# R*3.2 Show Exactness or determine IFM to make exact

## Given

1st order ODE:

$\left(\frac{1}{3}x^{3}+d_1\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}+sin(x)+d_2\right)=0$

$\displaystyle (1) p. 13-4$

From lecture notes Mtg 13

## Find

Show that the above is exact or find the integrating factor $h$ to make it exact

## Solution

 Solved on our own

For the the 1st condition of exactness the given equation must be in the form of:

$N(x,y)y' + M(x,y) = 0 \!$

$\displaystyle (2) p. 7-6$

Mtg 7 (b,x)

It has been shown in R3.1 that the integration functions must be constant in order to determine the integration factor. Thus the integration functions and constants will be set to zero for this problem, and $\displaystyle (1) p. 13-4$, becomes:

$\underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$

which is in the form of the first exactness condition $\displaystyle (2) p. 7-6$ so the 1st condition is satisfied.

For the 2nd exactness condition the partial derivatives must be equal per meeting lecture notes 9, Mtg 9

$M_y = N_x\!$

$\displaystyle (2) p. 9-3$

.

It can be seen that:

$M_y \neq N_x$

An integrating factor will be found such that the given ODE can be integrated exactly.

The strategy is to get the following equation from lecture notes 13:

$\underbrace{\bar b(x,y)c(y)}_{N(x,y)}\ y' + \underbrace{a(x) \bar c(x,y)}_{M(x,y)}=0 \!$

$\displaystyle (1) p. 13-2$

.

into the form of equtaion $\displaystyle (2) p. 12-2$

$y'+a_o(x)y=0 \!$

$\displaystyle (2) p. 12-2$

.

So first lets calculate the the inputs to $\displaystyle (1) p. 13-2$

$\bar b(x)= \frac{x^3}{3}\!$
$\bar c(y)= \frac{y^5}{3}\!$

We also know that

$c(y)= y^4\!$
$a(x)= 5x^3+2\!$

Plugging these values into $\displaystyle (1) p. 13-2$ yields:

$(\frac{x^3}{3} y^4) y'+ (5x^3+2) y^5=0\!$

After some algebra to get in the form of $\displaystyle (2) p. 12-2$ the result is:

$\underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y =0\!$

From equation $\displaystyle (3) p. 11-4$ the integrating factor can be found as follows:

$h = \exp\int^{x}a_0(s)ds \!$

with the integration constant equal to zero

Substituting $a_0\!$ yields

$h = \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \!$

Now the integral can be evaluated, which yields:

$h = \exp\left(3x - \frac{3}{5x^2}\right)\!$

This is the integrating factor to make the ODE given exact.

# R*3.12 Linear time-variant system, Rocket Roll Control in matrix notation

## Given

$\displaystyle \dot{\phi}=\omega$

$\displaystyle (1) p. 16-2$

$\displaystyle \dot{\omega}=-\frac{1}{\tau }\omega+\frac{Q}{\tau }\delta$

$\displaystyle (2) p. 16-2$

$\displaystyle \dot{\delta}=u$

$\displaystyle (3) p. 16-2$

See lecture notes Mtg 16 (b) for problem description.

$\displaystyle \mathbf{\dot{x}}(t)=\mathbf{A}(t)\, \mathbf{x}(t)+\mathbf{B}(t)\, \mathbf{u}(t)$

$\displaystyle (1) p. 14-4$

Reference lecture notesMtg 14 for form of equation.

## Find

Put (1-3)p.16-2 in the form of (1)p.14-4

## Solve

 Solved on our own

$\displaystyle \begin{bmatrix} \dot{\phi }\\ \dot{\omega }\\ \dot{\delta } \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \phi \\ \omega \\ \delta \end{bmatrix} + \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \mathbf {u}(t)$

# R*3.14 Discuss the search for the solution with out making h constant

## Given

${h}_{x}+{h}_{y}\cdot P =0$

$\displaystyle (1) p. 17-1$

## Find

$h(x,y)$ with out assuming that h=constant, and discuss the search for the solution

## Solution

This problem is assigned in Mtg 18 (b)

