * b.	International Law

 * a.	Samuel von Pufendorf (1632 – 1694) & Of the Law of Nature and Nations (1672)

 * c.	Wars to Maintain Balance of Power

 * a.	Louis XIV & France’s “Natural Borders”

 * b.	War of the League of Augsburg (1688-1697)

 * i.	Also Called King William’s War (North America), Nine Years War, War of the Grand Alliance, & War of Palatine Succession

 * ii.	England/Dutch United Provinces/Austria/Spain vs. France

 * iii.	Treaty of Ryswick (1697)

 * c.	War of Spanish Succession (1701-1714)

 * i.	Spain, Charles II, & Succession (1700)

 * ii.	Philip of Anjou (Louis XIV’s grandson)

 * iii.	Philip V, King of Spain (1700)

 * iv.	Philip V, Heir to the French Throne (1701)

 * v.	England/Dutch United Provinces/Austria/Prussia vs. Spain/France

 * vi.	Peace of Utrecht (1713)

 * a.	It Said

 * b.	Territorial Exchanges for Austria

 * c.	Territorial Exchanges for England

 * vii.	 Louis XV (1715-1774)

 * d.	Creation of the Modern State

VII. Unit 3 I.D. Terms:

1)	Absolutism:

2)	Reasons of the State:

3)	Baroque:

4)	“One Law, One King, One Faith”:

5)	Louis XIV:

6)	Fronde:

7)	Parlement:

8)	Palace at Versailles:

9)	Mercantilism:

10)	Jean-Baptiste Colbert:

11)	Metropolitan Monarchy:

12)	Siege of Vienna:

13)	Leopold I:

14)	War of the Holy League:

15)	Junkers:

16)	Frederick William, the Great Elector:

17)	Frederick William I:

18)	Ivan IV:

19)	Time of Troubles:

20)	Michael Romanov:

21)	Peter I:

22)	Modernization:

23)	Great Northern War:

24)	Treaty of Nystadt:

25)	European Balance of Power:

26)	War of the League of Augsburg:

27)	Treaty of Ryswick:

28)	War of Spanish Succession:

29)	Philip of Anjou:

30)	Peace of Utrecht: