User:EGM6321.F11.TEAM1.R*2.16-2.18 Bens

Given
The solution of

$$\displaystyle a_1(x) \cdot y^{(1)}+a_0(x)y^{(0)}=b(x)$$  $$\displaystyle (3) p. 11-3 $$

is

$$\displaystyle y(x) = \frac{1}{h(x)} \left[\int^x h(s)b(s)ds + k_2 \right] $$ $$\displaystyle (1) p. 11-5 $$

From lecture notes [[media:pea1.f11.mtg11.djvu|Mtg 11 (b)]]

Find
Show that $$\displaystyle (1) P. 11-5 $$ agrees with $$\displaystyle King p. 512 $$ shown below


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

y = A{y}_{H}\left(x \right)+{y}_{P}\left(x \right) $$ $$
 * $$\displaystyle (2.16.1)
 * style= |
 * }

Solution
Start by using the integration factor defined by $$\displaystyle (3) p. 11-4$$

$$ h(x) = \exp \left[ \int^{x} {a}_{0} \left( s\right)ds + k_1 \right]$$ $$\displaystyle (3) p. 11-4 $$

Using the properties for exponential functions $$ \exp \left[ a+b\right] = \exp \left[ a\right]\exp \left[b\right]$$

Now $$\displaystyle (3) p. 11-4$$ becomes


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

\exp \left[ \int {a}_{0} \left( x\right)dx \right]\underbrace{\exp \left[k\right]}_{=: k_1} $$ $$
 * $$\displaystyle (2.16.2)
 * style= |
 * }

Plugging this back into $$\displaystyle (3) p. 11-4 $$ gives us

$$ h \left( x\right)= k_{1}\exp \left[ \int {a}_{0} \left( x\right)dx\right] $$ $$\displaystyle (Eq 2.16.3) $$

To find $$ {k}_{2}$$ we will evaluate the integral in $$\displaystyle (1) p. 11-5$$

$$\int^{x} h\left( s\right)b\left( s\right)ds= \int h\left( x\right)b\left( x\right)dx + k_{2}$$ $$\displaystyle (Eq 2.16.4) $$

Next plug equations $$\displaystyle Eq 2.16.3 $$ and $$\displaystyle Eq 2.16.4 $$ into $$\displaystyle (1) p. 11-5 $$ and canceling $$\displaystyle k_1 $$

$$y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]}\left[ \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\right]$$ $$\displaystyle (Eq 2.16.5) $$

From $$\displaystyle K p. 512 $$ the equation for the solution of the homogeneous differential equation and the particular integral solution of the inhomogeneous differential equation are given below, respectively


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A{y}_{H}\left(x \right)= A\exp \left\{ - \int ^{x}{P} \left( t\right)dt\right\} $$ $$
 * $$\displaystyle (Eq 2.16.6)
 * style= |
 * }


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{y}_{P}\left(x \right)= \exp \left\{ - \int ^{x}{P} \left( t\right)dt\right\}\int ^{x}{Q} \left( s\right)\exp \left\{  \int ^{x}{P} \left(t\right)dt\right\}ds $$ $$
 * $$\displaystyle (Eq 2.16.7)
 * style= |
 * }

Now, using $$\displaystyle (Eq 2.16.5) $$ which is the expanded form of $$\displaystyle (3) p. 11-4 $$ we can further expand and rearrange to get


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y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]} \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +\frac{k_2}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]} $$ $$
 * $$\displaystyle (Eq 2.16.8)
 * style= |
 * }


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

y \left( x\right)= \exp \left[  -\int {a}_{0} \left( x\right)dx\right] \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\exp \left[  -\int {a}_{0} \left( x\right)dx\right] $$ $$
 * $$\displaystyle (Eq 2.16.9)
 * style= |
 * }

Now we have $$\displaystyle (3) p. 11-4 $$ in the form of $$\displaystyle (Eq 2.16.1)$$ where


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

A{y}_{H}\left(x \right)= k_2\exp \left[ -\int {a}_{0} \left( x\right)dx\right] $$ $$
 * $$\displaystyle (Eq 2.16.10)
 * style= |
 * }


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

{y}_{P}\left(x \right)= \exp \left[ -\int {a}_{0} \left( x\right)dx\right] \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx $$ $$
 * $$\displaystyle (Eq 2.16.11)
 * style= |
 * }

By comparing $$\displaystyle (Eq 2.16.6) $$ with $$\displaystyle (Eq 2.16.10)$$ and $$\displaystyle (Eq 2.16.7) $$ with $$\displaystyle (Eq 2.16.11) $$ we can see that these equations are in agreement

This problem was adapted from http://en.wikiversity.org/w/index.php?title=User:Egm6321.f10.team03/Hwk2&oldid=615983#Problem_8

The equations have been updated, the flow improved and, the wording made more clear. Addtionally everything was updated to be consistent with the termonology of the Fall 2011 lecture notes and adapted to answer the question more exactly.

The source location is http://en.wikiversity.org/w/index.php?title=User:EGM6321.F11.TEAM1.R*2.16-2.18_Bens

This solution was updated by Ben Neri, Checked by, and compiled into the complete report by.

Given

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

$$ From lecture notes [[media:pea1.f11.mtg11.djvu|Mtg 11 (b)]]
 * $$\displaystyle {{y}^{'}}+{{a}_{0}(x)}y=0$$
 * $$\displaystyle (2) p. 12-2
 * $$\displaystyle (2) p. 12-2
 * }
 * }

Find

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


 * Find $$\displaystyle y_H$$
 * style= |
 * }
 * }

Solution
Since $$\displaystyle (2) p. 12-2 $$ is the homogeneous portion of

$$\displaystyle a_1(x) \cdot y^{(1)}+a_0(x)y^{(0)}=b(x)$$ with $$\displaystyle  a_1(x)=1$$ <p style="text-align:right;">$$\displaystyle (3) p. 11-3 $$

Equation $$\displaystyle (2) p. 12-2 $$  is seperable, So we can solve directly for $$ y_H$$

Equation $$\displaystyle (2) p. 12-2 $$ can be rearranged as follows

<p style="text-align:left;"> $$\displaystyle \frac{y'}{y} = -{a}_{0} $$ <p style="text-align:right;">$$\displaystyle (Eq 2.17.1) $$

Integration of both sides of this equation gives

<p style="text-align:left;"> $$ \ln \left( y\right)= -\int^{x} {a}_{0} \left( s\right)ds + C$$   <p style="text-align:right;">$$\displaystyle (Eq 2.17.2)$$

Solving for $$y$$ gives

$$ y= \exp\left[ -\int^{x} {a}_{0} \left( s\right)ds + C \right]= A\exp\left[ -\int^{x} {a}_{0} \left( s\right)ds\right]= y_H $$<p style="text-align:right;">$$\displaystyle (Eq 2.17.3) $$

This problem was adapted from http://en.wikiversity.org/w/index.php?title=User:Egm6321.f10.team03/Hwk2&oldid=615983#Problem_8

The equations have been updated, the flow improved and, the wording made more clear. Addtionally everything was updated to be consistent with the termonology of the Fall 2011 lecture notes and adapted to answer the question more exactly.

The source location is http://en.wikiversity.org/w/index.php?title=User:EGM6321.F11.TEAM1.R*2.16-2.18_Bens

This solution was updated by Ben Neri, Checked by, and compiled into the complete report by.

