User:Egm6321.f11.team1.steele.m

http://en.wikiversity.org/wiki/User:egm6321.f11.team1.steele.m/Lecture 1

R2.11 Question from 11-1: Why is it usually difficult to solve for the integrating factor h(x,y) in the following equation? $$h_xN - h_yM + h(N_x - M_y) = 0$$ The solution requires an understanding of how the Euler Integrating Factor Method (IFM) is derived with the reduction of order method. The solution expands upon concepts introduced in meetings 7-11.

First, the general nonlinear ODE of order n without the missing function y(x) is:

$$G(y^{(n)},y^{(n-1)},...,y'',y',y,x) = 0 $$

If we remove y, then the general nonlinear ODE without the function is:

$$G(y^{(n)},y^{(n-1)},...,y'',y',y,x) = 0 $$

Reduce the order by substituting p(x) := y'(x) below

$$G(p^{(n-1)},p^{(n-2)},...,p',p,x) = 0$$

The substitution gives the following solution for y(x).

$$y(x) = \int p(x)dx + k$$

where k is the integration constant. The key is in the dummy integration variable derivation below.

$$\int_a^x p(z)dz = \int_a^x y'(z)dz = y(x) - y(a)$$

Rearrangment gives

$$y(x) = \int_a^x p(z)dz + y(a) = P(x) - P(a) + y(a) = P(x) + k$$

The derivation requires the astute reader to note the distinction beween capital "P" and small "p." In addition, k = -P(a) + y(a)

Now, the Euler IFM applies to G(y',y,x)=0, and for a particular class of N1 ordinary differential equations, a solution linear in y' is

M(x,y) + N(x,y)y' = 0 where

$$ y' = \frac{dy}{dx}$$

So a particular solution is M(x,y) + N(x,y)dy = 0

The derivations to this point are from meeting 7 A more general solution for f(y') of the particular case is

$$\bar{M}(x,y) + \bar{N}(x,y)F(y') = 0$$

for the nonlinear case. But the assumption is that F(*) has an inverse. This leads to the first exactness conditions of N1 ordinary differential equations.

First Exactness Condition of N1-ODEs

If an N1-ODE is exact then it has a particular form of

M(x,y) + N(x,y)y' = 0 where

$$ y' = \frac{dy}{dx}$$

The appended derivations to this point are from meeting 8 Second Exactness Condition of N1-ODEs

Now, we must derive the 2nd exactness condition. A general ND-ODE is exact if for G the following relations exist:

$$\frac{d\phi}{dx} = G$$

$$\phi(x,y) = k$$

$$\frac{d\phi(x,y(x))}{dx} = G(y',y,x) = 0$$

Let $$\phi(x,y)$$ be a smooth function, then for the 2nd exactness condition we have

$$M_y(x,y) = \frac {\partial^2\phi(x,y)}{\partial x\partial y} = N_x(x,y) = \frac{\partial^2\phi(x,y)}{\partial y\partial x}$$ The appended derivations to this point are from meeting 9 If an N1-ODE satisfies the 1st exactness condition but not the 2nd, then the N1-ODE can be transformed by using the Euler Integrating Factor Method (IFM). Assume the conditions below.

$$M(x,y) + N(x,y)y' = 0 \And M_y(x,y) \ne N_x(x,y)$$

The transformation of the given N1-ODE to a form that fulfills both exactness conditions begins by finding an integrating factor h(x,y).

For condition 1, the substitution of h gives

$$(hM) + (hN)y' = \bar{M} + \bar{N}y' = 0$$

The key is to derive a solution using

$$\bar{M} = hM \And \bar{N} = hN$$

The derivations are as follows:

$$\bar{M}_y = h_yM + hM_y = \bar{N}_x = h_xN + hN_x$$

So the Euler IFM gives the following solution using h(x,y) as the transformation.

$$h_xN - h_yM + h(N_x - M_y) = 0 $$

The cumulative derivations to this point are from meetings 10-11 The solution h(x,y) is difficult to find since a separation of variables may not be possible for x and y. It is difficult to solve for h(x,y) in the equation when N(x,y) and M(x,y) are multivariate.