As a starting point I looked at this:

To find $h(x,y)$ solve equation $(1) p. 17-1$ for $\displaystyle {h}_{x}$ as shown below

 $\displaystyle {h}_{x} = - {h}_{y}P$ $\displaystyle (3.14.1)$

Use the 2nd exactness condition, which states

 $\displaystyle {h}_{xy} = {h}_{yx}$ $\displaystyle (3.14.2)$

to determine $h(x,y)$. Differentiate both sides of $\displaystyle (3.14.1)$ to determine ${h}_{xy}$ using the chain rule (http://en.wikipedia.org/wiki/Chain_rule)

 ${h}_{x}= - {h}_{y} \underbrace{{P}}_{:= {y}^{'}} \Rightarrow :::::::::::{h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''}$ $\displaystyle (3.14.3)$

Substituting $(3.14.3)$ into $(3.14.2)$ yields:

 ${h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} \neq {h}_{yx}$ $\displaystyle (3.14.4)$

By comparimg $\displaystyle(3.14.1)$ to $\displaystyle(3.14.4)$, and realizing $\displaystyle y' = P$ ,it can be see that the last term must be zero to satify the given equation.

Since $\displaystyle(3.14.1)$ is just $\displaystyle (1) p. 17-1$ rearranged, in order for $(1) p. 17-1$ to be true, $\displaystyle {h}_{x} = {h}_{y} = 0$.

The next step is to integrate both sides to get $\displaystyle h(x,y) = constant$. When integrating math>\displaystyle {h}_{x} [/itex] and math>\displaystyle {h}_{y} [/itex] it must be considered that in general a function of of integration is also possible.

For reference a similar answer was obtained by Team 6 Fall 2010, for convienience that solution is shown in its entirety with a slight adaptation of equation numbering only for completeness from http://en.wikiversity.org/w/index.php?title=User:EGM6321.f10.team6.cook/hw3&oldid=621925#solution_of_h.28x.2Cy.29

 \begin{align} \underbrace{x{(y')}^{2}+y \cdot y'}_{g(x,y,y')}+\underbrace{(xy)}_{f(x,y,y')}y''=0 \end{align} (3.14.5)

we can get eqn(3.14.6) using definition of \begin{align} g(x,y,y') \end{align}

 \begin{align} {h}_{x}+{h}_{y}\cdot P =0 \end{align} (3.14.6)

where \begin{align} P:=y' \end{align}

• without assuming h=const. find the solution of eqn(3.14.6)

We can rewrite the eqn(3.14.5) as (3.14.6)

 \begin{align} \frac{\partial h}{\partial x} + \frac{\partial h}{\partial y}\cdot \frac{dy}{dx}=0 \end{align} (3.14.7)

We are familiar with this equation, as we learned already. Total derivative - Egm6321.f10_HW1_prob#1_team6

 \begin{align} \frac{d}{dx}h(x,y)=\frac{\partial h}{\partial x} + \frac{\partial h}{\partial y}\cdot \frac{dy}{dx}=0 \end{align} (3.14.8)

As \begin{align} \frac{d}{dx}h(x,y)=0 \end{align}, we know that \begin{align} h(x,y)=f(y) \end{align} only.
It means \begin{align} {h}_{x}=0 \end{align}

Hence, eqn(3.14.6) becomes,

 \begin{align} {h}_{y}\cdot P=0 \end{align} (3.14.9)

There are two possible solutions.

1) \begin{align} P=y'=0 \end{align}

2) \begin{align} {h}_{y}=0 \end{align}

If 1) were satisfied, whole problems became zero, which is trivial. We can conclude that 2) is the solution.

As \begin{align} {h}_{x}=0 \end{align} and \begin{align} {h}_{y}=0 \end{align},

 \begin{align} h={k}_{1} = \mathrm{constant} \end{align} (3.14.10)

Based on the first clarified solution which has been improved for clarity and correctness amd the 2nd solution which was copied almost directly it appears that the search for $\displaystyle h$ with out assuming h is constant yiled s the result that $\displaystyle h$ is indeed constant to satisfy the exactness conditions.