= R*2.18 - Find the integrating factors to make L1-ODE-VC exact =

Given
$$\displaystyle \left[ x^4 y + 10 \right] + \frac{1}{2} x^2 y' = 0 $$ <p style="text-align:right;">$$\displaystyle (4) p. 12.4 $$

From lecture notes [[media:pea1.f11.mtg12.djvu|Mtg 12]]

Find
Determine if $$\displaystyle (4) p. 12-4 $$ is exact, if it is not find the integrating factor $$\displaystyle h $$ to make it exact

Solution
The first exactness condition for an N1-ODE is

$$\displaystyle M(x,y) + N(x,y) y'= 0$$ <p style="text-align:right;">$$\displaystyle (2) p. 7-6 $$

Since the equation is already in this form the first exactness condition is fulfilled

The second exactness condition requires that

$$\displaystyle M_y(x,y)=N_y(x,y)$$ <p style="text-align:right;">$$\displaystyle (3) p. 9-3 $$

Where:

$$\displaystyle M(x,y) = \left[x^4y+10\right]$$ <p style="text-align:right;">$$\displaystyle (2.18.1) $$

And:

$$\displaystyle N(x,y) = \left (\frac{1}{2}x^2\right)$$ <p style="text-align:right;">$$\displaystyle (2.18.2) $$

Taking the derivatives:

$$\displaystyle M_y(x,y)=x^4$$ <p style="text-align:right;">$$\displaystyle (2.18.3) $$

$$\displaystyle N_x(x,y)= x$$ <p style="text-align:right;">$$\displaystyle (2.18.4) $$

Since $$\displaystyle M_y(x,y) \ \ne N_y(x,y) $$ <p style="text-align:right;">$$\displaystyle (2.18.5) $$

Therefore equation $$\displaystyle (4) p. 12-4 $$ is not exact

So we re-write the ODE


 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at the first exactness condition for the integrating factor we obtain:

$$\displaystyle \frac{1}{N}\left(N_x - M_y\right) = -f(x) $$<p style="text-align:right;">$$\displaystyle (2.18.6) $$ replacing values found above $$\displaystyle \frac{2}{x^2}\left(x - x^4\right) = -f(x) $$<p style="text-align:right;">$$\displaystyle (2.18.7) $$ applying knowledge from integrating factor we know $$\displaystyle h(x)=\exp \int^{x}f(s)ds $$<p style="text-align:right;">$$\displaystyle (2.18.8) $$ replacing f(x) into the integral and evaluating it, obtain $$\displaystyle h(x) = \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right)$$<p style="text-align:right;">$$\displaystyle (2.18.9) $$

Lets verify that the integrating factor has made the ODE exact by checking the second condition of exactness $$\displaystyle \bar{M}_y = \bar{N}_x $$<p style="text-align:right;">$$\displaystyle (2.18.10) $$ $$\displaystyle\underbrace{(hM)}_{\bar{M}}dx = \underbrace{(hN)}_{\bar{N}}dy$$ <p style="text-align:right;">$$\displaystyle (2.18.11) $$

Calculating the respective values we obtain

$$\displaystyle \bar{N}_x = x^2 \exp\left(\frac{2x^3}{3}\right)$$<p style="text-align:right;">$$\displaystyle (2.18.12) $$ $$\displaystyle \bar{M}_y =x^2 \exp\left(\frac{2x^3}{3}\right)$$<p style="text-align:right;">$$\displaystyle (2.18.13) $$

Since $$\displaystyle (2.18.10)$$ is satisfied(i.e. $$\displaystyle ((2.18.12) = (2.19.13))$$ the non-homogenous L1_ODE_VC has been made exact by using the appropriate integrating factor $$\displaystyle h $$.

This solution is based on http://en.wikiversity.org/w/index.php?title=User:Egm6321.f09.team3/HW2&oldid=484226|#Problem_3

The equations have been updated, the flow improved and, the wording made more clear. Addtionally everything was updated to be consistent with the termonology of the Fall 2011 lecture notes and adapted to answer the question more exactly.

The source location is http://en.wikiversity.org/w/index.php?title=User:EGM6321.F11.TEAM1.R*2.16-2.18_Bens

This solution was updated by Ben Neri, Checked by, and compiled into the complete report by.