In addition, the following variables may have non-zero values, which complicates integrations to solve for h. The non-zero conditions complicate integration for h

$$h_x \ne 0$$ and $$h_y \ne 0$$

$$N_x \ne 0$$ and $$M_y \ne 0$$

So the solution of h is often cumbersome to solve because of the multivariate dependencies of h(x,y), M(x,y), and N(x,y) in the following equation.

$$\frac {h_x}{h} = -\frac{1}{N}(N_x - M_y)$$

The integration of this equation above is difficult with dependencies on both x and y and the non-zero conditions listed.

References for R2.11 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 7) 5 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 8) 6 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 9) 8 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 10) 8 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) 13 Sep 2011.

R2.12 Question from 11-3: Find h for the following equation where h is the Euler Integration Factor Method. $$\frac{h_y}{h} = \frac{1}{M}(N_x - M_y) =: m(y)$$ The solution provides a condition in which simplification allows easier integration for one variable y

$$h_x(x,y) = 0$$

The condition above simplifies the integral for the Euler Integration Factor Method and the transformation involving h(x,y).

$$h_xN - h_yM + h(N_x - M_y) = -h_y + h(N_x - M_y) = 0$$

Rearrangement gives

$$h_yM = h(N_x - M_y) = 0$$

$$\frac{h_y}{h} = \frac{1}{M}(N_x - M_y) =: m(y)$$

$$\int \frac{h_y}{h} = - \int \frac {1}{N}(N_x - M_y)$$

Substitution of m(s) for the right side of the equation gives

$$ \log h(y) = \int\limits^{y} m(s)ds + k$$

So the solution is

$$h(y) = \exp {[\int\limits^{y} m(s)ds + k]}$$

The derivations used notes from meeting 11 References for R2.12 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) 13 Sep 2011.

R2.13 Question from 11-5: If the following equation is given $$y' + \frac{1}{x} = x^2$$

then show that

$$h(x) = x \And y(x) = \frac{x^3}{4} + \frac{k}{x}$$ The equation does not fulfill the 2nd exactness condition, so an integration factor h can be derived by rearranging the equation.

$$ a_0(x)y + y' = M(x,y) + N(x,y)y' = b(x)$$

Recall that h is defined as follows:

$$If h_y(x,y) = 0 then$$

$$\frac{h_x}{h} = \frac{1}{M}(N_x - M_y) =: n(x) = +a_0(x)$$

This gives h(x) as

$$h(x) = \exp[\int\limits^{x} a_0(s)ds + k_1]$$

Recall that

$$a_0(x) = \frac{1}{x}$$

So that $$h(x) = \exp[\int\limits^{x} \frac{1}{s}] = \exp(\ln {x}) = x $$

Now that we have solved for h(x), the next step is to solve for y(x). The transformation with h(x) is done here.

$$h(y' + a_0y) = hb$$

$$hy' + h'y = hb = (hy)'$$

Integration (hy)' to get

$$\int\limits^{x}(hy)' = \int\limits^{x}h(s)b(s)ds + k_2$$

Divide both sides by h(x) to get y(x).

$$y(x) = \frac{1}{h(x)}[\int\limits^{x}h(s)b(s)ds + k_2]$$

Recall that the given equation gives b as shown below.

$$h(x) = x \And b(x) = x^2$$

So substitute the functions back into the integral for y(x).

$$y(x) = \frac{1}{x}[\int\limits^{x} x^2 ] = \frac{1}{x}[\frac{x^4}{4} + k]$$ Simplification gives the solution for y(x) below.

$$y(x) = \frac{x^3}{4} + \frac{k}{x}$$

References for R2.13 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) 13 Sep 2011.

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$$\mathbf{K} = \begin{bmatrix} k1 & 0 {-k1} & 0 \\ 0 & k2 & {-k2} & 0 \\ {-k1} & {-k2} & {k1+k2+k3} & {-k3} \\ 0 & 0 & {-k3} & {k3+k4+k5} \end{bmatrix}$$

$$\mathbf{P} = \begin{bmatrix} P1 \\ P2 \\ P3 \\ P4 \end{bmatrix}$$

$$\mathbf{F} = \begin{bmatrix} f1 \\ f2 \\ 0 \\ 0 \end{bmatrix}$$